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stringlengths 2
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| description
stringlengths 29
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| source
int64 1
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| difficulty
int64 0
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| solution
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788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.util.Arrays;
import java.util.Scanner;
public class Main{
public static long getAns(int[] s,int n){
long maxSum = 0;
long thisSum = 0;
for(int i = 1;i<n;i++){
thisSum+=s[i];
if(thisSum>maxSum){
maxSum = thisSum;
} else if(thisSum<0){
thisSum = 0;
}
}
return maxSum;
}
public static void init(int[] a,int[] s,int[] sr,int n){
for(int i = 1;i<n;i++){
int num = Math.abs(a[i] - a[i + 1]) * ((i - 1) % 2 == 0 ? 1 : -1);
s[i]= num;
sr[i]= -num;
}
}
public static void main(String[] args) throws FileNotFoundException {
// File file = new File("C:/oy/input.txt");
// FileInputStream in = new FileInputStream(file);
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] a = new int[n+1];
for(int i = 1;i<=n;i++){
a[i] = scanner.nextInt();
}
int[] s = new int[n];
int[] sr = new int[n];
init(a,s,sr,n);
System.out.println(Math.max(getAns(s,n),getAns(sr,n)));
}
}
| JAVA |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.io.Writer;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.InputStream;
/**
* Built using CHelper plug-in Actual solution is at the top
*
* @author MaxHeap
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
AFunctionsAgain solver = new AFunctionsAgain();
solver.solve(1, in, out);
out.close();
}
static class AFunctionsAgain {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int n = in.nextInt();
int[] a = in.nextIntArray(n);
long[] diffs1 = new long[n - 1];
long[] diffs2 = new long[n - 1];
int pt1 = 0, pt2 = 0;
int sig = -1;
for (int i = 0; i + 1 < n; ++i) {
diffs1[pt1++] = Math.abs(a[i] - a[i + 1]) * sig;
diffs2[pt2++] = Math.abs(a[i] - a[i + 1]) * sig * -1;
sig *= -1;
}
out.println(Math.max(
maxIn(diffs1, pt1),
maxIn(diffs2, pt2)
));
}
private long maxIn(long[] a, int pt1) {
long best = 0;
long cur = 0;
for (int i = 0; i < pt1; ++i) {
cur += a[i];
if (cur < 0) {
cur = 0;
}
best = Math.max(cur, best);
}
return best;
}
}
static class OutputWriter extends PrintWriter {
public OutputWriter(OutputStream out) {
super(out);
}
public OutputWriter(Writer out) {
super(out);
}
public OutputWriter(File file) throws FileNotFoundException {
super(file);
}
public void close() {
super.close();
}
}
static class InputReader {
private InputStream stream;
private static final int DEFAULT_BUFFER_SIZE = 1 << 16;
private static final int EOF = -1;
private byte[] buf = new byte[DEFAULT_BUFFER_SIZE];
private int curChar;
private int numChars;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (this.numChars == EOF) {
throw new UnknownError();
} else {
if (this.curChar >= this.numChars) {
this.curChar = 0;
try {
this.numChars = this.stream.read(this.buf);
} catch (IOException ex) {
throw new InputMismatchException();
}
if (this.numChars <= 0) {
return EOF;
}
}
return this.buf[this.curChar++];
}
}
public int nextInt() {
int c;
for (c = this.read(); isSpaceChar(c); c = this.read()) {
}
byte sgn = 1;
if (c == 45) {
sgn = -1;
c = this.read();
}
int res = 0;
while (c >= 48 && c <= 57) {
res *= 10;
res += c - 48;
c = this.read();
if (isSpaceChar(c)) {
return res * sgn;
}
}
throw new InputMismatchException();
}
public static boolean isSpaceChar(int c) {
return c == 32 || c == 10 || c == 13 || c == 9 || c == EOF;
}
public int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
}
}
| JAVA |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | import java.util.Scanner;
import java.util.stream.IntStream;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0; i < a.length; i++) {
a[i] = sc.nextInt();
}
System.out.println(solve(a));
sc.close();
}
static long solve(int[] a) {
int[] diffs = IntStream.range(0, a.length - 1).map(i -> Math.abs(a[i] - a[i + 1])).toArray();
return Math.max(computeSubArrayMaxSum(negate(diffs, 0)), computeSubArrayMaxSum(negate(diffs, 1)));
}
static int[] negate(int[] diffs, int beginIndex) {
return IntStream.range(0, diffs.length).map(i -> diffs[i] * (i % 2 == beginIndex ? -1 : 1)).toArray();
}
static long computeSubArrayMaxSum(int[] p) {
long result = 0;
long sum = 0;
for (int pi : p) {
sum = Math.max(0, sum + pi);
result = Math.max(result, sum);
}
return result;
}
}
| JAVA |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | input()
numbers = [int(x) for x in input().split()]
diffs = []
sum_odd, sum_even = 0, 0
for x in range(len(numbers)-1):
diffs.append(abs(numbers[x] - numbers[x+1]))
_max = 0
for x in range(len(diffs)):
aux = diffs[x]
if x % 2 == 1: aux *= -1
sum_even += aux
sum_odd -= aux
if sum_even < 0: sum_even = 0
if sum_odd < 0: sum_odd = 0
_max = max([sum_even, sum_odd, _max])
print(_max)
| PYTHON3 |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | #include <bits/stdc++.h>
const int N = 1000005;
using namespace std;
long long n, a[N];
long long b[N], c[N];
int main() {
ios::sync_with_stdio(0);
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
for (int i = 1; i < n; i++) {
int t = abs(a[i] - a[i + 1]);
if (i & 1) {
b[i] = t;
c[i] = -t;
} else {
b[i] = -t;
c[i] = t;
}
}
long long ans = 0, now = 0;
for (int i = 1; i < n; i++) {
now += b[i];
if (now > ans) ans = now;
if (now < 0) now = 0;
}
now = 0;
for (int i = 1; i < n; i++) {
now += c[i];
if (now > ans) ans = now;
if (now < 0) now = 0;
}
cout << ans << '\n';
return 0;
}
| CPP |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 110000;
int n;
long long arr[maxn];
long long pre1[maxn], pre2[maxn];
long long dp1[maxn], dp2[maxn];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%lld", &arr[i]);
for (int i = 1; i < n; i++) {
pre1[i] = abs(arr[i] - arr[i + 1]) * ((i & 1) ? 1 : -1);
pre2[i] = -pre1[i];
}
pre2[1] = 0;
long long ans = 0;
for (int i = 1; i < n; i++) {
if (i & 1)
dp1[i] = max(dp1[i - 1] + pre1[i], pre1[i]);
else
dp1[i] = max(0LL, dp1[i - 1] + pre1[i]);
ans = max(ans, dp1[i]);
}
for (int i = 2; i < n; i++) {
if (!(i & 1))
dp2[i] = max(dp2[i - 1] + pre2[i], pre2[i]);
else
dp2[i] = max(0LL, dp2[i - 1] + pre2[i]);
ans = max(ans, dp2[i]);
}
cout << ans << endl;
return 0;
}
| CPP |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 |
import java.util.Arrays;
import java.util.Scanner;
public class FunctionsAgain {
static long [][] dp;
static long maxVal;
static void max(int index, int [] mods) {
//System.out.println("index:"+index+", len:"+mods.length);
if(index >= mods.length ) return;
if (dp[0][index+1] == Integer.MAX_VALUE)
max(index+1,mods);
long _max = Math.max(mods[index] - dp[0][index+1],mods[index]);
dp[1][index] = _max;
maxVal = Math.max(maxVal,_max);
dp[0][index]= Math.min(mods[index] - dp[1][index+1],mods[index]);
//System.out.println("index:"+index+","+dp[0][index]+","+dp[1][index]);
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int [] nums = new int[n];
int [] mods = new int[n-1];
for (int i=0;i<n;i++){
nums[i] = scanner.nextInt();
if (i > 0) {
mods[i-1] = Math.abs(nums[i]-nums[i-1]);
//System.out.println(mods[i-1]);
}
}
dp = new long[2][n];
Arrays.fill(dp[0],Integer.MAX_VALUE);
Arrays.fill(dp[1],Integer.MIN_VALUE);
dp[0][n-2] = dp[1][n-2]= mods[mods.length - 1];
dp[0][n-1] = dp[1][n-1] = 0;
maxVal = mods[mods.length - 1];
max(0,mods);
System.out.println(maxVal);
scanner.close();
}
}
| JAVA |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
int n;
long long a[100005], t, ans, tot, sum, x[100005], y[100005];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) cin >> a[i];
t = -1;
for (int i = 1; i <= n; i++) {
t *= -1;
x[i] = abs(a[i] - a[i + 1]) * t;
}
t = 1;
for (int i = 1; i <= n; i++) {
t *= -1;
y[i] = abs(a[i] - a[i + 1]) * t;
}
sum = ans = tot = 0;
for (int i = 1; i < n; i++) {
if (sum <= 0)
sum = x[i];
else
sum += x[i];
if (tot <= 0)
tot = y[i];
else
tot += y[i];
if (sum > ans) ans = sum;
if (tot > ans) ans = tot;
}
cout << ans;
}
| CPP |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | n = int(input())
a = [int(i) for i in input().split()]
b = [abs(a[i] - a[i+1]) for i in range(n-1)]
n -= 1
def fn(i0):
f0 = 0
ans=0
for i in range(i0, n):
if (i-i0)&1:
f0 -= b[i]
if f0 < 0: f0 = 0
else:
f0 += b[i]
ans = max(ans, f0)
return ans
ans = max(fn(0), fn(1))
print(ans)
| PYTHON3 |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class C{
public static void main(String[] args) {
FastScanner scan = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int n = scan.nextInt();
long[] a = scan.nextLongArray(n);
long[] d = new long[n-1], dd = new long[n-1];
for(int i = 0; i < n-1; i++){
d[i] = Math.abs(a[i]-a[i+1]);
if(i%2==1) d[i] = -d[i];
dd[i] = -d[i];
}
long res = Math.max(go(d, 0), go(dd, 1));
out.println(res);
out.close();
}
static long go(long[] d, int s){
long max = Long.MIN_VALUE, min = 0, sum = 0;
for(int i = s; i < d.length; i++){
sum += d[i];
if(sum < min){
min = sum;
max = Math.max(max, d[i]);
} else max = Math.max(max, sum-min);
}
return max;
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner() {
try {
br = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer(br.readLine());
} catch (Exception e){e.printStackTrace();}
}
public String next() {
if (st.hasMoreTokens()) return st.nextToken();
try {st = new StringTokenizer(br.readLine());}
catch (Exception e) {e.printStackTrace();}
return st.nextToken();
}
public int nextInt() {return Integer.parseInt(next());}
public long nextLong() {return Long.parseLong(next());}
public double nextDouble() {return Double.parseDouble(next());}
public String nextLine() {
String line = "";
if(st.hasMoreTokens()) line = st.nextToken();
else try {return br.readLine();}catch(IOException e){e.printStackTrace();}
while(st.hasMoreTokens()) line += " "+st.nextToken();
return line;
}
public int[] nextIntArray(int n) {
int[] a = new int[n];
for(int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
public long[] nextLongArray(int n){
long[] a = new long[n];
for(int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
public double[] nextDoubleArray(int n){
double[] a = new double[n];
for(int i = 0; i < n; i++) a[i] = nextDouble();
return a;
}
public char[][] nextGrid(int n, int m){
char[][] grid = new char[n][m];
for(int i = 0; i < n; i++) grid[i] = next().toCharArray();
return grid;
}
}
} | JAVA |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
void Solve() {
long long n;
cin >> n;
long long a[n + 1];
for (int i = 1; i <= n; i++) cin >> a[i];
long long b[n];
for (int i = 1; i < n; i++) b[i] = abs(a[i] - a[i + 1]);
long long mx[n + 1], mn[n + 1];
mx[n] = 0;
mn[n] = 0;
long long ans = LLONG_MIN;
for (int i = n - 1; i >= 1; i--) {
mx[i] = b[i] - mn[i + 1];
mn[i] = b[i] - mx[i + 1];
mx[i] = max(mx[i], b[i]);
ans = max(ans, mx[i]);
}
cout << ans;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
while (t--) {
Solve();
}
return 0;
}
| CPP |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | //189301019.akshay
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.Random;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Collections;
public class C
{
public static void main(String[] args)
{
FastReader sc=new FastReader();
StringBuffer ans=new StringBuffer();
int test=1;
while(test-->0)
{
int n=sc.nextInt();
long brr[]=new long[n+1];
for(int i=0;i<n;i++) brr[i+1]=sc.nextLong();
long [] arr=new long[n+1];
for(int i=0;i<n-1;i++) arr[i+1]=Math.abs(brr[i+1]-brr[i+2]);
long dp[][]=new long[n+1][2];
/*
* 0: +ve
* 1: -ve
*/
long maxm =Long.MIN_VALUE;
for(int i=1;i<=n;i++) {
dp[i][0] = Math.max(dp[i][0], dp[i-1][1])+arr[i];
dp[i][1] = Math.max(dp[i][1], dp[i-1][0])-arr[i];
maxm= Math.max(Math.max(dp[i][0], dp[i][1]), maxm);
// System.out.println(dp[i][0]+" "+dp[i][1]);
}
ans.append(maxm+"\n");
}
System.out.print(ans);
}
static final Random random=new Random();
static void ruffleSort(int[] a) {
int n=a.length;//shuffle, then sort
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
int n=a.length;//shuffle, then sort
for (int i=0; i<n; i++) {
int oi=random.nextInt(n);
long temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
}
| JAVA |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false), cin.tie(NULL);
int n;
while (cin >> n) {
long long a[n];
for (int i = 0; i < n; i++) cin >> a[i];
if (n == 2) {
cout << abs(a[0] - a[1]) << endl;
continue;
}
long long ar1[n - 1], ar2[n - 2];
for (int i = 0; i < n - 1; i++)
ar1[i] = abs(a[i] - a[i + 1]) * (i & 1 ? -1 : 1);
for (int i = 1; i < n - 1; i++)
ar2[i - 1] = abs(a[i] - a[i + 1]) * (i & 1 ? 1 : -1);
long long maxi1 = ar1[0], c1 = maxi1, maxi2 = ar2[0], c2 = maxi2;
for (int i = 1; i < n - 1; i++) {
c1 = max(c1 + ar1[i], ar1[i]);
maxi1 = max(maxi1, c1);
}
for (int i = 1; i < n - 2; i++) {
c2 = max(c2 + ar2[i], ar2[i]);
maxi2 = max(maxi2, c2);
}
cout << max(maxi1, maxi2) << endl;
}
return 0;
}
| CPP |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | #------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=998244353
EPS=1e-6
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def Answer(a):
till=0
ans=-inf
for i in a:
till+=i
ans=max(till,ans)
till=max(0,till)
return ans
n=Int()
a=array()
a=[abs(a[i]-a[i+1]) for i in range(n-1)]
# print(*a)
n-=1
a=[a[i]*(-1)**i for i in range(n)]
b=[-i for i in a]
print(max(Answer(a),Answer(b)))
| PYTHON3 |
788_A. Functions again | Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array. | 2 | 7 | n = int(input())
arr = list(map(int, input().split()))
def function(n, arr):
dp = {}
for i in range(1, n):
temp, dp[i] = abs(arr[i] - arr[i - 1]), {}
if i == 1:
dp[i][0], dp[i][1] = temp, 0
else:
dp[i][0] = max(dp[i - 1][1], 0) + temp
dp[i][1] = dp[i - 1][0] - temp
Max = dp[1][0]
for i in range(2, n):
Max = max(Max, dp[i][0], dp[i][1])
return Max
print(function(n, arr))
| PYTHON3 |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | import java.util.*;
public class C{
final int MOD = (int)1e9 + 7;
long get(int x, int y, int lim){
if (Math.min(x,y) < 0) return 0L;
long[][][][] d = new long[32][2][2][2], sm = new long[32][2][2][2];
d[31][1][1][1] = 1;
for (int w = 30; w >= 0; w--)
for (int a = 0; a < 2; a++)
for (int b = 0; b < 2; b++)
for (int c = 0; c < 2; c++)
for (int aa = 0; aa <= (a==0?1:x>>w&1); aa++)
for (int bb = 0; bb <= (b==0?1:y>>w&1); bb++)
if (c == 0 || (aa^bb) <= (lim>>w&1)){
d[w][(a==0||aa<(x>>w&1))?0:1][(b==0||bb<(y>>w&1))?0:1][(c==0||(aa^bb)<(lim>>w&1))?0:1]
+= d[w+1][a][b][c];
sm[w][(a==0||aa<(x>>w&1))?0:1][(b==0||bb<(y>>w&1))?0:1][(c==0||(aa^bb)<(lim>>w&1))?0:1]
+= sm[w+1][a][b][c] + (1L<<w)*(aa^bb)*d[w+1][a][b][c];
sm[w][(a==0||aa<(x>>w&1))?0:1][(b==0||bb<(y>>w&1))?0:1][(c==0||(aa^bb)<(lim>>w&1))?0:1]
%= MOD;
}
long ret = 0;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int w = 0; w < 2; w++){
ret += d[0][i][j][w];
ret += sm[0][i][j][w];
ret %= MOD;
}
return ret;
}
void solve(){
// System.out.println(get(4, 3, 4));
Scanner sc = new Scanner(System.in);
int q = sc.nextInt();
while (q-->0){
int x1 = sc.nextInt(), y1 = sc.nextInt();
int x2 = sc.nextInt(), y2 = sc.nextInt();
int k = sc.nextInt();
x1--; x2--; y1--; y2--; k--;
System.out.printf("%d\n", (get(x2, y2, k)-get(x2,y1-1, k)-get(x1-1, y2, k)+get(x1-1,y1-1, k)+MOD*10L)%MOD);
}
}
public static void main(String[] args){
C sol = new C();
sol.solve();
}
}
| JAVA |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void addmod(long long int &a, long long int b) {
a += b;
if (a >= 1000000007) a -= 1000000007;
if (a < 0) a += 1000000007;
}
void multmod(long long int &a, long long int b) {
a *= b;
a %= 1000000007;
}
long long int solve(int x, int y, int k) {
if (x < 0 || y < 0 || k < 0) {
return 0;
}
long long int dp[31 + 1][2][2][2];
long long int sum[31 + 1][2][2][2];
memset(dp, 0, sizeof(dp));
memset(sum, 0, sizeof(sum));
int xb[31], yb[31], kb[31];
for (int i = 0; i < 31; i++) {
xb[i] = x & 1;
yb[i] = y & 1;
kb[i] = k & 1;
x >>= 1;
y >>= 1;
k >>= 1;
}
reverse(xb, xb + 31);
reverse(yb, yb + 31);
reverse(kb, kb + 31);
dp[0][1][1][1] = 1;
sum[0][1][1][1] = 0;
for (int i = 0; i < 31; i++) {
for (int preI = 0; preI < 2; preI++) {
for (int preJ = 0; preJ < 2; preJ++) {
for (int preK = 0; preK < 2; preK++) {
for (int xBit = 0; xBit < 2; xBit++) {
for (int yBit = 0; yBit < 2; yBit++) {
int val = xBit ^ yBit;
if (preI == 1 && xb[i] == 0 && xBit == 1) continue;
if (preJ == 1 && yb[i] == 0 && yBit == 1) continue;
if (preK == 1 && kb[i] == 0 && val == 1) continue;
int npreI = preI & (xb[i] == xBit);
int npreJ = preJ & (yb[i] == yBit);
int npreK = preK & (kb[i] == val);
addmod(dp[i + 1][npreI][npreJ][npreK], dp[i][preI][preJ][preK]);
addmod(sum[i + 1][npreI][npreJ][npreK], sum[i][preI][preJ][preK]);
long long int bmult = val << (31 - i - 1);
multmod(bmult, dp[i][preI][preJ][preK]);
addmod(sum[i + 1][npreI][npreJ][npreK], bmult);
}
}
}
}
}
}
long long int ans = 0;
for (int a = 0; a < 2; a++) {
for (int b = 0; b < 2; b++) {
for (int c = 0; c < 2; c++) {
addmod(ans, sum[31][a][b][c]);
addmod(ans, dp[31][a][b][c]);
}
}
}
return ans;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int q;
cin >> q;
while (q--) {
int x1, y1, x2, y2, k;
cin >> x1 >> y1 >> x2 >> y2 >> k;
x1--;
y1--;
x2--;
y2--;
k--;
long long int ans = solve(x2, y2, k);
addmod(ans, -solve(x1 - 1, y2, k));
addmod(ans, -solve(x2, y1 - 1, k));
addmod(ans, solve(x1 - 1, y1 - 1, k));
cout << ans << endl;
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int nx, ny, nz;
bool g[32][2][2][2];
pair<int, int> f[32][2][2][2];
void add(pair<int, int> &x, pair<int, int> y) {
x.first = (x.first + y.first) % mod;
x.second = (x.second + y.second) % mod;
}
pair<int, int> F(int p, int x, int y, int z) {
pair<int, int> &ret = f[p][x][y][z];
bool &visit = g[p][x][y][z];
if (visit) return ret;
if (!p) return pair<int, int>(1, 0);
visit = 1, ret = pair<int, int>(0, 0);
int tx = nx >> (p - 1) & 1;
int ty = ny >> (p - 1) & 1;
int tz = nz >> (p - 1) & 1;
tx |= x, ty |= y, tz |= z;
if ((tx ^ ty) <= tz) {
pair<int, int> tmp = F(p - 1, x, y, z | ((tx ^ ty) < tz));
add(ret, tmp);
ret.second = (ret.second + 1LL * tmp.first * ((tx ^ ty) << (p - 1))) % mod;
}
if (tx) {
if (ty <= tz) {
pair<int, int> tmp = F(p - 1, 1, y, z | (ty < tz));
add(ret, tmp);
ret.second = (ret.second + 1LL * tmp.first * (ty << (p - 1))) % mod;
}
}
if (ty) {
if (tx <= tz) {
pair<int, int> tmp = F(p - 1, x, 1, z | (tx < tz));
add(ret, tmp);
ret.second = (ret.second + 1LL * tmp.first * (tx << (p - 1))) % mod;
}
}
if (tx && ty) {
pair<int, int> tmp = F(p - 1, 1, 1, z | tz);
add(ret, tmp);
}
return ret;
}
int cal(int x, int y, int z) {
if (x == -1 || y == -1) return 0;
nx = x, ny = y, nz = z;
memset(g, 0, sizeof g);
pair<int, int> ret = F(31, 0, 0, 0);
return (ret.first + ret.second) % mod;
}
void solve() {
int a, b, c, d, e;
cin >> a >> b >> c >> d >> e;
a--, b--, c--, d--, e--, a--, b--;
long long res = cal(c, d, e) - cal(a, d, e) - cal(c, b, e) + cal(a, b, e);
res %= mod;
if (res < 0) res += mod;
cout << res << '\n';
}
int main() {
ios::sync_with_stdio(false);
int q;
cin >> q;
while (q--) solve();
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long mo = 1e9 + 7;
long long _;
long long x1, kjhjk, x2, dfdfd, k;
long long a[40], b[40], c[40];
long long dp1[40][2][2][2], dp2[40][2][2][2];
long long solve(long long x, long long y) {
if (x < 0 || y < 0 || k < 0) return 0;
long long i;
memset(dp1, 0, sizeof(dp1));
memset(dp2, 0, sizeof(dp2));
long long xx = x, yy = y, kk = k;
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(c, 0, sizeof(c));
for (i = 0; i < 31; i++) {
a[i] = xx % 2;
xx /= 2;
b[i] = yy % 2;
yy /= 2;
c[i] = kk % 2;
kk /= 2;
}
reverse(a, a + 31);
reverse(b, b + 31);
reverse(c, c + 31);
long long A, B, C, X, Y, Z;
dp1[0][1][1][1] = 1;
for (i = 0; i < 31; i++)
for (A = 0; A < 2; A++)
for (B = 0; B < 2; B++)
for (C = 0; C < 2; C++)
for (X = 0; X < 2; X++)
for (Y = 0; Y < 2; Y++) {
Z = X ^ Y;
if (A == 1 && X == 1 && a[i] == 0) continue;
if (B == 1 && Y == 1 && b[i] == 0) continue;
if (C == 1 && Z == 1 && c[i] == 0) continue;
(dp1[i + 1][A & (a[i] == X)][B & (b[i] == Y)][C & (c[i] == Z)] +=
dp1[i][A][B][C]) %= mo;
(dp2[i + 1][A & (a[i] == X)][B & (b[i] == Y)][C & (c[i] == Z)] +=
dp2[i][A][B][C]) %= mo;
(dp2[i + 1][A & (a[i] == X)][B & (b[i] == Y)][C & (c[i] == Z)] +=
dp1[i][A][B][C] % mo * ((Z << (30 - i)) % mo) % mo) %= mo;
}
long long sum = 0;
for (A = 0; A < 2; A++)
for (B = 0; B < 2; B++)
for (C = 0; C < 2; C++) {
(sum += dp1[31][A][B][C]) %= mo;
(sum += dp2[31][A][B][C]) %= mo;
}
return sum;
}
int main() {
scanf("%lld", &_);
while (_--) {
scanf("%lld%lld%lld%lld%lld", &x1, &kjhjk, &x2, &dfdfd, &k);
x1--;
kjhjk--;
x2--;
dfdfd--;
k--;
printf("%lld\n",
(solve(x2, dfdfd) - solve(x1 - 1, dfdfd) - solve(x2, kjhjk - 1) +
solve(x1 - 1, kjhjk - 1) + mo * 10) %
mo);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline long long read() {
long long x = 0, f = 1;
char a = getchar();
while (a < '0' || a > '9') {
if (a == '-') f = -1;
a = getchar();
}
while (a >= '0' && a <= '9') x = x * 10 + a - '0', a = getchar();
return x * f;
}
int solve(int x, int y, int k, int add) {
if (x > y) swap(x, y);
if (!x) return 0;
if (k <= 0) return 0;
int z = 1 << (31 - __builtin_clz(y)), kk = min(k, z), ret = 0;
if (x < z) {
ret += (((1LL * (kk + 1) * kk / 2 % 1000000007) * x) % 1000000007 +
((1LL * kk * add) % 1000000007 * x) % 1000000007) %
1000000007;
(ret += solve(x, y - z, k - z, add + z)) %= 1000000007;
} else {
ret += (((1LL * (kk + 1) * kk / 2 % 1000000007) * z) % 1000000007 +
((1LL * kk * add) % 1000000007 * z) % 1000000007) %
1000000007;
(ret += solve(z, y - z, k - z, add + z)) %= 1000000007;
(ret += solve(x - z, z, k - z, add + z)) %= 1000000007;
(ret += solve(x - z, y - z, k, add)) %= 1000000007;
}
return ret;
}
int main() {
int q, x1, x2, y1, y2, k, ans;
q = read();
while (q--) {
x1 = read() - 1;
y1 = read() - 1;
x2 = read();
y2 = read();
k = read();
ans = 0;
(ans += (solve(x1, y1, k, 0) + solve(x2, y2, k, 0)) % 1000000007) %=
1000000007;
(ans -= (solve(x1, y2, k, 0) + solve(x2, y1, k, 0)) % 1000000007) %=
1000000007;
printf("%d\n", (ans + 1000000007) % 1000000007);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAX_N = 33;
const long long MOD = 1000000007;
long long dp[MAX_N][2][2][2];
long long numb[MAX_N][2][2][2];
int query;
int X1, Y1, X2, Y2, k;
vector<int> change(int r) {
vector<int> res;
for (int i = 0; i < 31; i++) {
if ((long long)r & (1LL << (long long)i)) {
res.push_back(1);
} else {
res.push_back(0);
}
}
return res;
}
long long calc(int x, int y) {
if (x == 0 || y == 0) {
return 0;
}
memset(dp, 0LL, sizeof(dp));
memset(numb, 0LL, sizeof(numb));
vector<int> X, Y, K;
X = change(x - 1);
Y = change(y - 1);
K = change(k - 1);
long long res = 0LL;
numb[31][1][1][1] = 1;
for (int i = 31; i >= 1; i--) {
for (int flagX = 0; flagX <= 1; flagX++) {
for (int flagY = 0; flagY <= 1; flagY++) {
for (int flagK = 0; flagK <= 1; flagK++) {
if (numb[i][flagX][flagY][flagK] == 0) {
continue;
}
for (int b1 = 0; b1 <= 1; b1++) {
for (int b2 = 0; b2 <= 1; b2++) {
if ((flagX == 1 && b1 > X[i - 1]) ||
(flagY == 1 && b2 > Y[i - 1]) ||
(flagK == 1 && ((b1 ^ b2) > K[i - 1]))) {
continue;
}
int nflagX = (flagX == 1 && b1 == X[i - 1]);
int nflagY = (flagY == 1 && b2 == Y[i - 1]);
int nflagK = (flagK == 1 && ((b1 ^ b2) == K[i - 1]));
(numb[i - 1][nflagX][nflagY][nflagK] +=
numb[i][flagX][flagY][flagK]) %= MOD;
(dp[i - 1][nflagX][nflagY][nflagK] +=
((dp[i][flagX][flagY][flagK] * 2LL) % MOD +
numb[i][flagX][flagY][flagK] * (b1 ^ b2)) %
MOD) %= MOD;
}
}
}
}
}
}
for (int flagX = 0; flagX <= 1; flagX++) {
for (int flagY = 0; flagY <= 1; flagY++) {
for (int flagK = 0; flagK <= 1; flagK++) {
(res += dp[0][flagX][flagY][flagK]) %= MOD;
(res += numb[0][flagX][flagY][flagK]) %= MOD;
}
}
}
return res;
}
int main() {
scanf("%d", &query);
while (query--) {
scanf("%d %d %d %d %d", &X1, &Y1, &X2, &Y2, &k);
long long res = calc(X2, Y2);
res -= calc(X1 - 1, Y2);
res -= calc(X2, Y1 - 1);
(res += 3LL * MOD) %= MOD;
(res += calc(X1 - 1, Y1 - 1)) %= MOD;
printf("%I64d\n", res);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int f[32][2][2][2], g[32][2][2][2], q;
inline int rd() {
int x = 0;
char ch = getchar();
for (; ch < '0' || ch > '9'; ch = getchar())
;
for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
return x;
}
inline void print(int x) {
static char s[233];
if (!x) {
putchar('0');
putchar('\n');
return;
}
int tot = 0;
for (; x; x /= 10) s[++tot] = x % 10 + '0';
for (; tot; tot--) putchar(s[tot]);
putchar('\n');
}
inline int pls(const int &x, const int &y) {
return (x + y < 1000000007) ? x + y : x + y - 1000000007;
}
inline int mns(const int &x, const int &y) {
return (x - y < 0) ? x - y + 1000000007 : x - y;
}
inline int gao(int x, int y, int z) {
if (x < 0 || y < 0) return 0;
memset(f, 0, sizeof(f));
memset(g, 0, sizeof(g));
f[31][1][1][1] = 1;
for (int i = 30; ~i; i--)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
for (int l = 0; l < 2; l++)
if (f[i + 1][j][k][l]) {
for (int p = 0; p < 2; p++)
for (int q = 0; q < 2; q++) {
if (j && !((1 << i) & x) && p) continue;
if (k && !((1 << i) & y) && q) continue;
if (l && !((1 << i) & z) && (p ^ q)) continue;
int j1 = (j && (p == ((x >> i) & 1))),
k1 = (k && (q == ((y >> i) & 1))),
l1 = (l && (p ^ q) == ((z >> i) & 1));
f[i][j1][k1][l1] = pls(f[i][j1][k1][l1], f[i + 1][j][k][l]);
g[i][j1][k1][l1] = pls(g[i][j1][k1][l1], g[i + 1][j][k][l]);
if (p ^ q)
g[i][j1][k1][l1] =
pls(g[i][j1][k1][l1],
(long long)(1 << i) * f[i + 1][j][k][l] % 1000000007);
}
}
int ans = 0;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
ans = pls(ans, pls(f[0][i][j][k], g[0][i][j][k]));
return ans;
}
int main() {
q = rd();
while (q--) {
int x1 = rd() - 1, DCXISSOHANDSOME = rd() - 1, x2 = rd() - 1, y2 = rd() - 1,
k = rd() - 1;
print(mns(pls(gao(x2, y2, k), gao(x1 - 1, DCXISSOHANDSOME - 1, k)),
pls(gao(x1 - 1, y2, k), gao(x2, DCXISSOHANDSOME - 1, k))));
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Solution {
private static String inputFilename = "input.txt";
private static String outputFilename = "output.txt";
private BufferedReader in;
private StringTokenizer line;
private PrintWriter out;
private boolean isDebug;
public Solution(boolean isDebug) {
this.isDebug = isDebug;
}
private int mm = 1000000007;
// private int ccc = 0;
public void solve() throws IOException {
int x1 = nextInt();
int y1 = nextInt();
int x2 = nextInt();
int y2 = nextInt();
int k = nextInt();
long res = f(x2, y2, k)
- f(x1 - 1, y2, k)
- f(x2, y1 - 1, k)
+ f(x1 - 1, y1 - 1, k);
res = (res % mm + mm) % mm;
out.println(res);
// out.println(ccc);
}
private long f(long x, long y, long k) {
if (x <= 0 || y <= 0) return 0;
long t = 1;
while (x > t || y > t) t *= 2;
if (t == 1) {
return 1;
}
t /= 2;
long res = f(x, y, t, k, 0)
+ f(x - t, y - t, t, k, 0)
+ f(x - t, y, t, k, t)
+ f(x, y - t, t, k, t);
res = (res % mm + mm) % mm;
return res;
}
private long f(long x, long y, long mx, long k, long add) {
if (x <= 0 || y <= 0 || k <= add) return 0;
// ccc++;
if (x > mx) x = mx;
if (y > mx) y = mx;
if (x == mx || y == mx) {
long cnt = Math.min(mx, k - add);
long rows = Math.min(x, y);
return rows % mm * (add * cnt % mm + cnt * (cnt + 1) / 2 % mm) % mm;
} else {
mx /= 2;
long res = f(x, y, mx, k, add)
+ f(x - mx, y - mx, mx, k, add)
+ f(x - mx, y, mx, k, add + mx)
+ f(x, y - mx, mx, k, add + mx);
res = (res % mm + mm) % mm;
return res;
}
}
public static void main(String[] args) throws IOException {
new Solution(Arrays.asList(args).contains("DEBUG_MODE")).run(args);
}
public void run(String[] args) throws IOException {
if (isDebug) {
in = new BufferedReader(new InputStreamReader(new FileInputStream(inputFilename)));
// in = new BufferedReader(new InputStreamReader(System.in));
} else {
in = new BufferedReader(new InputStreamReader(System.in));
}
out = new PrintWriter(System.out);
// out = new PrintWriter(outputFilename);
int t = nextInt();
// int t = 1;
for (int i = 0; i < t; i++) {
// out.print("Case #" + (i + 1) + ": ");
solve();
}
in.close();
out.flush();
out.close();
}
private int[] nextIntArray(int n) throws IOException {
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = nextInt();
}
return res;
}
private long[] nextLongArray(int n) throws IOException {
long[] res = new long[n];
for (int i = 0; i < n; i++) {
res[i] = nextLong();
}
return res;
}
private int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
private long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
private double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
private String nextToken() throws IOException {
while (line == null || !line.hasMoreTokens()) {
line = new StringTokenizer(in.readLine());
}
return line.nextToken();
}
}
| JAVA |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline int M(int x) { return (x >= 1000000007) ? (x - 1000000007) : x; }
inline int M1(int x) { return (x < 0) ? (x + 1000000007) : x; }
int n, m, K, len;
bool vis[32][2][2][2];
pair<int, int> f[32][2][2][2];
pair<int, int> dfs(int x, bool f1, bool f2, bool f3) {
if (x > len) return make_pair(0, 1);
bool &v = vis[x][f1][f2][f3];
if (v) return f[x][f1][f2][f3];
v = true;
const int n1 = (int)(n >> (len - x)) & 1, u1 = f1 ? n1 : 1;
const int m1 = (int)(m >> (len - x)) & 1, u2 = f2 ? m1 : 1;
const int k1 = (int)(K >> (len - x)) & 1;
const int base = (1 << (len - x)) % 1000000007;
pair<int, int> res = make_pair(0, 0);
int i, j, p1;
pair<int, int> p;
for (i = 0; i <= u1; ++i)
for (j = 0; j <= u2; ++j) {
if (f3 && ((i ^ j) > k1)) continue;
p = dfs(x + 1, f1 && (i == n1), f2 && (j == m1), f3 && ((i ^ j) == k1));
p1 = p.second;
res.first =
M(M(res.first + (1ll * (i ^ j) * p1 * base % 1000000007)) + p.first);
res.second = M(res.second + p1);
}
return f[x][f1][f2][f3] = res;
}
inline int Get(int X, int Y, int P) {
if ((!X) || (!Y)) return 0;
n = X - 1;
m = Y - 1;
K = P - 1;
memset(vis, 0, sizeof(vis));
memset(f, 0, sizeof(f));
int cn, p;
len = 0;
p = n;
cn = 0;
while (p) {
++cn;
p >>= 1;
}
len = max(len, cn);
p = m;
cn = 0;
while (p) {
++cn;
p >>= 1;
}
len = max(len, cn);
p = K;
cn = 0;
while (p) {
++cn;
p >>= 1;
}
len = max(len, cn);
pair<int, int> k = dfs(1, 1, 1, 1);
return M(k.first + k.second);
}
inline int calc(int x1, int y1, int x2, int y2, int K) {
return M(M1(M1(Get(x2, y2, K) - Get(x1 - 1, y2, K)) - Get(x2, y1 - 1, K)) +
Get(x1 - 1, y1 - 1, K));
}
int main() {
int T, x1, y1, x2, y2, K;
scanf("%d", &T);
while (T--) {
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &K);
printf("%d\n", calc(x1, y1, x2, y2, K));
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
const long long int MAX = 1e9 + 7;
void array_show(int *array, int array_n, char middle = ' ') {
for (int i = 0; i < array_n; i++)
printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(long long int *array, int array_n, char middle = ' ') {
for (int i = 0; i < array_n; i++)
printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(vector<int> &vec_s, int vec_n = -1, char middle = ' ') {
if (vec_n == -1) vec_n = vec_s.size();
for (int i = 0; i < vec_n; i++)
printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
void array_show(vector<long long int> &vec_s, int vec_n = -1,
char middle = ' ') {
if (vec_n == -1) vec_n = vec_s.size();
for (int i = 0; i < vec_n; i++)
printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
pair<long long int, long long int> calc(long long int x, long long int y,
long long int t) {
if (x > y) swap(x, y);
if (x == 0) return make_pair(0, 0);
int i, j, k;
long long int a, b, c;
pair<long long int, long long int> pa;
for (i = 0; (1 << i) <= y; i++)
;
i--;
a = 1 << i;
if (t < a) {
b = t * (t + 1) / 2 % MAX * min(x, a) % MAX;
if (a < x) pa = calc(x - a, y - a, t);
return make_pair((b + pa.first) % MAX, (t * min(x, a) + pa.second) % MAX);
}
if (a < x) {
b = a * (a + 1) / 2 % MAX * a % MAX;
if (t < 2 * a) {
b += (t + a + 1) * (t - a) / 2 % MAX * (x + y - 2 * a) % MAX;
c = a * a + (x + y - 2 * a) * (t - a);
} else {
b += (3 * a + 1) * a / 2 % MAX * (x + y - 2 * a) % MAX;
c = (x + y - a) * a;
}
pa = calc(x - a, y - a, t);
b += pa.first;
pa.second %= MAX;
return make_pair(b % MAX, (c + pa.second) % MAX);
}
b = a * (a + 1) / 2 % MAX * x % MAX;
pa = calc(x, y - a, t - a);
b += pa.first;
pa.second %= MAX;
b += a * pa.second;
return make_pair(b % MAX, (x * a + pa.second) % MAX);
}
int main() {
int n, m;
int i, j, k;
long long int a, b, c, d, e;
long long int s = 0;
cin >> n;
for (i = 0; i < n; i++) {
cin >> a >> b >> c >> d >> e;
s = calc(c, d, e).first - calc(a - 1, d, e).first -
calc(c, b - 1, e).first + calc(a - 1, b - 1, e).first;
if (s < 0) s += 2 * MAX;
s %= MAX;
cout << s << endl;
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 40;
const long long mod = 1000000007;
long long f[maxn][2][2][2];
long long sum[maxn][2][2][2];
long long solve(int a, int b, int k) {
if (a < 0 || b < 0 || k < 0) return 0;
memset(f, 0, sizeof(f));
memset(sum, 0, sizeof(sum));
for (int s1 = 0; s1 <= 1; s1++)
for (int s2 = 0; s2 <= 1; s2++)
for (int s3 = 0; s3 <= 1; s3++) f[0][s1][s2][s3] = 1;
for (int i = 1; i <= 31; i++) {
int dx = (a >> (i - 1)) & 1;
int dy = (b >> (i - 1)) & 1;
int dxor = (k >> (i - 1)) & 1;
for (int s1 = 0; s1 <= 1; s1++) {
for (int s2 = 0; s2 <= 1; s2++) {
for (int s3 = 0; s3 <= 1; s3++) {
for (int x = 0; x <= 1; x++) {
for (int y = 0; y <= 1; y++) {
int t = x ^ y;
if (s1 == 0 && x > dx) continue;
if (s2 == 0 && y > dy) continue;
if (s3 == 0 && t > dxor) continue;
int news1 = s1, news2 = s2, news3 = s3;
if (dx == 1 && x == 0) news1 = 1;
if (dy == 1 && y == 0) news2 = 1;
if (dxor == 1 && t == 0) news3 = 1;
f[i][s1][s2][s3] += f[i - 1][news1][news2][news3];
f[i][s1][s2][s3] %= mod;
sum[i][s1][s2][s3] += (sum[i - 1][news1][news2][news3] +
f[i - 1][news1][news2][news3] *
(((1 << (i - 1)) * t) % mod)) %
mod;
sum[i][s1][s2][s3] %= mod;
}
}
}
}
}
}
long long ans = (sum[31][0][0][0] + f[31][0][0][0]) % mod;
return ans;
}
int main() {
int q;
scanf("%d", &q);
while (q--) {
int x0, y0, x1, y1, k;
scanf("%d%d%d%d%d", &x0, &y0, &x1, &y1, &k);
x0--;
y0--;
x1--;
y1--;
k--;
long long ans =
(solve(x1, y1, k) - solve(x0 - 1, y1, k) - solve(x1, y0 - 1, k) +
solve(x0 - 1, y0 - 1, k) + mod + mod) %
mod;
printf("%I64d\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
/**
* @author Don Li
*/
public class FindACar {
int N = 30;
int MOD = (int) 1e9 + 7;
int a, b, c;
int[][][][] dp = new int[N + 1][2][2][2];
int[][][][] cnt = new int[N + 1][2][2][2];
void solve() {
int q = in.nextInt();
while (q-- > 0) {
int x1 = in.nextInt() - 1, y1 = in.nextInt() - 1, x2 = in.nextInt() - 1, y2 = in.nextInt() - 1, k = in.nextInt() - 1;
long ans = 0;
ans = (ans + calc(x2, y2, k)) % MOD;
ans = (ans - calc(x1 - 1, y2, k) + MOD) % MOD;
ans = (ans - calc(x2, y1 - 1, k) + MOD) % MOD;
ans = (ans + calc(x1 - 1, y1 - 1, k)) % MOD;
out.println(ans);
}
}
int calc(int x, int y, int k) {
if (x < 0 || y < 0) return 0;
for (int i = 0; i <= N; i++)
for (int j = 0; j < 2; j++)
for (int u = 0; u < 2; u++)
for (int v = 0; v < 2; v++)
dp[i][j][u][v] = cnt[i][j][u][v] = -1;
a = x;
b = y;
c = k;
return dfs1(N, 1, 1, 1);
}
int dfs1(int i, int ea, int eb, int ec) {
if (i < 0) return 1;
if (dp[i][ea][eb][ec] >= 0) return dp[i][ea][eb][ec];
int bit_a = (a >> i) & 1;
int bit_b = (b >> i) & 1;
int bit_c = (c >> i) & 1;
long res = 0;
int nec = ec == 1 && bit_c == 0 ? 1 : 0;
int nea = ea == 1 && bit_a == 0 ? 1 : 0;
int neb = eb == 1 && bit_b == 0 ? 1 : 0;
res = (res + dfs1(i - 1, nea, neb, nec)) % MOD;
if ((eb == 0 || bit_b == 1) && (ec == 0 || bit_c == 1)) {
nec = ec == 1 && bit_c == 1 ? 1 : 0;
nea = ea == 1 && bit_a == 0 ? 1 : 0;
neb = eb == 1 && bit_b == 1 ? 1 : 0;
res = (res + (1L << i) * dfs2(i - 1, nea, neb, nec) % MOD + dfs1(i - 1, nea, neb, nec)) % MOD;
}
if ((ea == 0 || bit_a == 1) && (ec == 0 || bit_c == 1)) {
nec = ec == 1 && bit_c == 1 ? 1 : 0;
nea = ea == 1 && bit_a == 1 ? 1 : 0;
neb = eb == 1 && bit_b == 0 ? 1 : 0;
res = (res + (1L << i) * dfs2(i - 1, nea, neb, nec) % MOD + dfs1(i - 1, nea, neb, nec)) % MOD;
}
if ((ea == 0 || bit_a == 1) && (eb == 0 || bit_b == 1)) {
nec = ec == 1 && bit_c == 0 ? 1 : 0;
nea = ea == 1 && bit_a == 1 ? 1 : 0;
neb = eb == 1 && bit_b == 1 ? 1 : 0;
res = (res + dfs1(i - 1, nea, neb, nec)) % MOD;
}
return dp[i][ea][eb][ec] = (int) res;
}
int dfs2(int i, int ea, int eb, int ec) {
if (i < 0) return 1;
if (cnt[i][ea][eb][ec] >= 0) return cnt[i][ea][eb][ec];
int bit_a = (a >> i) & 1;
int bit_b = (b >> i) & 1;
int bit_c = (c >> i) & 1;
long res = 0;
int nec = ec == 1 && bit_c == 0 ? 1 : 0;
int nea = ea == 1 && bit_a == 0 ? 1 : 0;
int neb = eb == 1 && bit_b == 0 ? 1 : 0;
res = (res + dfs2(i - 1, nea, neb, nec)) % MOD;
if ((eb == 0 || bit_b == 1) && (ec == 0 || bit_c == 1)) {
nec = ec == 1 && bit_c == 1 ? 1 : 0;
nea = ea == 1 && bit_a == 0 ? 1 : 0;
neb = eb == 1 && bit_b == 1 ? 1 : 0;
res = (res + dfs2(i - 1, nea, neb, nec)) % MOD;
}
if ((ea == 0 || bit_a == 1) && (ec == 0 || bit_c == 1)) {
nec = ec == 1 && bit_c == 1 ? 1 : 0;
nea = ea == 1 && bit_a == 1 ? 1 : 0;
neb = eb == 1 && bit_b == 0 ? 1 : 0;
res = (res + dfs2(i - 1, nea, neb, nec)) % MOD;
}
if ((ea == 0 || bit_a == 1) && (eb == 0 || bit_b == 1)) {
nec = ec == 1 && bit_c == 0 ? 1 : 0;
nea = ea == 1 && bit_a == 1 ? 1 : 0;
neb = eb == 1 && bit_b == 1 ? 1 : 0;
res = (res + dfs2(i - 1, nea, neb, nec)) % MOD;
}
return cnt[i][ea][eb][ec] = (int) res;
}
public static void main(String[] args) {
in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
out = new PrintWriter(System.out);
new FindACar().solve();
out.close();
}
static FastScanner in;
static PrintWriter out;
static class FastScanner {
BufferedReader in;
StringTokenizer st;
public FastScanner(BufferedReader in) {
this.in = in;
}
public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(in.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(nextToken());
}
public long nextLong() {
return Long.parseLong(nextToken());
}
public double nextDouble() {
return Double.parseDouble(nextToken());
}
}
}
| JAVA |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
int q;
long long a, b, u, v, k;
long long calc(long long u, long long v, long long k, long long o = 0) {
if (u > v) swap(u, v);
if (u < 1 || k <= 0) return 0;
long long p = 1, res = 0;
while (p * 2LL <= v) p *= 2LL;
long long tmp = min(p, k);
if (u <= p) {
res = (tmp * (tmp + 1) / 2 % mod) * u % mod;
res += (o * tmp % mod) * u % mod + calc(u, v - p, k - p, o + p);
res %= mod;
} else {
res = (tmp * (tmp + 1) / 2 % mod) * p % mod;
res += (o * tmp % mod) * p % mod;
res += calc(u - p, v - p, k, o) + calc(p, v - p, k - p, o + p) +
calc(u - p, p, k - p, o + p);
}
return res;
}
int main() {
scanf("%d", &q);
while (q-- > 0) {
scanf("%lld %lld %lld %lld %lld", &a, &b, &u, &v, &k);
long long res = (calc(u, v, k) + calc(a - 1, b - 1, k)) % mod;
res = (res - calc(a - 1, v, k) - calc(u, b - 1, k) + mod * mod) % mod;
printf("%lld\n", res);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline int read() {
char c = getchar();
int p = 1, ret = 0;
while ((c < '0') || (c > '9')) {
if (c == '-') p = -1;
c = getchar();
}
while ((c >= '0') && (c <= '9'))
ret = (ret << 1) + (ret << 3) + c - '0', c = getchar();
return ret * p;
}
int n;
int dp[32][2][2][2], Dp[32][2][2][2];
int a[31], b[31], c[31];
void add(int &x, int y) { x = (x + y) % 1000000007; }
inline long long query(int x, int y, int k) {
if (x < 0 || (y < 0)) return 0;
for (int i = 0; i < 31; i++)
a[i] = x >> i & 1, b[i] = y >> i & 1, c[i] = k >> i & 1;
dp[31][1][1][1] = 1;
Dp[31][1][1][1] = 0;
for (int i = 0; i < 31; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
for (int l = 0; l < 2; l++) dp[i][j][k][l] = Dp[i][j][k][l] = 0;
for (int i = 30; i >= 0; i--)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
for (int l = 0; l < 2; l++)
for (int p = 0; p < 2; p++)
for (int q = 0; q < 2; q++) {
int v = p ^ q;
if ((j) && (a[i] == 0) && (p)) continue;
if ((k) && (b[i] == 0) && (q)) continue;
if ((l) && (c[i] == 0) && (v)) continue;
add(dp[i][j & (p >= a[i])][k & (q >= b[i])][l & (v >= c[i])],
dp[i + 1][j][k][l]);
add(Dp[i][j & (p >= a[i])][k & (q >= b[i])][l & (v >= c[i])],
Dp[i + 1][j][k][l]);
if (v)
add(Dp[i][j & (p >= a[i])][k & (q >= b[i])][l & (v >= c[i])],
1LL * (1 << i) * dp[i + 1][j][k][l] % 1000000007);
}
int ans = 0;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
add(ans, Dp[0][i][j][k]), add(ans, dp[0][i][j][k]);
return ans;
}
int main() {
n = read();
while (n--) {
int x0 = read() - 2, y0 = read() - 2, x1 = read() - 1, y1 = read() - 1,
k = read() - 1;
int ans = 0;
add(ans, query(x1, y1, k));
add(ans, -query(x0, y1, k));
add(ans, -query(x1, y0, k));
add(ans, query(x0, y0, k));
if (ans < 0) ans += 1000000007;
printf("%d\n", ans);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mo = 1000000007;
int lx[55], ly[55], rx[55], ry[55];
int v[55], vis[55][2][2][2][2][2];
pair<int, int> f[55][2][2][2][2][2];
pair<int, int> F(int d, int v1, int v2, int v3, int v4, int v5) {
if (d == -1) return pair<int, int>(0, 1);
if (vis[d][v1][v2][v3][v4][v5]) return f[d][v1][v2][v3][v4][v5];
pair<int, int> ans(0, 0);
for (int vx = (int)(v1 ? lx[d] : 0); vx <= (int)(v2 ? rx[d] : 1); vx++)
for (int vy = (int)(v3 ? ly[d] : 0); vy <= (int)(v4 ? ry[d] : 1); vy++)
if (!v5 || ((vx ^ vy) <= v[d])) {
pair<int, int> tmp =
F(d - 1, v1 & (vx == lx[d]), v2 & (vx == rx[d]), v3 & (vy == ly[d]),
v4 & (vy == ry[d]), v5 & ((vx ^ vy) == v[d]));
if (vx ^ vy) tmp.first = (tmp.first + (1ll << d) * tmp.second) % mo;
ans.first = (ans.first + tmp.first) % mo;
ans.second = (ans.second + tmp.second) % mo;
}
vis[d][v1][v2][v3][v4][v5] = 1;
return f[d][v1][v2][v3][v4][v5] = ans;
}
void decode(int *a, int d) {
for (int i = (int)(0); i <= (int)(29); i++) a[i] = (d >> i) & 1;
}
void solve() {
int a, b, c, d, e;
scanf("%d%d%d%d%d", &a, &c, &b, &d, &e);
decode(lx, a - 1);
decode(rx, b - 1);
decode(ly, c - 1);
decode(ry, d - 1);
decode(v, min(e, 1 << 30) - 1);
memset(f, 0, sizeof(f));
memset(vis, 0, sizeof(vis));
pair<int, int> ans = F(29, 1, 1, 1, 1, 1);
printf("%d\n", (ans.first + ans.second) % mo);
}
int main() {
int T;
scanf("%d", &T);
while (T--) solve();
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long p = 1e9 + 7;
long long read() {
long long s = 0;
char c = getchar(), lc = '+';
while (c < '0' || '9' < c) lc = c, c = getchar();
while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
return lc == '-' ? -s : s;
}
void write(long long x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x < 10)
putchar(x + '0');
else {
write(x / 10);
putchar(x % 10 + '0');
}
}
void print(long long x, char c = '\n') {
write(x);
putchar(c);
}
inline void add(long long &x, long long y) {
x += y;
if (x >= p) x -= p;
}
inline long long lowbit(long long x) { return x & -x; }
inline long long calc(long long x1, long long x2, long long y1, long long y2,
long long k) {
if (x2 - x1 < y2 - y1) swap(x1, y1), swap(x2, y2);
long long tmp = x2 - x1;
x2 = (x1 = (x1 ^ y1) & ~(tmp - 1)) + tmp;
x1++;
x2 = min(x2, k);
if (x1 > x2 || y1 > y2) return 0;
return (x1 + x2) * (x2 - x1 + 1) / 2 % p * (y2 - y1) % p;
}
inline long long solve(long long x, long long y, long long k) {
long long ret = 0;
for (long long i = x; i; i &= i - 1)
for (long long j = y; j; j &= j - 1)
add(ret, calc(i - lowbit(i), i, j - lowbit(j), j, k));
return ret;
}
signed main() {
long long q = read();
while (q--) {
long long x1 = read(), y1 = read(), x2 = read(), y2 = read(), k = read();
print((solve(x2, y2, k) - solve(x1 - 1, y2, k) - solve(x2, y1 - 1, k) +
solve(x1 - 1, y1 - 1, k) + p + p) %
p);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int cond = 0, multi = 1, gcj = 0;
const int MOD = (int)1e9 + 7;
unordered_map<long long, array<long long, 2> > cache;
array<long long, 2> dp(int r, int c, int k, int len) {
if (r <= 0 || c <= 0 || k <= 0) return array<long long, 2>();
k = min(k, len);
r = min(r, len);
c = min(c, len);
long long state = (((((long long)r * MOD) + c) * MOD) + k) * MOD + len;
auto it = cache.find(state);
if (it != cache.end()) return it->second;
if (len == 1) {
return array<long long, 2>({{1, 1}});
;
}
r = min(r, len);
c = min(c, len);
array<long long, 2> res = array<long long, 2>();
for (auto dr = (0); dr < (2); ++dr)
for (auto dc = (0); dc < (2); ++dc) {
long long nk = (dr == dc) ? k : (k - len / 2);
array<long long, 2> next =
dp(r - (len / 2) * dr, c - (len / 2) * dc, nk, len / 2);
res[1] = (res[1] + next[1]) % MOD;
res[0] = (res[0] + next[0]) % MOD;
if (dr != dc) res[0] = (res[0] + next[1] * (long long)(len / 2)) % MOD;
}
return cache[state] = res;
}
array<long long, 2> operator+(array<long long, 2> le, array<long long, 2> ri) {
return array<long long, 2>({{le[0] + ri[0], le[1] + ri[1]}});
}
array<long long, 2> operator-(array<long long, 2> le, array<long long, 2> ri) {
return array<long long, 2>({{le[0] - ri[0], le[1] - ri[1]}});
}
struct solver {
void solve() {
function<int(int)> dfs = [&](int x) { return x; };
int x1, y1, x2, y2, k;
cin >> x1 >> y1 >> x2 >> y2 >> k;
array<long long, 2> res =
(dp(x2, y2, k, 1 << 30) - dp(x2, y1 - 1, k, 1 << 30) -
dp(x1 - 1, y2, k, 1 << 30) + dp(x1 - 1, y1 - 1, k, 1 << 30));
cout << ((res[0] % MOD + MOD) % MOD) << endl;
}
};
int main(int argc, char** argv) {
ios::sync_with_stdio(false);
cond = argc >= 2 && argv[1][0] == 'q' ? 1 << 30 : 0;
cout.setf(ios::fixed);
cout.precision(10);
int t;
if (multi || gcj)
cin >> t;
else
t = 1;
for (auto i = (1); i <= (t); ++i) {
if (cond) cerr << 82 << " " << i << endl;
if (gcj) cout << "Case #" << i << ": ";
solver s;
s.solve();
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline int read() {
int x = 0, f = 0;
char ch = getchar();
for (; ch < '0' || ch > '9'; ch = getchar())
if (ch == '-') f = 1;
for (; ch >= '0' && ch <= '9'; ch = getchar())
x = (x << 1) + (x << 3) + ch - 48;
return f ? -x : x;
}
inline void write(long long x) {
if (x < 10)
putchar(x + 48);
else
write(x / 10), putchar(x % 10 + 48);
}
inline void writeln(long long x) {
if (x < 0) x = -x, putchar('-');
write(x);
putchar('\n');
}
long long k;
const long long mod = 1e9 + 7;
long long work(long long x, long long y, long long p) {
if (!x || !y || p >= k) return 0;
if (x < y) swap(x, y);
long long t = x;
for (; t != (t & -t); t -= (t & -t))
;
long long w = min(k, p + t);
if (t >= y)
return ((w - p) * (p + w + 1) / 2 % mod * y % mod + work(x - t, y, p + t)) %
mod;
else
return ((w - p) * (p + w + 1) / 2 % mod * t % mod +
work(x - t, min(y, t), p + t) + work(min(t, x), y - t, p + t) +
work(x - t, y - t, p)) %
mod;
}
int main() {
long long Q = read();
while (Q--) {
long long x1 = read(), y1 = read(), x2 = read(), y2 = read();
k = read();
writeln((work(x2, y2, 0) + work(x1 - 1, y1 - 1, 0) - work(x1 - 1, y2, 0) -
work(x2, y1 - 1, 0) + mod + mod) %
mod);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, mod = 1e9 + 7, M = 1e9 + 5;
int inx1, iny1, inx2, iny2, k;
bool dInc(int x, int y, int len) {
int x2 = x + len - 1, y2 = y + len - 1;
return (x2 < inx1 || x > inx2 || y2 < iny1 || y > iny2);
}
int getSum(unsigned long long x) { return ((x * (x + 1)) / 2) % mod; }
int sum(int len, int start, int len2) {
int d = start + len - 1;
d = min(d, k);
long long S = (getSum(d) - getSum(start - 1)) % mod;
S = (S + mod) % mod;
return S * len2 % mod;
}
int solve(int x, int y, int len, int start) {
if (start > k || dInc(x, y, len)) return 0;
if (x >= inx1 && x + len - 1 <= inx2)
return sum(len, start, max(0, min(y + len - 1, iny2) - max(iny1, y) + 1));
if (y >= iny1 && y + len - 1 <= iny2)
return sum(len, start, max(0, min(x + len - 1, inx2) - max(inx1, x) + 1));
long long ret = 0;
int n_x, n_y;
n_x = x, n_y = y;
int n_len = len >> 1;
ret += solve(n_x, n_y, n_len, start);
n_x = x, n_y = y + n_len;
ret += solve(n_x, n_y, n_len, start + n_len);
n_x = x + n_len, n_y = y;
ret += solve(n_x, n_y, n_len, start + n_len);
n_x = x + n_len, n_y = y + n_len;
ret += solve(n_x, n_y, n_len, start);
ret %= mod;
return ret;
}
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int q;
cin >> q;
while (q--) {
cin >> inx1 >> iny1 >> inx2 >> iny2 >> k;
cout << solve(1, 1, (1 << 30), 1) << "\n";
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
using ld = long double;
const int mod = 1000000007;
struct Mod {
public:
int num;
Mod() : Mod(0) { ; }
Mod(long long int n) : num((n % mod + mod) % mod) {
static_assert(mod < INT_MAX / 2,
"mod is too big, please make num 'long long int' from 'int'");
}
Mod(int n) : Mod(static_cast<long long int>(n)) { ; }
operator int() { return num; }
};
Mod operator+(const Mod a, const Mod b) { return Mod((a.num + b.num) % mod); }
Mod operator+(const long long int a, const Mod b) { return Mod(a) + b; }
Mod operator+(const Mod a, const long long int b) { return b + a; }
Mod operator++(Mod &a) { return a + Mod(1); }
Mod operator-(const Mod a, const Mod b) {
return Mod((mod + a.num - b.num) % mod);
}
Mod operator-(const long long int a, const Mod b) { return Mod(a) - b; }
Mod operator--(Mod &a) { return a - Mod(1); }
Mod operator*(const Mod a, const Mod b) {
return Mod(((long long)a.num * b.num) % mod);
}
Mod operator*(const long long int a, const Mod b) { return Mod(a) * b; }
Mod operator*(const Mod a, const long long int b) { return Mod(b) * a; }
Mod operator*(const Mod a, const int b) { return Mod(b) * a; }
Mod operator+=(Mod &a, const Mod b) { return a = a + b; }
Mod operator+=(long long int &a, const Mod b) { return a = a + b; }
Mod operator-=(Mod &a, const Mod b) { return a = a - b; }
Mod operator-=(long long int &a, const Mod b) { return a = a - b; }
Mod operator*=(Mod &a, const Mod b) { return a = a * b; }
Mod operator*=(long long int &a, const Mod b) { return a = a * b; }
Mod operator*=(Mod &a, const long long int &b) { return a = a * b; }
Mod operator^(const Mod a, const int n) {
if (n == 0) return Mod(1);
Mod res = (a * a) ^ (n / 2);
if (n % 2) res = res * a;
return res;
}
Mod mod_pow(const Mod a, const long long int n) {
if (n == 0) return Mod(1);
Mod res = mod_pow((a * a), (n / 2));
if (n % 2) res = res * a;
return res;
}
Mod inv(const Mod a) { return a ^ (mod - 2); }
Mod operator/(const Mod a, const Mod b) {
assert(b.num != 0);
return a * inv(b);
}
Mod operator/(const long long int a, const Mod b) { return Mod(a) / b; }
Mod operator/=(Mod &a, const Mod b) { return a = a / b; }
Mod fact[1024000], factinv[1024000];
void init(const int amax = 1024000) {
fact[0] = Mod(1);
factinv[0] = 1;
for (int i = 0; i < amax - 1; ++i) {
fact[i + 1] = fact[i] * Mod(i + 1);
factinv[i + 1] = factinv[i] / Mod(i + 1);
}
}
Mod comb(const int a, const int b) {
return fact[a] * factinv[b] * factinv[a - b];
}
int x1, x2, yy, y2;
int K;
int cnt = 0;
long long solve(int lu_num, long long int l, long long int r, long long int u,
long long int d) {
cnt++;
if (r <= x1 || x2 <= l || d <= yy || y2 <= u) return Mod(0);
long long ans = 0;
bool lr_break = x1 <= l && r <= x2;
bool ud_break = yy <= u && d <= y2;
if (lr_break && ud_break) {
long long int sz = max((r - l), (d - u));
long long int set_num = min((r - l), (d - u));
long long int sum = (lu_num + lu_num + sz - 1) * sz / 2;
sum %= mod;
long long int n_sum = 0;
if (lu_num > K) {
n_sum = 0;
} else if (lu_num + sz - 1 > K) {
long long int minus =
(K + 1 + lu_num + sz - 1) * (lu_num + sz - 1 - K - 1 + 1) / 2;
minus %= mod;
n_sum = (mod + sum - minus) % mod;
} else {
n_sum = sum;
}
return (n_sum * set_num) % mod;
} else if (lr_break) {
ans += solve(lu_num, l, r, u, (u + d) / 2);
ans += solve(lu_num, l, r, (u + d) / 2, d);
} else if (ud_break) {
ans += solve(lu_num, l, (l + r) / 2, u, d);
ans += solve(lu_num, (l + r) / 2, r, u, d);
} else {
ans += solve(lu_num, l, (l + r) / 2, u, (u + d) / 2);
ans += solve(lu_num + (r - l) / 2, (l + r) / 2, r, u, (u + d) / 2);
ans += solve(lu_num + (r - l) / 2, l, (l + r) / 2, (u + d) / 2, d);
ans += solve(lu_num, (l + r) / 2, r, (u + d) / 2, d);
}
ans %= mod;
return ans;
}
int main() {
int Q;
cin >> Q;
while (Q--) {
scanf("%d %d %d %d %d", &x1, &yy, &x2, &y2, &K);
x1--;
yy--;
long long int answer = solve(1, 0, (1 << 30), 0, (1 << 30));
int aans = answer % mod;
const char ch = '/' + 1;
printf("%d\n", aans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline long long read() {
long long s = 0, t = 1;
char c = getchar();
while (!isdigit(c)) {
if (c == '-') t = -1;
c = getchar();
}
while (isdigit(c)) s = s * 10 + c - 48, c = getchar();
return s * t;
}
const int N = 3e5 + 5, p = 1e9 + 7;
inline void Max(int &x, int v) {
if (v > x) x = v;
}
inline void Min(int &x, int v) {
if (x > v) x = v;
}
inline void M(long long &x) {
if (x >= p) x -= p;
}
long long k;
inline long long solve(long long d, long long x1, long long y1, long long x2,
long long y2, long long f) {
if (k < f) return 0;
int D = d / 2;
if (x1 == x2 && x1 == 1) {
long long lf = f + y1 - 1, rf = min(f + y2 - 1, k);
if (lf <= rf) return (lf + rf) * (rf - lf + 1) / 2 % p;
}
if (y1 == y2 && y1 == 1) {
long long lf = f + x1 - 1, rf = min(f + x2 - 1, k);
if (lf <= rf) return (lf + rf) * (rf - lf + 1) / 2 % p;
}
if (x1 == 1 && x2 == d) {
long long lf = f, rf = min(f + d - 1, k);
if (lf <= rf) return (lf + rf) * (rf - lf + 1) / 2 % p * (y2 - y1 + 1) % p;
}
if (y1 == 1 && y2 == d) {
long long lf = f, rf = min(f + d - 1, k);
if (lf <= rf) return (lf + rf) * (rf - lf + 1) / 2 % p * (x2 - x1 + 1) % p;
}
if (x1 == 1 && y1 == 1 && x2 == d && y2 == d) {
long long lf = f, rf = min(f + d - 1, k);
return (lf + rf) * (rf - lf + 1) / 2 % p;
} else {
long long ret = 0, lx = 1, rx = D, ly = 1, ry = D, qlx, qrx, qly, qry;
for (register int i = 0; i <= 1; ++i) {
qlx = max(lx, x1), qrx = min(rx, x2);
qly = max(ly, y1), qry = min(ry, y2);
if (qlx <= qrx && qly <= qry)
M(ret += solve(D, qlx - i * D, qly, qrx - i * D, qry, f + i * D));
lx += D, rx += D;
}
lx = 1, rx = D, ly = D + 1, ry = d;
for (register int i = 0; i <= 1; ++i) {
qlx = max(lx, x1), qrx = min(rx, x2);
qly = max(ly, y1), qry = min(ry, y2);
if (qlx <= qrx && qly <= qry)
M(ret +=
solve(D, qlx - i * D, qly - D, qrx - i * D, qry - D, f + (!i) * D));
lx += D, rx += D;
}
return ret;
}
}
int main() {
int n = read();
while (n--) {
int x1 = read(), y1 = read(), x2 = read(), y2 = read(), g = 1;
while (g < x2 || g < y2) g <<= 1;
k = read();
cout << solve(g, x1, y1, x2, y2, 1) << endl;
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | mod = 1000000007
def sum(x,y,k,add) :
if k<add:return 0
up=x+add
if up>k:up=k
add=add+1
return y*(((add+up)*(up-add+1)//2)%mod)%mod
def solve(x,y,k,add=0) :
if x==0 or y==0:return 0
if x>y:x,y=y,x
pw = 1
while (pw<<1)<=y:pw<<=1
if pw<=x:return (sum(pw,pw,k,add)+sum(pw,x+y-pw-pw,k,add+pw)+solve(x-pw,y-pw,k,add))%mod
else:return (sum(pw,x,k,add)+solve(x,y-pw,k,add+pw))%mod
q=int(input())
for i in range(0,q):
x1,y1,x2,y2,k=list(map(int,input().split()))
ans=(solve(x2, y2, k)-solve(x1 - 1, y2, k)-solve(x2, y1 - 1, k)+solve(x1-1,y1-1,k))%mod
if ans<0:ans+=mod
print(ans) | PYTHON3 |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int q;
long long int x1;
long long int my1;
long long int x2;
long long int y2;
long long int k;
long long int stage[32];
long long int width[32];
long long int height[32];
long long int dp[32][2][4];
long long int num[32][2][4];
bool exist[32][2][4];
long long int calc(int x, int y) {
if (x == 0 || y == 0) return 0;
int t = max(x, y);
int cnt = 1;
while (t > 0) {
cnt++;
t /= 2;
}
long long int k_ = k;
int cnt2 = 1;
while (1) {
k_ = k_ % 2 ? (k_ + 1) / 2 : k_ / 2;
cnt2++;
if (k_ == 1) break;
}
cnt = max(cnt, cnt2);
width[cnt - 1] = x;
height[cnt - 1] = y;
stage[cnt - 1] = k;
for (int i = cnt - 2; i >= 0; i--) {
stage[i] = stage[i + 1] % 2 ? (stage[i + 1] + 1) / 2 : stage[i + 1] / 2;
width[i] = width[i + 1] / 2 + width[i + 1] % 2;
height[i] = height[i + 1] / 2 + height[i + 1] % 2;
}
fill(num[0][0], num[cnt][0], 0);
fill(dp[0][0], dp[cnt][0], 0);
fill(exist[0][0], exist[cnt][0], false);
num[0][1][3] = 1;
dp[0][1][3] = 1;
exist[0][1][3] = true;
for (int i = 0; i < cnt - 1; i++) {
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 4; k++) {
if (!exist[i][j][k]) continue;
for (int s = 0; s < 2; s++) {
for (int t = 0; t < 2; t++) {
long long int p = (s + t) % 2;
long long int v =
(1000000007 +
((dp[i][j][k] * 2 + (p - 1) * num[i][j][k]) % 1000000007)) %
1000000007;
long long int n = num[i][j][k];
if (j == 1 && p && stage[i] * 2 > stage[i + 1]) continue;
if ((k & 2) && s && width[i] * 2 > width[i + 1]) continue;
if ((k & 1) && t && height[i] * 2 > height[i + 1]) continue;
int nj = 0;
if (j == 1 && stage[i] * 2 - 1 + p == stage[i + 1]) nj = 1;
int nk = 0;
if ((k & 1) && height[i] * 2 - 1 + t == height[i + 1]) nk += 1;
if ((k & 2) && width[i] * 2 - 1 + s == width[i + 1]) nk += 2;
exist[i + 1][nj][nk] = true;
dp[i + 1][nj][nk] = (dp[i + 1][nj][nk] + v) % 1000000007;
num[i + 1][nj][nk] = (num[i + 1][nj][nk] + n) % 1000000007;
}
}
}
}
}
long long int ret = 0;
for (int i = 0; i < 4; i++) {
ret += dp[cnt - 1][0][i];
ret += dp[cnt - 1][1][i];
ret %= 1000000007;
}
return ret;
}
int main() {
scanf("%d", &q);
while (q--) {
scanf("%I64d%I64d%I64d%I64d%I64d", &x1, &my1, &x2, &y2, &k);
long long int ans = calc(x2, y2) - calc(x2, my1 - 1) - calc(x1 - 1, y2) +
calc(x1 - 1, my1 - 1);
printf("%I64d\n", (1000000007 + ans % 1000000007) % 1000000007);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
using namespace placeholders;
template <class T>
void mini(T &l, T r) {
l = min(l, r);
}
template <class T>
void maxi(T &l, T r) {
l = max(l, r);
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
while (*sdbg != ',') cerr << *sdbg++;
cerr << "=" << h << ",";
_dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> v) {
os << "[";
for (auto vv : v) os << vv << ",";
return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> p) {
return os << "(" << p.first << "," << p.second << ")";
}
using llint = long long;
const int B = 1e9 + 7;
int a[40][40];
llint mem[40], cnt[40];
void add(llint &a, llint b) {
a = (a + b) % B;
if (a < 0) a += B;
}
void init() {
mem[0] = 1;
cnt[0] = 1;
for (int i = 1; i < 40; ++i) {
int c = 1 << (i - 1);
cnt[i] = cnt[i - 1] * 4 % B;
add(mem[i], mem[i - 1] * 4 % B + 1ll * c * c % B * c % B * 2 % B);
}
}
llint calc(llint a, llint b) {
a %= B, b %= B;
return ((a + b) * (b - a + 1) / 2) % B;
}
pair<llint, llint> get(int x, int y, int k, int d) {
llint ret = 0;
int sz = 1 << d;
if (x < y) swap(x, y);
if (x <= 0 || y <= 0 || k <= 0) return make_pair(0, 0);
if (x >= sz) {
llint a = min(k, sz);
llint b = min(y, sz);
ret = calc(1, a) * b % B;
;
return make_pair(b * calc(1, a) % B, b * a % B);
}
if (d == 0) return make_pair(1 <= k, 1 <= k);
int t = 1 << (d - 1);
auto t1 = get(x, y, k, d - 1);
auto t2 = get(x - t, y, k - t, d - 1);
auto t3 = get(y, y - t, k - t, d - 1);
auto t4 = get(x - t, y - t, k, d - 1);
;
add(ret, t1.first);
add(ret, t2.first + t2.second * t % B);
add(ret, t3.first + t3.second * t % B);
add(ret, t4.first);
llint c = 0;
add(c, t1.second);
add(c, t2.second);
add(c, t3.second);
add(c, t4.second);
;
return make_pair(ret, c);
}
llint go(int x1, int y1, int x2, int y2, int k) {
llint ret = 0;
add(ret, get(x2, y2, k, 30).first);
add(ret, -get(x1 - 1, y2, k, 30).first);
add(ret, -get(x2, y1 - 1, k, 30).first);
add(ret, get(x1 - 1, y1 - 1, k, 30).first);
return ret;
}
void run() {
init();
;
int q;
scanf("%d", &q);
for (int i = 1; i <= q; ++i) {
int x1, y1, x2, y2, k;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
printf("%lld\n", go(x1, y1, x2, y2, k));
}
}
int main() {
run();
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int md = 1e9 + 7;
void Rd(int &res) {
res = 0;
static char c;
while (c = getchar(), c < 48)
;
do res = (res << 1) + (res << 3) + (c ^ 48);
while (c = getchar(), c > 47);
}
void Add(int &x, int y) {
x += y;
if (x >= md) x -= md;
}
int dp[2][2][2][2], Sum[2][2][2][2];
int Calc(int x, int y, int K) {
if (!x || !y) return 0;
for (int i = (0), i_END_ = (1); i <= i_END_; ++i)
for (int j = (0), i_END_ = (1); j <= i_END_; ++j)
for (int k = (0), i_END_ = (1); k <= i_END_; ++k)
for (int h = (0), i_END_ = (1); h <= i_END_; ++h)
Sum[i][j][k][h] = dp[i][j][k][h] = 0;
int cur = 0;
dp[cur][0][0][0] = 1;
x--;
y--;
for (int i = (30), i_END_ = (0); i >= i_END_; --i) {
cur ^= 1;
int a = (x >> i) & 1, b = (y >> i) & 1, c = (K >> i) & 1;
int res = (1 << i) % md;
for (int j = (0), i_END_ = (1); j <= i_END_; ++j)
for (int k = (0), i_END_ = (1); k <= i_END_; ++k)
for (int h = (0), i_END_ = (1); h <= i_END_; ++h)
Sum[cur][j][k][h] = dp[cur][j][k][h] = 0;
for (int j = (0), i_END_ = (1); j <= i_END_; ++j)
for (int k = (0), i_END_ = (1); k <= i_END_; ++k)
for (int h = (0), i_END_ = (1); h <= i_END_; ++h) {
Add(dp[cur][j | a][k | b][h | c], dp[cur ^ 1][j][k][h]);
Add(Sum[cur][j | a][k | b][h | c], Sum[cur ^ 1][j][k][h]);
if ((j || a) && (k || b)) {
Add(dp[cur][j][k][h | c], dp[cur ^ 1][j][k][h]);
Add(Sum[cur][j][k][h | c], Sum[cur ^ 1][j][k][h]);
}
if (h || c) {
if (j || a) {
Add(dp[cur][j][k | b][h], dp[cur ^ 1][j][k][h]);
int &t = Sum[cur][j][k | b][h];
Add(t, Sum[cur ^ 1][j][k][h]);
Add(t, 1ll * res * dp[cur ^ 1][j][k][h] % md);
}
if (k || b) {
Add(dp[cur][j | a][k][h], dp[cur ^ 1][j][k][h]);
int &t = Sum[cur][j | a][k][h];
Add(t, Sum[cur ^ 1][j][k][h]);
Add(t, 1ll * res * dp[cur ^ 1][j][k][h] % md);
}
}
}
}
int ans = 0;
for (int i = (0), i_END_ = (1); i <= i_END_; ++i)
for (int j = (0), i_END_ = (1); j <= i_END_; ++j)
Add(ans, (dp[cur][i][j][1] + Sum[cur][i][j][1]) % md);
return ans;
}
int main() {
int _;
int x1, y1, x2, y2, K;
for (Rd(_); _; _--) {
Rd(x1), Rd(y1), Rd(x2), Rd(y2), Rd(K);
int ans = (Calc(x2, y2, K) + Calc(x1 - 1, y1 - 1, K) - Calc(x1 - 1, y2, K) -
Calc(x2, y1 - 1, K)) %
md;
if (ans < 0) ans += md;
printf("%d\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
void add(int &x, int y) {
x += y;
if (x >= mod) x -= mod;
}
int mult(int x, int y) { return (1ll * x * y) % mod; }
int dp1[55][2][2][2];
int dp2[55][2][2][2];
int query(int x, int y, int k) {
if (k < 0 || x < 0 || y < 0) return 0ll;
memset(dp1, 0, sizeof(dp1));
memset(dp2, 0, sizeof(dp2));
dp1[31][1][1][1] = 1;
dp2[31][1][1][1] = 0;
for (int i = 31; i; --i)
for (int a = 0; a < 2; ++a)
for (int b = 0; b < 2; ++b)
for (int c = 0; c < 2; ++c)
for (int p1 = 0; p1 < 2; ++p1)
for (int p2 = 0; p2 < 2; ++p2) {
const int p3 = p1 ^ p2;
const int x1 = ((x >> (i - 1)) & 1);
const int y1 = ((y >> (i - 1)) & 1);
const int k1 = ((k >> (i - 1)) & 1);
if (a && x1 < p1) continue;
if (b && y1 < p2) continue;
if (c && k1 < p3) continue;
add(dp1[i - 1][a & (p1 == x1)][b & (p2 == y1)][c & (p3 == k1)],
dp1[i][a][b][c]);
add(dp2[i - 1][a & (p1 == x1)][b & (p2 == y1)][c & (p3 == k1)],
dp2[i][a][b][c]);
add(dp2[i - 1][a & (p1 == x1)][b & (p2 == y1)][c & (p3 == k1)],
mult(p3 << (i - 1), dp1[i][a][b][c]));
}
int ans = 0;
for (int a = 0; a < 2; ++a)
for (int b = 0; b < 2; ++b)
for (int c = 0; c < 2; ++c)
add(ans, dp1[0][a][b][c]), add(ans, dp2[0][a][b][c]);
return ans;
}
int main(void) {
int t;
cin >> t;
while (t--) {
int x1, y1, x2, y2, k;
cin >> x1 >> y1 >> x2 >> y2 >> k;
--x1;
--y1;
--x2;
--y2;
--k;
int ans = 0;
add(ans, query(x2, y2, k));
add(ans, query(x1 - 1, y1 - 1, k));
add(ans, mod - query(x1 - 1, y2, k));
add(ans, mod - query(x2, y1 - 1, k));
cout << ans << '\n';
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int dp[2][2][2], oldDp[2][2][2], cnt[2][2][2], oldCnt[2][2][2];
const int mod = 1e9 + 7;
int add(int x, int y) {
int ans = x + y;
if (ans >= mod) ans -= mod;
return ans;
}
int subtract(int x, int y) {
if (x >= y) return x - y;
return x - y + mod;
}
int mul(int x, int y) { return 1LL * x * y % mod; }
void adto(int &x, int y) {
x += y;
if (x >= mod) x -= mod;
}
int solve(int n, int m, int K) {
if (n < 0 || m < 0) return 0;
n++, m++;
memset(dp, 0, sizeof(dp));
memset(cnt, 0, sizeof(cnt));
cnt[0][0][(K >> 30) & 1] = 1;
for (int p = 29; p >= 0; p--) {
int bN = (n >> p) & 1, bM = (m >> p) & 1, bK = (K >> p) & 1;
memcpy(oldCnt, cnt, sizeof(cnt));
memcpy(oldDp, dp, sizeof(dp));
memset(dp, 0, sizeof(dp));
memset(cnt, 0, sizeof(cnt));
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
if (oldCnt[i][j][k])
for (int b1 = 0; b1 < 2; b1++)
for (int b2 = 0; b2 < 2; b2++) {
if (b1 > bN && i == 0) continue;
if (b2 > bM && j == 0) continue;
if ((b1 ^ b2) > bK && k == 0) continue;
int ni = i | (b1 < bN), nj = j | (b2 < bM),
nk = k | ((b1 ^ b2) < bK);
adto(cnt[ni][nj][nk], oldCnt[i][j][k]);
if (b1 ^ b2)
adto(dp[ni][nj][nk],
add(oldDp[i][j][k], mul(oldCnt[i][j][k], 1 << p)));
else
adto(dp[ni][nj][nk], oldDp[i][j][k]);
}
}
return add(dp[1][1][1], cnt[1][1][1]);
}
int main() {
int Tests;
scanf("%d", &Tests);
while (Tests--) {
int a1, b1, a2, b2, k;
scanf("%d %d %d %d %d", &a1, &b1, &a2, &b2, &k), a1--, b1--, a2--, b2--;
int ans = subtract(add(solve(a2, b2, k), solve(a1 - 1, b1 - 1, k)),
add(solve(a1 - 1, b2, k), solve(a2, b1 - 1, k)));
printf("%d\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int l, r, u, d, k;
int bin(int a, int i) { return a & ((1 << i) - 1); }
int sum(int k) { return 1LL * (1 + k) * k / 2 % MOD; }
void add(int &a, int b) {
if ((a += b) >= MOD) a -= MOD;
}
int dfs(int i, int a, int b, int k, int d = 0) {
if (i < 0) return (d + 1) % MOD;
if (a > b) swap(a, b);
if (b + 1 == (1LL << (i + 1))) {
return (1LL * (a + 1) * sum(k + 1) % MOD +
1LL * (a + 1) * (k + 1) % MOD * d % MOD) %
MOD;
} else {
int ret = 0, mv = (1LL << i) - 1;
int ai = (a >> i & 1), bi = (b >> i & 1), ki = (k >> i & 1);
for (int aa = 0; aa <= ai; aa++) {
for (int bb = 0; bb <= bi; bb++) {
if ((aa ^ bb) > ki) continue;
add(ret, dfs(i - 1, aa < ai ? mv : a - (ai << i),
bb < bi ? mv : b - (bi << i),
(aa ^ bb) < ki ? mv : k - (ki << i),
(d + ((aa ^ bb) << i)) % MOD));
}
}
return ret;
}
}
int solve(int a, int b, int k) {
if (a > b) swap(a, b);
if (a < 0) return 0;
return dfs(31, a, b, k);
}
int main() {
int q;
scanf("%d", &q);
while (q--) {
scanf("%d %d %d %d %d", &l, &d, &r, &u, &k);
l--, d--, r--, u--, k--;
int ans = (solve(r, u, k) - solve(l - 1, u, k) + MOD) % MOD;
ans = (ans - solve(r, d - 1, k) + MOD) % MOD;
ans = (ans + solve(l - 1, d - 1, k)) % MOD;
cout << ans << endl;
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long M = 1000000007;
long long S(long long n, long long k) {
long long mi = min(n, k);
return (mi * (mi + 1) / 2) % M;
}
long long Q2(long long y, long long x, long long k, long long delta) {
if (k - delta <= 0) return 0;
if (x > y) swap(x, y);
if (x == 0 || y == 0) return 0;
assert(y >= x);
assert(x >= 0);
assert(y >= 0);
long long p = 1;
while (p * 2 <= y) p *= 2;
long long ret = min(p, x) * ((S(p + delta, k) - S(delta, k) + M) % M) % M;
long long newX = x - min(p, x);
long long newY = y - p;
ret = (ret + Q2(newY, x - newX, k, delta + p) +
Q2(y - newY, newX, k, delta + p) + Q2(newY, newX, k, delta)) %
M;
return ret;
}
long long Q1(long long y1, long long x1, long long y2, long long x2,
long long k) {
long long ret = (Q2(y2, x2, k, 0) + Q2(y1 - 1, x1 - 1, k, 0)) % M;
long long ret2 = (Q2(y1 - 1, x2, k, 0) + Q2(y2, x1 - 1, k, 0)) % M;
return (ret - ret2 + M) % M;
}
int main() {
std::ios::sync_with_stdio(false);
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
long long y1, x1, y2, x2, k;
cin >> y1 >> x1 >> y2 >> x2 >> k;
cout << Q1(y1, x1, y2, x2, k) << endl;
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int mod_plus(int a, int b) {
a += b;
if (a >= 1000000007) a -= 1000000007;
return a;
}
int mod_mul(long long a, long long b) { return (a * b) % 1000000007; }
int pw2[55];
int dp[55][4][4][2], sum[55][4][4][2];
long long X1, X2, Y1, Y2, val;
int solve(int bit, int xst, int yst, int valst) {
if (bit < 0) {
sum[bit + 1][xst][yst][valst] = 0;
return 1;
}
int &ret = dp[bit][xst][yst][valst];
if (ret != -1) return ret;
ret = 0;
int &s = sum[bit + 1][xst][yst][valst];
s = 0;
int X1bit = !!(X1 & (1ll << bit)), X2bit = !!(X2 & (1ll << bit)),
Y1bit = !!(Y1 & (1ll << bit)), Y2bit = !!(Y2 & (1ll << bit));
int xlo = 0, xhi = 1, ylo = 0, yhi = 1;
int valbit = !!(val & (1ll << bit));
if (!(xst & 1) && X1bit) xlo++;
if (!(xst & 2) && !X2bit) xhi--;
if (!(yst & 1) && Y1bit) ylo++;
if (!(yst & 2) && !Y2bit) yhi--;
for (int i = xlo; i <= xhi; i++) {
for (int j = ylo; j <= yhi; j++) {
int kval = i ^ j;
if (!valst && kval > valbit) continue;
int tmp = solve(bit - 1, xst | (1 * (i > X1bit)) | (2 * (i < X2bit)),
yst | (1 * (j > Y1bit)) | (2 * (j < Y2bit)),
valst | (kval < valbit));
int jog = sum[bit][xst | (1 * (i > X1bit)) | (2 * (i < X2bit))]
[yst | (1 * (j > Y1bit)) | (2 * (j < Y2bit))]
[valst | (kval < valbit)];
if (kval) s = mod_plus(s, mod_mul(tmp, pw2[bit]));
ret = mod_plus(ret, tmp);
s = mod_plus(s, jog);
}
}
return ret;
}
int main() {
int q;
pw2[0] = 1;
for (q = 1; q < 55; q++) pw2[q] = mod_mul(pw2[q - 1], 2);
scanf("%d", &q);
while (q--) {
scanf("%lld%lld%lld%lld%lld", &X1, &Y1, &X2, &Y2, &val);
X1--;
X2--;
Y1--;
Y2--;
val--;
memset(dp, -1, sizeof(dp));
int ttt = solve(50, 0, 0, 0);
printf("%d\n", mod_plus(sum[51][0][0][0], ttt));
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline long long read() {
long long x = 0, f = 1;
char c = getchar();
while ((c < '0' || c > '9') && (c != '-')) c = getchar();
if (c == '-') f = -1, c = getchar();
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
const int mod = 1e9 + 7;
int dp[32][2][2][2], g[32][2][2][2];
inline void upd(int &x, int y) {
x += y;
if (x >= mod) x -= mod;
}
inline int calc(int l, int r, int k) {
if (l < 0 || r < 0) return 0;
memset((dp), (0), sizeof(dp)), memset((g), (0), sizeof(g));
dp[31][1][1][1] = 1;
for (register int i = (30); i >= (0); i--) {
int a = (l >> i & 1), b = (r >> i & 1), c = (k >> i & 1);
for (register int j = (0); j <= (1); j++)
for (register int k = (0); k <= (1); k++)
for (register int l = (0); l <= (1); l++) {
for (register int x = (0); x <= (j ? a : 1); x++)
for (register int y = (0); y <= (k ? b : 1); y++) {
if (l == 0) {
upd(dp[i][j && x == a][k && y == b][l], dp[i + 1][j][k][l]);
upd(g[i][j && x == a][k && y == b][l],
(g[i + 1][j][k][l] +
1ll * dp[i + 1][j][k][l] * ((x ^ y) << i)) %
mod);
} else if ((x ^ y) <= c) {
upd(dp[i][j && x == a][k && y == b][l && ((x ^ y) == c)],
dp[i + 1][j][k][l]);
upd(g[i][j && x == a][k && y == b][l && ((x ^ y) == c)],
(g[i + 1][j][k][l] +
1ll * dp[i + 1][j][k][l] * ((x ^ y) << i)) %
mod);
}
}
}
}
int ans = 0;
for (register int i = (0); i <= (1); i++)
for (register int j = (0); j <= (1); j++)
for (register int k = (0); k <= (1); k++)
upd(ans, dp[0][i][j][k]), upd(ans, g[0][i][j][k]);
return ans;
}
inline void solve() {
int x1 = read() - 1, y1 = read() - 1, x2 = read() - 1, y2 = read() - 1,
k = read() - 1;
int ans = calc(x2, y2, k);
ans = (ans + calc(x1 - 1, y1 - 1, k)) % mod;
ans = (ans + mod - calc(x1 - 1, y2, k)) % mod;
ans = (ans + mod - calc(x2, y1 - 1, k)) % mod;
printf("%d\n", ans);
}
int main() {
int T = read();
while (T--) solve();
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 500010;
const int mod = 1000000007;
int k, a, b;
int memo1[40][2][2][2];
int count(int i, bool limk, bool lima, bool limb) {
if (k < 0 || a < 0 || b < 0) return 0;
if (i == -1) return 1;
int &ret = memo1[i][limk][lima][limb];
if (ret != -1) return ret;
ret = 0;
int bita = ((1 << i) & a) > 0;
int bitb = ((1 << i) & b) > 0;
int bitk = ((1 << i) & k) > 0;
bool nlimk = !bitk && limk;
bool nlima = !bita && lima;
bool nlimb = !bitb && limb;
ret += count(i - 1, nlimk, nlima, nlimb);
if (!(limk && !bitk) && !(lima && !bita)) {
nlimk = bitk && limk;
nlima = bita && lima;
nlimb = !bitb && limb;
ret += count(i - 1, nlimk, nlima, nlimb);
ret %= mod;
}
if (!(limk && !bitk) && !(limb && !bitb)) {
nlimk = bitk && limk;
nlima = !bita && lima;
nlimb = bitb && limb;
ret += count(i - 1, nlimk, nlima, nlimb);
ret %= mod;
}
if (!(lima && !bita) && !(limb && !bitb)) {
nlimk = !bitk && limk;
nlima = bita && lima;
nlimb = bitb && limb;
ret += count(i - 1, nlimk, nlima, nlimb);
ret %= mod;
}
return ret;
}
int memo[40][2][2][2];
int bt(int i, bool limk, bool lima, bool limb) {
if (k < 0 || a < 0 || b < 0) return 0;
if (i == -1) return 1;
int &ret = memo[i][limk][lima][limb];
if (ret != -1) return ret;
ret = 0;
int bita = ((1 << i) & a) > 0;
int bitb = ((1 << i) & b) > 0;
int bitk = ((1 << i) & k) > 0;
bool nlimk = !bitk && limk;
bool nlima = !bita && lima;
bool nlimb = !bitb && limb;
ret += bt(i - 1, nlimk, nlima, nlimb);
if (!(limk && !bitk) && !(lima && !bita)) {
nlimk = bitk && limk;
nlima = bita && lima;
nlimb = !bitb && limb;
ret += (1LL * count(i - 1, nlimk, nlima, nlimb) * (1 << i) +
bt(i - 1, nlimk, nlima, nlimb)) %
mod;
ret %= mod;
}
if (!(limk && !bitk) && !(limb && !bitb)) {
nlimk = bitk && limk;
nlima = !bita && lima;
nlimb = bitb && limb;
ret += (1LL * count(i - 1, nlimk, nlima, nlimb) * (1 << i) +
bt(i - 1, nlimk, nlima, nlimb)) %
mod;
ret %= mod;
}
if (!(lima && !bita) && !(limb && !bitb)) {
nlimk = !bitk && limk;
nlima = bita && lima;
nlimb = bitb && limb;
ret += bt(i - 1, nlimk, nlima, nlimb);
ret %= mod;
}
return ret;
}
int calc(int i, int j, int inp) {
a = i - 1;
b = j - 1;
k = inp - 1;
memset(memo, -1, sizeof memo);
memset(memo1, -1, sizeof memo1);
int ret = bt(30, 1, 1, 1);
return ret;
}
int main() {
int q;
cin >> q;
while (q--) {
int x1, y1, x2, y2, inp;
cin >> x1 >> y1 >> x2 >> y2 >> inp;
long long ans = 0;
ans += calc(x2, y2, inp);
ans -= calc(x2, y1 - 1, inp);
ans -= calc(x1 - 1, y2, inp);
ans += calc(x1 - 1, y1 - 1, inp);
ans %= mod;
if (ans < 0) ans += mod;
cout << ans << "\n";
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int max_n = 3e5 + 5;
const int mod = 1e9 + 7;
int maska[35];
int maska2[35];
int rob2(int a, int b, int k) {
if (b < a) return rob2(b, a, k);
int ans = 0;
for (int i = (0); i < (30); ++i) {
if (!(a & (1 << i))) continue;
for (int j = (0); j < (30); ++j) {
if (!(b & (1 << j))) continue;
int dowolnych = max(i, j);
int pocz = ((a ^ (1 << i)) & maska2[dowolnych]) ^
((b ^ (1 << j)) & maska2[dowolnych]);
pocz = pocz & maska2[dowolnych];
int mam = 1 << min(i, j);
if (pocz < (k & maska2[dowolnych])) {
int kutas = ((1 << dowolnych) * 1LL * mam) % mod;
ans += (kutas * 1LL * pocz) % mod;
ans %= mod;
kutas =
((((1 << (dowolnych)) * 1LL * ((1 << dowolnych) - 1))) / 2) % mod;
ans += (kutas * 1LL * mam) % mod;
ans %= mod;
ans += ((1 << dowolnych) * 1LL * mam) % mod;
ans %= mod;
} else {
if ((k & maska2[dowolnych]) == pocz) {
int reszta = k & maska[dowolnych];
int kutas = ((reszta + 1) * 1LL * mam) % mod;
ans += (kutas * 1LL * pocz) % mod;
ans %= mod;
kutas = ((reszta * 1LL * (reszta + 1)) / 2) % mod;
ans += (kutas * 1LL * mam) % mod;
ans %= mod;
ans += ((reszta + 1) * 1LL * mam) % mod;
ans %= mod;
}
}
}
}
return ans;
}
int rob(int a, int b, int k) {
if (a == 0 || b == 0) return 0;
return rob2(a, b, k - 1);
}
int main() {
maska[0] = 0;
for (int i = (1); i < (32); ++i) maska[i] = maska[i - 1] + (1 << (i - 1));
for (int i = (0); i < (31); ++i) maska2[i] = maska[i] ^ maska[31];
int(q);
scanf("%d", &(q));
;
while (q--) {
int(x1), (y1);
scanf("%d%d", &(x1), &(y1));
;
int(x2), (y2);
scanf("%d%d", &(x2), &(y2));
;
int(k);
scanf("%d", &(k));
;
int ans1 = (rob(x2, y2, k) + rob(x1 - 1, y1 - 1, k)) % mod;
int ans2 = (rob(x1 - 1, y2, k) + rob(x2, y1 - 1, k)) % mod;
if (ans1 < 0) ans1 += mod;
printf("%d\n", (ans1 + mod - ans2) % mod);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
struct Typ {
int sum, g;
Typ(int a = 0, int b = 0) { sum = a, g = b; }
};
Typ dp[35][2][2][2];
bool vis[35][2][2][2];
inline void add(int &a, int b) {
a += b;
if (a >= mod) a -= mod;
}
Typ dfs(int w, int px, int py, int pk, int ux, int uy, int x, int y, int k) {
if (w == -1) return Typ(0, 1);
if (vis[w][px][py][pk]) return dp[w][px][py][pk];
int sum = 0, g = 0;
int wx = x >> w & 1;
int wy = y >> w & 1;
int wk = k >> w & 1;
if (wx) ux = 1;
if (wy) uy = 1;
for (int i = (0), _i = (3); i <= _i; i++) {
int rx = i >> 1 & 1;
int ry = i & 1;
if (rx > ux || ry > uy) continue;
if ((px && rx > wx) || (py && ry > wy) || (pk && (rx ^ ry) > wk)) continue;
Typ ret = dfs(w - 1, px && rx == wx, py && ry == wy, pk && (rx ^ ry) == wk,
ux, uy, x, y, k);
add(sum, ret.sum);
add(g, ret.g);
if (rx ^ ry) {
int tt = (1LL << w) * ret.g % mod;
add(sum, tt);
}
}
vis[w][px][py][pk] = 1;
return dp[w][px][py][pk] = Typ(sum, g);
}
int solve(int x, int y, int k) {
if (x == 0 || y == 0) return 0;
--x, --y, --k;
memset(vis, 0, sizeof(vis));
Typ ans = dfs(31, 1, 1, 1, 0, 0, x, y, k);
add(ans.sum, ans.g);
return ans.sum;
}
int main() {
int q;
scanf("%d", &q);
for (int i = (1), _i = (q); i <= _i; i++) {
int x1, y1, x2, y2, k;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
long long ans = (long long)solve(x2, y2, k) - solve(x2, y1 - 1, k) -
solve(x1 - 1, y2, k) + solve(x1 - 1, y1 - 1, k);
ans %= mod;
if (ans < 0) ans += mod;
printf("%d\n", (int)ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
int q, dp[31][8], sum[30][8], k;
void add(int &a, int b) {
a += b;
if (a >= M) a -= M;
}
void sub(int &a, int b) {
a -= b;
if (a < 0) a += M;
}
int find(int x, int y) {
if (x < 0 || y < 0) return 0;
int ret = 0;
memset(dp, 0, sizeof dp);
memset(sum, 0, sizeof sum);
dp[30][0] = 1;
for (int i = 30; i; i--)
for (int j = 0; j < 8; j++)
if (dp[i][j])
for (int at = 0; at < 4; at++) {
bool bitx = at & 1, bity = at & 2;
if (bitx && !(x & (1 << i)) && !(j & 4)) continue;
if (bity && !(y & (1 << i)) && !(j & 2)) continue;
if ((bitx ^ bity) && !(k & (1 << i)) && !(j & 1)) continue;
int to = j;
to |= 4 * (bitx < !!(x & (1 << i)));
to |= 2 * (bity < !!(y & (1 << i)));
to |= (bitx ^ bity) < !!(k & (1 << i));
add(dp[i - 1][to], dp[i][j]);
add(sum[i - 1][to], sum[i][j]);
add(sum[i - 1][to], ((bitx ^ bity) << i) * 1LL * dp[i][j] % M);
bitset<3> a(j), b(to);
}
int i = 0;
for (int j = 0; j < 8; j++)
for (int at = 0; at < 4; at++) {
bool bitx = at & 1, bity = at & 2;
if (bitx && !(x & (1 << i)) && !(j & 4)) continue;
if (bity && !(y & (1 << i)) && !(j & 2)) continue;
if ((bitx ^ bity) && !(k & (1 << i)) && !(j & 1)) continue;
add(ret, (bitx ^ bity) * dp[i][j]);
add(ret, dp[i][j]);
add(ret, sum[i][j]);
}
return ret;
}
int main() {
scanf("%d", &q);
while (q--) {
int x, x1, y, y1;
scanf("%d%d%d%d%d", &x, &y, &x1, &y1, &k);
x--;
y--;
x1--;
y1--;
k--;
int ans = 0;
add(ans, find(x1, y1));
sub(ans, find(x - 1, y1));
sub(ans, find(x1, y - 1));
add(ans, find(x - 1, y - 1));
printf("%d\n", ans);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int q, k;
int sum(int l, int r) {
if (l > r) return 0;
return (long long)(l + r) * (r - l + 1) / 2 % mod;
}
int solve(int x, int y, int h) {
if (x <= 0 || y <= 0) return 0;
int mx = max(x, y), t = 1, tmp = 0;
if (y > x) swap(x, y);
while (2 * t <= mx) t = 2 * t;
if (y <= t) {
tmp += (long long)sum(h + 1, min(k, t + h)) * y % mod;
tmp %= mod;
tmp += solve(x - t, y, h + t);
tmp %= mod;
return tmp;
} else {
tmp += (long long)sum(h + 1, min(k, t + h)) * t % mod;
tmp %= mod;
tmp += solve(x - t, y - t, h);
tmp %= mod;
tmp += solve(x - t, t, h + t);
tmp %= mod;
tmp += solve(t, y - t, h + t);
tmp %= mod;
return tmp;
}
}
int main() {
scanf("%d", &q);
while (q--) {
int x1, x2, y1, y2;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
int ans = solve(x1 - 1, y1 - 1, 0) + solve(x2, y2, 0) -
solve(x1 - 1, y2, 0) - solve(x2, y1 - 1, 0);
ans %= mod;
ans = (ans + mod) % mod;
printf("%d\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int q, i, f[40][2][2][2], g[40][2][2][2];
int read() {
char c = getchar();
int w = 0;
while (c < '0' || c > '9') c = getchar();
while (c <= '9' && c >= '0') {
w = w * 10 + c - '0';
c = getchar();
}
return w;
}
void add(int &x, int y) {
x += y;
if (x >= mod) x -= mod;
}
int cal(int x, int y, int z) {
if (x < 0 || y < 0) return 0;
memset(f, 0, sizeof(f));
memset(g, 0, sizeof(g));
f[31][1][1][1] = 1;
for (int i = 31; i >= 1; i--) {
for (int j = 0; j <= 1; j++) {
for (int k = 0; k <= 1; k++) {
for (int l = 0; l <= 1; l++) {
if (!f[i][j][k][l]) continue;
int la = j ? ((x >> (i - 1)) & 1) : 1;
int lb = k ? ((y >> (i - 1)) & 1) : 1;
int lc = l ? ((z >> (i - 1)) & 1) : 1;
for (int a = 0; a <= la; a++) {
for (int b = 0; b <= lb; b++) {
if ((a ^ b) > lc) continue;
add(f[i - 1][j & (a == la)][k & (b == lb)][l & ((a ^ b) == lc)],
f[i][j][k][l]);
add(g[i - 1][j & (a == la)][k & (b == lb)][l & ((a ^ b) == lc)],
g[i][j][k][l]);
add(g[i - 1][j & (a == la)][k & (b == lb)][l & ((a ^ b) == lc)],
1LL * f[i][j][k][l] * ((a ^ b) << (i - 1)) % mod);
}
}
}
}
}
}
int ans = 0;
for (int i = 0; i <= 1; i++) {
for (int j = 0; j <= 1; j++) {
for (int k = 0; k <= 1; k++) {
ans = (ans + f[0][i][j][k]) % mod;
ans = (ans + g[0][i][j][k]) % mod;
}
}
}
return ans;
}
int main() {
q = read();
while (q--) {
int x1 = read() - 1, y1 = read() - 1, x2 = read() - 1, y2 = read() - 1,
k = read() - 1;
long long ans = cal(x2, y2, k) - cal(x1 - 1, y2, k) - cal(x2, y1 - 1, k) +
cal(x1 - 1, y1 - 1, k);
ans = (ans + 2 * mod) % mod;
printf("%lld\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
long long int n, m, k;
long long int dp_cnt[35][2][2][2] = {};
long long int dp_sum[35][2][2][2] = {};
void reset_dp_cnt() {
for (int i = 0; i < 35; i++) {
dp_cnt[i][0][0][0] = -1;
dp_cnt[i][0][0][1] = -1;
dp_cnt[i][0][1][0] = -1;
dp_cnt[i][0][1][1] = -1;
dp_cnt[i][1][0][0] = -1;
dp_cnt[i][1][0][1] = -1;
dp_cnt[i][1][1][0] = -1;
dp_cnt[i][1][1][1] = -1;
}
}
void reset_dp_sum() {
for (int i = 0; i < 35; i++) {
dp_sum[i][0][0][0] = -1;
dp_sum[i][0][0][1] = -1;
dp_sum[i][0][1][0] = -1;
dp_sum[i][0][1][1] = -1;
dp_sum[i][1][0][0] = -1;
dp_sum[i][1][0][1] = -1;
dp_sum[i][1][1][0] = -1;
dp_sum[i][1][1][1] = -1;
}
}
long long int cnt(int bit, bool safe_n, bool safe_m, bool safe_k) {
if (bit == -1)
return 1;
else if (dp_cnt[bit][safe_n][safe_m][safe_k] == -1) {
dp_cnt[bit][safe_n][safe_m][safe_k] = 0;
dp_cnt[bit][safe_n][safe_m][safe_k] +=
cnt(bit - 1, (safe_n || (n & (1LL << bit))),
(safe_m || (m & (1LL << bit))), (safe_k || (k & (1LL << bit))));
if ((safe_k || (k & (1LL << bit))) && (safe_n || (n & (1LL << bit))))
dp_cnt[bit][safe_n][safe_m][safe_k] +=
cnt(bit - 1, safe_n, (safe_m || (m & (1LL << bit))), safe_k);
if ((safe_k || (k & (1LL << bit))) && (safe_m || (m & (1LL << bit))))
dp_cnt[bit][safe_n][safe_m][safe_k] +=
cnt(bit - 1, (safe_n || (n & (1LL << bit))), safe_m, safe_k);
if ((safe_n || (n & (1LL << bit))) && (safe_m || (m & (1LL << bit))))
dp_cnt[bit][safe_n][safe_m][safe_k] +=
cnt(bit - 1, safe_n, safe_m, (safe_k || (k & (1LL << bit))));
dp_cnt[bit][safe_n][safe_m][safe_k] %= 1000000007;
}
return dp_cnt[bit][safe_n][safe_m][safe_k];
}
long long int sum(int bit, bool safe_n, bool safe_m, bool safe_k) {
if (bit == -1)
return 0;
else if (dp_sum[bit][safe_n][safe_m][safe_k] == -1) {
dp_sum[bit][safe_n][safe_m][safe_k] = 0;
dp_sum[bit][safe_n][safe_m][safe_k] +=
sum(bit - 1, (safe_n || (n & (1LL << bit))),
(safe_m || (m & (1LL << bit))), (safe_k || (k & (1LL << bit))));
if ((safe_k || (k & (1LL << bit))) && (safe_n || (n & (1LL << bit))))
dp_sum[bit][safe_n][safe_m][safe_k] +=
(1LL << bit) *
cnt(bit - 1, safe_n, (safe_m || (m & (1LL << bit))), safe_k) +
sum(bit - 1, safe_n, (safe_m || (m & (1LL << bit))), safe_k);
if ((safe_k || (k & (1LL << bit))) && (safe_m || (m & (1LL << bit))))
dp_sum[bit][safe_n][safe_m][safe_k] +=
(1LL << bit) *
cnt(bit - 1, (safe_n || (n & (1LL << bit))), safe_m, safe_k) +
sum(bit - 1, (safe_n || (n & (1LL << bit))), safe_m, safe_k);
if ((safe_n || (n & (1LL << bit))) && (safe_m || (m & (1LL << bit))))
dp_sum[bit][safe_n][safe_m][safe_k] +=
sum(bit - 1, safe_n, safe_m, (safe_k || (k & (1LL << bit))));
dp_sum[bit][safe_n][safe_m][safe_k] %= 1000000007;
}
return dp_sum[bit][safe_n][safe_m][safe_k];
}
int main() {
int q;
long long int ans, x1, x2, y1, y2;
std::cin >> q;
while (q--) {
std::cin >> x1 >> y1 >> x2 >> y2 >> k;
x1--;
y1--;
x2--;
y2--;
k--;
ans = 0;
reset_dp_cnt();
reset_dp_sum();
n = x2;
m = y2;
ans += sum(34, 0, 0, 0);
ans += cnt(34, 0, 0, 0);
reset_dp_cnt();
reset_dp_sum();
n = x1 - 1;
m = y2;
if (x1 != 0) {
ans -= sum(34, 0, 0, 0);
ans -= cnt(34, 0, 0, 0);
}
reset_dp_cnt();
reset_dp_sum();
n = x2;
m = y1 - 1;
if (y1 != 0) {
ans -= sum(34, 0, 0, 0);
ans -= cnt(34, 0, 0, 0);
}
reset_dp_cnt();
reset_dp_sum();
n = x1 - 1;
m = y1 - 1;
if (x1 != 0 && y1 != 0) {
ans += sum(34, 0, 0, 0);
ans += cnt(34, 0, 0, 0);
}
std::cout << (ans % 1000000007 + 1000000007) % 1000000007 << '\n';
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
const int MAXN = 1e3 + 7;
const int MOD = 1e9 + 7;
int n;
int dp[34][2][2][2];
int sum[34][2][2][2];
int x, y, k;
int dfs(int pos, int limx, int limy, int limk, int &t) {
if (pos == -1) return 1;
int &st = sum[pos][limx][limy][limk];
if (~dp[pos][limx][limy][limk]) {
(t += st) %= MOD;
return dp[pos][limx][limy][limk];
}
st = 0;
int zx = 1, zy = 1;
if (limx) zx = x >> pos & 1;
if (limy) zy = y >> pos & 1;
long long res = 0, posk = k >> pos & 1;
long long ta;
for (int i = 0; i <= zx; i++)
for (int j = 0; j <= zy; j++) {
long long tb = ((i ^ j) << pos) % MOD;
if (limk) {
if ((i ^ j) <= posk) {
ta = dfs(pos - 1, limx && zx == i, limy && zy == j, (i ^ j) == posk,
st);
res += ta;
st = (ta * tb % MOD + st) % MOD;
}
} else {
ta = dfs(pos - 1, limx && zx == i, limy && zy == j, 0, st);
res += ta;
st = (ta * tb % MOD + st) % MOD;
}
}
res %= MOD;
(t += st) %= MOD;
dp[pos][limx][limy][limk] = res;
return res;
}
int solve(int a, int b) {
if (a < 0 || b < 0) return 0;
memset(dp, -1, sizeof(dp));
int ans = 0;
x = a, y = b;
long long tmp = dfs(30, 1, 1, 1, ans);
tmp = (tmp + ans) % MOD;
return tmp;
}
int main() {
int q, x1, y1, x2, y2;
scanf("%d", &q);
while (q--) {
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
x1--, y1--, x2--, y2--, k--;
long long ans = solve(x2, y2) - solve(x1 - 1, y2) - solve(x2, y1 - 1) +
solve(x1 - 1, y1 - 1);
ans = ans % MOD;
if (ans < 0) ans += MOD;
printf("%d\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
string print_iterable(T1 begin_iter, T2 end_iter, int counter) {
bool done_something = false;
stringstream res;
res << "[";
for (; begin_iter != end_iter and counter; ++begin_iter) {
done_something = true;
counter--;
res << *begin_iter << ", ";
}
string str = res.str();
if (done_something) {
str.pop_back();
str.pop_back();
}
str += "]";
return str;
}
vector<int> SortIndex(int size, std::function<bool(int, int)> compare) {
vector<int> ord(size);
for (int i = 0; i < size; i++) ord[i] = i;
sort(ord.begin(), ord.end(), compare);
return ord;
}
template <typename T>
bool MinPlace(T& a, const T& b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template <typename T>
bool MaxPlace(T& a, const T& b) {
if (a < b) {
a = b;
return true;
}
return false;
}
template <typename S, typename T>
ostream& operator<<(ostream& out, const pair<S, T>& p) {
out << "{" << p.first << ", " << p.second << "}";
return out;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
out << "[";
for (int i = 0; i < (int)v.size(); i++) {
out << v[i];
if (i != (int)v.size() - 1) out << ", ";
}
out << "]";
return out;
}
const long long mod = 1e9 + 7;
long long p[100];
long long ComputeNum(long long k, int i, int j) {
if (i < j) swap(i, j);
if (1 << i <= k) return p[i + j];
long long res = p[j] % mod;
for (int s = 0; 1 << s <= k; s++) {
if (k & 1 << s) {
res += p[s + j];
res %= mod;
}
}
return res;
}
long long ComputeSum(long long k, int i, int j) {
if (i < j) swap(i, j);
if (1 << i <= k) {
long long res1 = p[j];
long long res2 = (p[i] * (p[i] - 1ll)) / 2ll;
res1 %= mod;
res2 %= mod;
return (res1 * res2) % mod;
}
long long res = (p[j] * k) % mod;
for (int s = 0; (1 << s) <= k; s++) {
if (k & (1 << s)) {
long long ris1 = (k - k % (1 << (s + 1))) % mod;
long long ris2 = ((p[s] * (p[s] - 1ll)) / 2ll) % mod;
ris1 *= p[j + s];
ris2 *= p[j];
res += (ris1 + ris2) % mod;
res %= mod;
}
}
return res;
}
long long SimpleSum(long long k, int x, int y) {
long long ans = 0;
if (x <= 0 or y <= 0) return 0;
for (int i = 0; 1 << i <= x; i++) {
for (int j = 0; 1 << j <= y; j++) {
if (x & (1 << i) && y & (1 << j)) {
int x1 = x - x % (1 << max(i + 1, j));
int y1 = y - y % (1 << max(i, j + 1));
int v = x1 ^ y1;
if (k < v) continue;
ans += (long long)(1 + v) * ComputeNum(k - v, i, j) +
ComputeSum(k - v, i, j);
ans %= mod;
}
}
}
return ans;
}
long long GetSum(int x1, int y1, int x2, int y2, long long k) {
long long res = SimpleSum(k - 1, x2, y2) - SimpleSum(k - 1, x2, y1) -
SimpleSum(k - 1, x1, y2) + SimpleSum(k - 1, x1, y1);
res %= mod;
res += mod;
return res % mod;
}
int main() {
p[0] = 1;
for (int i = 1; i < 100; i++) p[i] = (p[i - 1] * 2ll) % mod;
ios::sync_with_stdio(false);
int q;
cin >> q;
for (int i = 0; i < q; i++) {
int x1, y1, x2, y2;
long long k;
cin >> x1 >> y1 >> x2 >> y2 >> k;
cout << GetSum(x1 - 1, y1 - 1, x2, y2, k) << "\n";
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
const int mu = 1e9 + 7;
void reduce(int &x) { x += x >> 31 & mu; }
int m, x, y, xx, yy, k, dp[33][2][2][2], sum[33][2][2][2];
int DP(int x, int y, int k) {
memset(dp, 0, sizeof(dp));
memset(sum, 0, sizeof(sum));
dp[31][1][1][1] = 1;
if (x < 0 || y < 0) return 0;
for (int i = 30; i >= 0; i--)
for (int a = 0; a <= 1; a++)
for (int b = 0; b <= 1; b++)
for (int c = 0; c <= 1; c++) {
int t = dp[i + 1][a][b][c], tsum = sum[i + 1][a][b][c];
if (!t) continue;
for (int A = 0; A <= 1; A++)
for (int B = 0; B <= 1; B++) {
int C = A ^ B;
if (c && C && !(k & (1 << i))) continue;
if (A && a && !(x & (1 << i))) continue;
if (B && b && !(y & (1 << i))) continue;
int na = a, nb = b, nc = c;
if ((k & (1 << i)) && !C) nc = 0;
if ((x & (1 << i)) && !A) na = 0;
if ((y & (1 << i)) && !B) nb = 0;
reduce(dp[i][na][nb][nc] += t - mu);
reduce(sum[i][na][nb][nc] += tsum - mu);
if (C)
sum[i][na][nb][nc] =
(sum[i][na][nb][nc] + 1ll * t * (1 << i)) % mu;
}
}
int ret = 0;
for (int i = 0; i <= 1; i++)
for (int j = 0; j <= 1; j++)
for (int h = 0; h <= 1; h++)
reduce(ret += sum[0][i][j][h] - mu), reduce(ret += dp[0][i][j][h] - mu);
return ret;
}
int main() {
scanf("%d", &m);
while (m--) {
scanf("%d%d%d%d%d", &x, &y, &xx, &yy, &k);
x--, y--;
printf("%d\n",
(1ll * DP(xx - 1, yy - 1, k - 1) - DP(x - 1, yy - 1, k - 1) -
DP(xx - 1, y - 1, k - 1) + DP(x - 1, y - 1, k - 1) + mu + mu) %
mu);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
if (x < y) x = y;
}
namespace fastio {
char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
inline char getc() {
return p1 == p2 &&
(p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
? -1
: *p1++;
}
inline void putc(char x) { (*p3++ = x); }
template <typename T>
void read(T &x) {
x = 0;
char c = getchar();
T neg = 0;
while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
if (neg) x = (~x) + 1;
}
template <typename T>
void recursive_print(T x) {
if (!x) return;
recursive_print(x / 10);
putc(x % 10 ^ 48);
}
template <typename T>
void print(T x) {
if (!x) putc('0');
if (x < 0) putc('-'), x = ~x + 1;
recursive_print(x);
}
void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
} // namespace fastio
const int MOD = 1e9 + 7;
const int INV2 = 5e8 + 4;
int sum(int l, int r) { return 1ll * (l + r) * (r - l + 1) % MOD * INV2 % MOD; }
int calc(int l1, int r1, int l2, int r2, int k) {
if ((r1 - l1 + 1) < (r2 - l2 + 1)) l1 ^= l2 ^= l1 ^= l2, r1 ^= r2 ^= r1 ^= r2;
int tl = (l1 ^ l2) & ~(r1 - l1), tr = min(tl + r1 - l1, k - 1);
if (tl > k - 1) return 0;
return 1ll * (r2 - l2 + 1) * sum(tl + 1, tr + 1) % MOD;
}
int solve(int x, int y, int k) {
int ans = 0;
for (int i = x; i; i &= (i - 1))
for (int j = y; j; j &= (j - 1))
ans = (ans + calc(i & (i - 1), i - 1, j & (j - 1), j - 1, k)) % MOD;
return ans;
}
int main() {
int qu;
scanf("%d", &qu);
while (qu--) {
int x1, y010101010101, x2, y2, k;
scanf("%d%d%d%d%d", &x1, &y010101010101, &x2, &y2, &k);
printf("%d\n",
((solve(x2, y2, k) - solve(x2, y010101010101 - 1, k) -
solve(x1 - 1, y2, k) + solve(x1 - 1, y010101010101 - 1, k)) %
MOD +
MOD) %
MOD);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
int q;
long long f[40][5][5][5], g[40][5][5][5];
inline long long calc(int x, int y, int n) {
if (x < 0 || y < 0) return 0;
long long ret = 0;
for (int i = 0; i <= 1; i++)
for (int j = 0; j <= 1; j++)
for (int k = 0; k <= 1; k++) f[32][i][j][k] = g[32][i][j][k] = 0;
f[32][1][1][1] = 1;
for (int i = 32; i; i--) {
for (int j = 0; j <= 1; j++)
for (int k = 0; k <= 1; k++)
for (int s = 0; s <= 1; s++) f[i - 1][j][k][s] = g[i - 1][j][k][s] = 0;
for (int j = 0; j <= 1; j++)
for (int k = 0; k <= 1; k++)
for (int s = 0; s <= 1; s++) {
if (!f[i][j][k][s]) continue;
for (int a = 0; a <= 1; a++)
for (int b = 0; b <= 1; b++) {
if (k && a > ((x >> i - 1) & 1)) continue;
if (s && b > ((y >> i - 1) & 1)) continue;
if (j && (a ^ b) > ((n >> i - 1) & 1)) continue;
int A = (j && (a ^ b) == ((n >> i - 1) & 1)),
B = (k && a == ((x >> i - 1) & 1)),
C = (s && b == ((y >> i - 1) & 1));
(f[i - 1][A][B][C] += f[i][j][k][s]) %= P,
(g[i - 1][A][B][C] +=
g[i][j][k][s] * 2 + (a ^ b) * f[i][j][k][s]) %= P;
}
}
}
for (int i = 0; i <= 1; i++)
for (int j = 0; j <= 1; j++)
for (int k = 0; k <= 1; k++) (ret += f[0][i][j][k] + g[0][i][j][k]) %= P;
return ret;
}
int main() {
scanf("%d", &q);
while (q--) {
int a1, b1, a2, b2, k;
scanf("%d%d%d%d%d", &a1, &b1, &a2, &b2, &k);
a1--;
b1--;
a2--;
b2--;
k--;
printf("%lld\n", (calc(a2, b2, k) + P - calc(a1 - 1, b2, k) + P -
calc(a2, b1 - 1, k) + calc(a1 - 1, b1 - 1, k)) %
P);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int MOD = 1000 * 1000 * 1000 + 7;
long long SUP = (long long)1024 * 1024 * 1024;
int nbReqs;
long long ax, ay, bx, by, k;
long long getSomme(long long x = 0, long long y = 0, long long taille = SUP,
long long deb = 0) {
if (bx <= x || by <= y || x + taille <= ax || y + taille <= ay) {
return 0;
}
if (ax <= x && x + taille <= bx) {
long long tailleInter = min(by, y + taille) - max(ay, y);
if (deb >= k) return 0;
long long fin = min(deb + taille, k);
long long totalTranche = (fin * (fin - 1) / 2 - deb * (deb - 1) / 2) % MOD;
return (tailleInter * totalTranche + tailleInter * (fin - deb)) % MOD;
}
if (ay <= y && y + taille <= by) {
long long tailleInter = min(bx, x + taille) - max(ax, x);
if (deb >= k) return 0;
long long fin = min(deb + taille, k);
long long totalTranche = (fin * (fin - 1) / 2 - deb * (deb - 1) / 2) % MOD;
return (tailleInter * totalTranche + tailleInter * (fin - deb)) % MOD;
}
return (getSomme(x, y, taille / 2, deb) +
getSomme(x + taille / 2, y, taille / 2, deb + taille / 2) +
getSomme(x, y + taille / 2, taille / 2, deb + taille / 2) +
getSomme(x + taille / 2, y + taille / 2, taille / 2, deb)) %
MOD;
}
int main() {
cin >> nbReqs;
for (int iReq = 0; iReq < nbReqs; iReq++) {
cin >> ax >> ay >> bx >> by >> k;
ax--;
ay--;
long long total = getSomme();
cout << total % MOD << endl;
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int q, x_1, y_1, x_2, y_2, k, len, a[35], b[35], c[35];
pair<int, int> dp[35][2][2][2];
int Add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
int Minus(int a, int b) { return a < b ? a - b + mod : a - b; }
pair<int, int> operator+(pair<int, int> a, pair<int, int> b) {
return make_pair(Add(a.first, b.first), Add(a.second, b.second));
}
pair<int, int> dfs(int now, int f1, int f2, int f3) {
if (now == -1) return make_pair(1, 0);
if (~dp[now][f1][f2][f3].first) return dp[now][f1][f2][f3];
pair<int, int> ret = make_pair(0, 0);
for (int i = 0; i <= (f1 ? a[now] : 1); i++)
for (int j = 0; j <= (f2 ? b[now] : 1); j++)
if ((i ^ j) <= (f3 ? c[now] : 1)) {
pair<int, int> tmp = dfs(now - 1, f1 && i == a[now], f2 && j == b[now],
f3 && (i ^ j) == c[now]);
ret = ret + tmp;
if (i ^ j) ret.second = (ret.second + (1ll << now) * tmp.first) % mod;
}
dp[now][f1][f2][f3] = ret;
return ret;
}
int calc(int x, int y, int k) {
if (x < 0 || y < 0) return 0;
memset(dp, -1, sizeof(dp));
len = 0;
while (x || y || k) {
a[len] = (x & 1);
b[len] = (y & 1);
c[len++] = (k & 1);
x >>= 1;
y >>= 1;
k >>= 1;
}
pair<int, int> ret = dfs(len - 1, 1, 1, 1);
return Add(ret.first, ret.second);
}
int main() {
scanf("%d", &q);
while (q--) {
scanf("%d%d%d%d%d", &x_1, &y_1, &x_2, &y_2, &k);
printf(
"%d\n",
Minus(
Add(calc(x_2 - 1, y_2 - 1, k - 1), calc(x_1 - 2, y_1 - 2, k - 1)),
Add(calc(x_1 - 2, y_2 - 1, k - 1), calc(x_2 - 1, y_1 - 2, k - 1))));
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int K = 40;
const long long md = 1e9 + 7;
pair<long long, long long> f[K][2][2][2];
bool us[K][2][2][2];
bool X[K], Y[K], KK[K];
pair<long long, long long> calc_prefix(long long t, bool eq_x, bool eq_y,
bool eq_k) {
if (t == 0) return {1, 0};
auto& p_f = f[t][eq_x][eq_y][eq_k];
if (us[t][eq_x][eq_y][eq_k]) return p_f;
p_f = {0, 0};
us[t][eq_x][eq_y][eq_k] = true;
long long t_sz = 1LL << (long long)t;
long long tp_sz = 1LL << ((long long)t - 1);
for (int i = 0; i < 2; i++) {
if (eq_x && i > X[t - 1]) continue;
for (int j = 0; j < 2; j++) {
if (eq_y && j > Y[t - 1]) continue;
bool bit = i ^ j;
if (eq_k && bit > KK[t - 1]) continue;
pair<long long, long long> pz{0, 0};
pz = calc_prefix(t - 1, eq_x && (i == X[t - 1]), eq_y && (j == Y[t - 1]),
eq_k && (bit == KK[t - 1]));
if (bit) {
pz.second += pz.first * tp_sz;
pz.second %= md;
}
p_f.first += pz.first;
p_f.second += pz.second;
}
}
p_f.first %= md;
p_f.second %= md;
return p_f;
}
long long calc(long long x, long long y, long long k) {
if (x < 0 || y < 0 || k < 0) return 0;
fill_n((bool*)us, sizeof(us) / sizeof(bool), false);
for (long long i = 0; i < K; i++) {
X[i] = (x >> i) & 1LL;
Y[i] = (y >> i) & 1LL;
KK[i] = (k >> i) & 1LL;
}
auto answer = calc_prefix(K - 1, 1, 1, 1);
return (answer.second + answer.first) % md;
}
long long calc(long long lx, long long rx, long long ly, long long ry,
long long k) {
--lx;
--rx;
--ly;
--ry;
--k;
long long answer = calc(rx, ry, k) + (-1) * calc(lx - 1, ry, k) +
(-1) * calc(rx, ly - 1, k) + calc(lx - 1, ly - 1, k);
answer += 10 * md;
return answer % md;
}
int main() {
ios::sync_with_stdio(false);
int q;
cin >> q;
while (q--) {
int lx, rx, ly, ry, k;
cin >> lx >> ly >> rx >> ry >> k;
long long answer = calc(lx, rx, ly, ry, k);
cout << answer << "\n";
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = (int)1e9 + 7;
pair<pair<int, int>, pair<int, int> > cut(
pair<pair<int, int>, pair<int, int> > a,
pair<pair<int, int>, pair<int, int> > b) {
return pair<pair<int, int>, pair<int, int> >(
pair<int, int>(max(a.first.first, b.first.first),
min(a.first.second, b.first.second)),
pair<int, int>(max(a.second.first, b.second.first),
min(a.second.second, b.second.second)));
}
bool fit(pair<pair<int, int>, pair<int, int> > a,
pair<pair<int, int>, pair<int, int> > b) {
return cut(a, b).first == b.first || cut(a, b).second == b.second;
}
int query(pair<pair<int, int>, pair<int, int> > cur,
pair<pair<int, int>, pair<int, int> > field, int lim, int rem) {
if (cur.first.first > cur.first.second ||
cur.second.first > cur.second.second)
return 0;
if (rem + 1 > lim) return 0;
if (fit(cur, field)) {
int len = min(cur.first.second - cur.first.first,
cur.second.second - cur.second.first) +
1;
int z = max(cur.first.second - cur.first.first,
cur.second.second - cur.second.first) +
1;
int okval = min(z, lim - rem);
return (1ll * rem * len % mod * okval +
(1ll * okval * (okval + 1) / 2) % mod * len) %
mod;
}
int dim = __builtin_ctz(field.first.second - field.first.first + 1);
int xmid = field.first.first + (1 << (dim - 1));
int ymid = field.second.first + (1 << (dim - 1));
int xout = field.first.first + (1 << (dim));
int yout = field.second.first + (1 << (dim));
pair<pair<int, int>, pair<int, int> > topleft = {
{field.first.first, xmid - 1}, {field.second.first, ymid - 1}};
pair<pair<int, int>, pair<int, int> > topright = {
{field.first.first, xmid - 1}, {ymid, yout - 1}};
pair<pair<int, int>, pair<int, int> > botleft = {
{xmid, xout - 1}, {field.second.first, ymid - 1}};
pair<pair<int, int>, pair<int, int> > botright = {{xmid, xout - 1},
{ymid, yout - 1}};
return (0ll + 1ll * query(cut(cur, topleft), topleft, lim, rem) +
1ll *
query(cut(cur, topright), topright, lim, rem + (1 << (dim - 1))) +
1ll * query(cut(cur, botleft), botleft, lim, rem + (1 << (dim - 1))) +
1ll * query(cut(cur, botright), botright, lim, rem)) %
mod;
}
void ngk() {
int q;
cin >> q;
while (q--) {
int x1, y1, x2, y2, k;
cin >> x1 >> y1 >> x2 >> y2 >> k;
pair<pair<int, int>, pair<int, int> > ask = {{x1, x2}, {y1, y2}};
pair<pair<int, int>, pair<int, int> > field = {{1, (1 << 30)},
{1, (1 << 30)}};
cout << query(ask, field, k, 0) << '\n';
}
}
int main(int argc, char *argv[]) {
iostream::sync_with_stdio(0);
int test = 1;
while (test--) {
ngk();
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long qpow(long long x, long long y) {
long long e = 1;
while (y) {
if (y % 2) e = (e * x) % 1000000007;
x = x * x % 1000000007;
y /= 2;
}
return e;
}
long long low(long long x) {
for (long long i = 1; i <= x; i <<= 1) {
if (i << 1 > x) return i;
}
}
long long getS(long long x, long long y, long long n) {
return (y - x + 1) * (x + y) % 1000000007 * qpow(2, 1000000007 - 2) %
1000000007 * n % 1000000007;
}
long long dfs(long long a, long long b, long long k, long long t) {
if (a <= 0 || b <= 0 || k <= 0) return 0;
long long l = min(a, b);
long long r = max(a, b);
long long x = low(l);
long long limit = min(r / x, k / x);
long long ans = 0;
if (l == r && l == x) {
ans = (ans + getS(t + 1, t + min(x, k), x)) % 1000000007;
} else {
if (limit % 2 == 0) {
long long rema = min(r - limit * x, x);
ans = (ans + dfs(x, rema, k - limit * x, t + limit * x)) % 1000000007;
ans = (ans + dfs(l - x, rema, k - limit * x - x, t + limit * x + x)) %
1000000007;
rema = min(r - limit * x - x, x);
ans = (ans + dfs(x, rema, k - limit * x - x, t + limit * x + x)) %
1000000007;
ans = (ans + dfs(l - x, rema, k - limit * x, t + limit * x)) % 1000000007;
} else {
limit--;
ans = (ans + dfs(x, x, k - limit * x, t + limit * x)) % 1000000007;
ans = (ans + dfs(l - x, x, k - limit * x - x, t + limit * x + x)) %
1000000007;
long long rema = min(r - limit * x - x, x);
ans = (ans + dfs(x, rema, k - limit * x - x, t + limit * x + x)) %
1000000007;
ans = (ans + dfs(l - x, rema, k - limit * x, t + limit * x)) % 1000000007;
}
ans = (ans + getS(t + 1, t + limit * x, l)) % 1000000007;
}
return ans;
}
int main() {
int N, x1, x2, y1, y2, k;
scanf("%d", &N);
for (int i = 1; i <= N; i++) {
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
long long ans = ((dfs(x2, y2, k, 0) - dfs(x1 - 1, y2, k, 0) -
dfs(x2, y1 - 1, k, 0) + dfs(x1 - 1, y1 - 1, k, 0)) %
1000000007 +
1000000007) %
1000000007;
printf("%lld\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline int readInt() {
static int n, ch;
n = 0, ch = getchar();
while (!isdigit(ch)) ch = getchar();
while (isdigit(ch)) n = n * 10 + ch - '0', ch = getchar();
return n;
}
const int MAX_N = 200 + 3, MOD = 1e9 + 7;
int n;
long long a[MAX_N][MAX_N];
int mex(int i, int j) {
static bool vis[MAX_N * 2];
memset(vis, 0, sizeof vis);
for (int k = 1; k < i; ++k) vis[a[k][j]] = true;
for (int k = 1; k < j; ++k) vis[a[i][k]] = true;
for (int k = 1; k < MAX_N * 2; ++k)
if (!vis[k]) return k;
}
void printBit(int x) {
for (int i = 15; i >= 0; --i) printf("%d", x >> i & 1);
cout << endl;
}
pair<long long, long long> operator+(const pair<long long, long long> &lhs,
const pair<long long, long long> &rhs) {
return make_pair((lhs.first + rhs.first) % MOD,
(lhs.second + rhs.second) % MOD);
}
pair<long long, long long> operator-(const pair<long long, long long> &lhs,
const pair<long long, long long> &rhs) {
return make_pair((lhs.first - rhs.first + MOD) % MOD,
(lhs.second - rhs.second + MOD) % MOD);
}
pair<long long, long long> operator*(const pair<long long, long long> &lhs,
const int &x) {
return make_pair((lhs.first + lhs.second * x % MOD) % MOD, lhs.second);
}
pair<long long, long long> operator^(const pair<long long, long long> &lhs,
const int &x) {
return make_pair(lhs.first * x % MOD, lhs.second * x % MOD);
}
pair<long long, long long> calc2(long long n, long long k,
long long start = 1) {
if (k < start) return make_pair(0, 0);
k = min(k, n + start - 1);
return make_pair((k + start) * (k - start + 1) / 2 % MOD,
(k - start + 1) % MOD);
}
pair<long long, long long> calc(long long n, long long k) {
return calc2(n, k) ^ n;
}
pair<long long, long long> solve(int x, int y, int k) {
if (x <= 0 || y <= 0 || k <= 0) return make_pair(0, 0);
if (x == 1 && y == 1) return k >= 1 ? make_pair(1, 1) : make_pair(0, 0);
if (x < y) swap(x, y);
int logX = ceil(log2(x)), n = 1 << logX, halfN = 1 << (logX - 1);
if (y <= halfN) {
return (calc2(halfN, k) ^ y) + solve(x - halfN, y, k - halfN) * halfN;
} else {
return calc(halfN, k) +
(calc2(halfN, k, halfN + 1) ^ (x - halfN + y - halfN)) +
solve(x - halfN, y - halfN, k);
}
}
int main() {
int q = readInt();
while (q--) {
int x1 = readInt(), y1 = readInt(), x2 = readInt(), y2 = readInt(),
k = readInt();
pair<long long, long long> ans = solve(x2, y2, k) - solve(x2, y1 - 1, k) -
solve(x1 - 1, y2, k) +
solve(x1 - 1, y1 - 1, k);
cout << ans.first << endl;
}
return 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) printf("%4d", a[i][j]);
cout << endl;
}
puts("---------------");
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) printf("%4d", ((i - 1) ^ (j - 1)) + 1);
cout << endl;
}
return 0;
for (int k = 1; k <= 16; ++k) {
puts(
"----------------------------------------------------------------------"
"---");
printf("%d ", k);
printBit(k);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (a[i][j] == k)
printf("x ");
else
printf(" ");
}
cout << endl;
}
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
if (a[i][j] == k)
cout << i << ' ' << j << ' ' << (i ^ j ^ abs(i - j)) << endl;
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
const int mod = 1000000007;
int f[33][2][2][2], g[33][2][2][2];
int calc(int x, int y, int k) {
for (int i = 0; i <= 31; i++)
for (int sx = 0; sx < 2; sx++)
for (int sy = 0; sy < 2; sy++)
for (int sk = 0; sk < 2; sk++) {
f[i][sx][sy][sk] = g[i][sx][sy][sk] = 0;
if (!i)
f[i][sx][sy][sk] = !sx && !sy && !sk;
else
for (int dx = 0; dx < 2; dx++)
for (int dy = 0; dy < 2; dy++) {
if (sx && dx > (x >> i - 1 & 1) ||
sy && dy > (y >> i - 1 & 1) ||
(sk && (dx ^ dy) > (k >> i - 1 & 1)))
continue;
int tx = sx && dx == (x >> i - 1 & 1),
ty = sy && dy == (y >> i - 1 & 1),
tk = sk && (dx ^ dy) == (k >> i - 1 & 1);
(f[i][sx][sy][sk] += f[i - 1][tx][ty][tk]) %= mod;
(g[i][sx][sy][sk] += g[i - 1][tx][ty][tk]) %= mod;
if (dx ^ dy)
(g[i][sx][sy][sk] +=
(1ll << i - 1) * f[i - 1][tx][ty][tk] % mod) %= mod;
}
}
return (f[31][1][1][1] + g[31][1][1][1]) % mod;
}
int main() {
int q;
scanf("%d", &q);
while (q--) {
int x1, y1, x2, y2, k, s;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
s = (calc(x2, y2, k) - calc(x2, y1 - 1, k) - calc(x1 - 1, y2, k) +
calc(x1 - 1, y1 - 1, k)) %
mod;
printf("%d\n", s < 0 ? s + mod : s);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline int read() {
int k = 0;
char ch = getchar();
while (ch < '0' || ch > '9') ch = getchar();
while (ch <= '9' && ch >= '0') {
k = k * 10 + ch - 48;
ch = getchar();
}
return k;
}
const int p = 1e9 + 7;
inline int Jia(int x) { return x >= p ? x - p : x; }
inline int Jian(int x) { return x >= 0 ? x : x + p; }
int n, f[32][2][2][2], g[32][2][2][2];
int Calc(int lim1, int lim2, int lim3) {
if (lim1 < 0 || lim2 < 0) return 0;
memset(f, 0, sizeof(f));
memset(g, 0, sizeof(g));
f[31][1][1][1] = 1;
for (int i = 30; i >= 0; --i) {
for (int k1 = 0; k1 <= 1; ++k1) {
for (int k2 = 0; k2 <= 1; ++k2) {
for (int k3 = 0; k3 <= 1; ++k3) {
if (!f[i + 1][k1][k2][k3]) continue;
for (int s1 = 0; s1 <= 1; ++s1) {
for (int s2 = 0; s2 <= 1; ++s2) {
if (k1 && s1 && !(lim1 & (1 << i))) continue;
if (k2 && s2 && !(lim2 & (1 << i))) continue;
if (k3 && (s1 ^ s2) && !(lim3 & (1 << i))) continue;
int n1 = k1, n2 = k2, n3 = k3;
if (k1 && s1 == ((lim1 >> i) & 1))
n1 = 1;
else
n1 = 0;
if (k2 && s2 == ((lim2 >> i) & 1))
n2 = 1;
else
n2 = 0;
if (k3 && (s1 ^ s2) == ((lim3 >> i) & 1))
n3 = 1;
else
n3 = 0;
f[i][n1][n2][n3] = Jia(f[i][n1][n2][n3] + f[i + 1][k1][k2][k3]);
g[i][n1][n2][n3] = Jia(g[i][n1][n2][n3] + g[i + 1][k1][k2][k3]);
if (s1 ^ s2)
g[i][n1][n2][n3] =
Jia(g[i][n1][n2][n3] +
1LL * (1 << i) * f[i + 1][k1][k2][k3] % p);
}
}
}
}
}
}
int ans = 0;
for (int k1 = 0; k1 <= 1; ++k1) {
for (int k2 = 0; k2 <= 1; ++k2) {
for (int k3 = 0; k3 <= 1; ++k3) {
ans = Jia(ans + f[0][k1][k2][k3]);
ans = Jia(ans + g[0][k1][k2][k3]);
}
}
}
return ans;
}
int main() {
int T = read();
while (T--) {
int x1 = read() - 1, y1 = read() - 1, x2 = read() - 1, y2 = read() - 1,
K = read() - 1;
int ans = 0;
ans = Jia(ans + Calc(x2, y2, K));
ans = Jian(ans - Calc(x1 - 1, y2, K));
ans = Jian(ans - Calc(x2, y1 - 1, K));
ans = Jia(ans + Calc(x1 - 1, y1 - 1, K));
cout << ans << '\n';
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int md = 1e9 + 7;
long long f[40][2][2][2], sum[40][2][2][2];
int q, k, d[3][40], tmp[40];
void split(int x, int m) {
for (int i = 1; i <= 31; i++) tmp[i] = (x >> (i - 1)) & 1;
for (int i = 1; i <= 31; i++) d[m][i] = tmp[32 - i];
}
long long work(int x, int y) {
if (x < 0 || y < 0) return 0;
memset(f, 0, sizeof(f));
memset(sum, 0, sizeof(sum));
split(x, 0);
split(y, 1);
split(k, 2);
f[0][0][0][0] = 1;
for (int i = 0; i < 31; i++)
for (int f1 = 0; f1 < 2; f1++)
for (int f2 = 0; f2 < 2; f2++)
for (int f3 = 0; f3 < 2; f3++)
if (f[i][f1][f2][f3])
for (int n1 = 0; n1 < 2; n1++)
for (int n2 = 0; n2 < 2; n2++) {
int nf1 = f1, nf2 = f2, nf3 = f3;
if (nf1 == 0 && n1 > d[0][i + 1]) continue;
if (n1 < d[0][i + 1]) nf1 = 1;
if (nf2 == 0 && n2 > d[1][i + 1]) continue;
if (n2 < d[1][i + 1]) nf2 = 1;
int val = n1 ^ n2;
if (nf3 == 0 && val > d[2][i + 1]) continue;
if (val < d[2][i + 1]) nf3 = 1;
f[i + 1][nf1][nf2][nf3] =
(f[i + 1][nf1][nf2][nf3] + f[i][f1][f2][f3]) % md;
sum[i + 1][nf1][nf2][nf3] =
(sum[i + 1][nf1][nf2][nf3] + sum[i][f1][f2][f3] +
f[i][f1][f2][f3] * (val * 1ll << (30 - i)) % md) %
md;
}
long long ans = 0;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
ans = (ans + sum[31][i][j][k] + f[31][i][j][k]) % md;
return ans;
}
int main() {
scanf("%d", &q);
while (q--) {
int x1, y1, x2, y2;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
x1--;
y1--;
x2--;
y2--;
k--;
printf("%I64d\n", (work(x2, y2) - work(x1 - 1, y2) - work(x2, y1 - 1) +
work(x1 - 1, y1 - 1) + 2ll * md) %
md);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long int cnt[33][3][3][3], dp[33][3][3][3];
pair<long long int, long long int> solve(long long int pos, long long int fx,
long long int fy, long long int fk,
long long int x, long long int y,
long long int k) {
if (cnt[pos][fx][fy][fk] != -1)
return make_pair(cnt[pos][fx][fy][fk], dp[pos][fx][fy][fk]);
long long int& ct = cnt[pos][fx][fy][fk];
long long int& sum = dp[pos][fx][fy][fk];
ct = 0;
sum = 0;
if (pos == 1) {
if (fx == 0 && fy == 0) {
if (fk == 0) {
ct = 4;
sum = 2;
} else {
if (k & 1) {
ct = 2;
sum = 0;
}
}
} else if (fx == 0) {
if (y & 1) {
if (fk == 0) {
ct = 2;
sum = 1;
} else {
if (k & 1) {
ct = 1;
sum = 0;
}
}
}
} else if (fy == 0) {
if (x & 1) {
if (fk == 0) {
ct = 2;
sum = 1;
} else {
if (k & 1) {
ct = 1;
sum = 0;
}
}
}
} else {
if ((x & 1) && (y & 1)) {
if (fk == 0) {
ct = 1;
sum = 0;
} else {
if (k & 1) {
ct = 1;
sum = 0;
}
}
}
}
return make_pair(ct, sum);
}
long long int vx = x & (1LL << (pos - 1));
long long int vy = y & (1LL << (pos - 1));
long long int vk = k & (1LL << (pos - 1));
vx = !vx;
vx = !vx;
vy = !vy;
vy = !vy;
vk = !vk;
vk = !vk;
for (long long int xbit = 0; xbit <= 1; ++xbit) {
for (long long int ybit = 0; ybit <= 1; ++ybit) {
if (fx == 1 && xbit > vx) continue;
if (fy == 1 && ybit > vy) continue;
if (fk == 1 && ((xbit ^ ybit) > vk)) continue;
long long int px = 0;
if (fx == 1 && xbit == vx) px = 1;
long long int py = 0;
if (fy == 1 && ybit == vy) py = 1;
long long int pk = 0;
if (fk == 1 && ((xbit ^ ybit) == vk)) pk = 1;
pair<long long int, long long int> xp1 =
solve(pos - 1, px, py, pk, x, y, k);
sum = (sum + xp1.second) % 1000000007;
ct = (ct + xp1.first) % 1000000007;
sum = (sum +
(xp1.first * ((xbit ^ ybit) * (1LL << (pos - 1)))) % 1000000007) %
1000000007;
}
}
if (pos != 31)
return make_pair(ct, sum);
else
return (make_pair(ct, (ct + sum) % 1000000007));
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
long long int q;
cin >> q;
while (q--) {
long long int x1, y1, x2, y2, kk;
cin >> x1 >> y1 >> x2 >> y2 >> kk;
for (long long int i = 0; i <= 32; ++i)
for (long long int j = 0; j <= 2; ++j)
for (long long int k = 0; k <= 2; ++k)
for (long long int l = 0; l <= 2; ++l)
dp[i][j][k][l] = 0, cnt[i][j][k][l] = -1;
long long int ans = solve(31, 1, 1, 1, x2, y2, kk).second;
for (long long int i = 0; i <= 32; ++i)
for (long long int j = 0; j <= 2; ++j)
for (long long int k = 0; k <= 2; ++k)
for (long long int l = 0; l <= 2; ++l)
dp[i][j][k][l] = 0, cnt[i][j][k][l] = -1;
if (x1 != 1)
ans = ((ans - solve(31, 1, 1, 1, x1 - 1, y2, kk).second) % 1000000007 +
1000000007) %
1000000007;
for (long long int i = 0; i <= 32; ++i)
for (long long int j = 0; j <= 2; ++j)
for (long long int k = 0; k <= 2; ++k)
for (long long int l = 0; l <= 2; ++l)
dp[i][j][k][l] = 0, cnt[i][j][k][l] = -1;
if (y1 != 1)
ans = ((ans - solve(31, 1, 1, 1, x2, y1 - 1, kk).second) % 1000000007 +
1000000007) %
1000000007;
for (long long int i = 0; i <= 32; ++i)
for (long long int j = 0; j <= 2; ++j)
for (long long int k = 0; k <= 2; ++k)
for (long long int l = 0; l <= 2; ++l)
dp[i][j][k][l] = 0, cnt[i][j][k][l] = -1;
if (y1 != 1 && x1 != 1)
ans = (ans + solve(31, 1, 1, 1, x1 - 1, y1 - 1, kk).second) % 1000000007;
cout << ans << '\n';
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
template <class T>
bool ckmin(T& a, const T& b) {
return a > b ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int B = 31, M = 1e9 + 7;
template <class T>
void add(T& x, int y) {
x += y;
if (M <= x) x -= M;
}
template <class T>
void sub(T& x, int y) {
x -= y;
if (x < 0) x += M;
}
int dp[B + 1][2][2][2], f[B + 1][2][2][2];
void add(int& x, int y) {
x += y;
if (M <= x) x -= M;
}
void sub(int& x, int y) {
x -= y;
if (x < 0) x += M;
}
int g(int x, int y, int k) {
for (auto i = (0); i < (B); ++i)
for (auto a = (0); a < (2); ++a)
for (auto b = (0); b < (2); ++b)
for (auto c = (0); c < (2); ++c) dp[i][a][b][c] = f[i][a][b][c] = 0;
f[B][0][0][0] = 1;
for (auto z = (B); z-- > (0);) {
int u = x >> z & 1, v = y >> z & 1, w = k >> z & 1;
for (auto a : {0, 1}) {
for (auto b : {0, 1}) {
for (auto c : {0, 1}) {
for (auto p : {0, 1}) {
for (auto q : {0, 1}) {
auto r = p ^ q;
if ((!a && p > u) || (!b && q > v) || (!c && r > w)) continue;
auto i = a || p != u, j = b || q != v, k = c || r != w;
add(dp[z][i][j][k], dp[z + 1][a][b][c]);
if (r) add(dp[z][i][j][k], (ll(f[z + 1][a][b][c]) << z) % M);
add(f[z][i][j][k], f[z + 1][a][b][c]);
}
}
}
}
}
}
return (dp[0][1][1][1] + f[0][1][1][1]) % M;
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin >> t;
while (t--) {
int a, b, c, d, k;
cin >> a >> b >> c >> d >> k;
--a;
--b;
int ans = 0;
add(ans, g(c, d, k));
sub(ans, g(a, d, k));
sub(ans, g(c, b, k));
add(ans, g(a, b, k));
cout << ans << '\n';
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
template <int MOD>
struct Integral {
int v_ = 0;
template <typename T>
Integral(T v) : v_(norm(v)) {
static_assert(std::is_integral<T>::value, "input should be an integral.");
}
Integral() = default;
~Integral() = default;
template <typename T>
T norm(T v) const {
if (v >= MOD) v -= MOD;
if (v < 0) v += MOD;
if (v >= MOD || v < 0) v = (v % MOD + MOD) % MOD;
return v;
}
int val() const { return v_; }
Integral operator+(const Integral& rhs) const {
return Integral(val() + rhs.val());
}
Integral operator-(const Integral& rhs) const {
return Integral(val() - rhs.val());
}
Integral operator*(const Integral& rhs) const {
return Integral(val() * 1LL * rhs.val());
}
Integral operator/(const Integral& rhs) const { return *this * rhs.inv(); }
Integral& operator+=(const Integral& rhs) { return *this = *this + rhs; }
Integral& operator-=(const Integral& rhs) { return *this = *this - rhs; }
Integral& operator*=(const Integral& rhs) { return *this = *this * rhs; }
Integral& operator/=(const Integral& rhs) { return *this = *this / rhs; }
bool operator==(const Integral& rhs) const { return val() == rhs.val(); }
bool operator!=(const Integral& rhs) const { return !(*this == rhs); }
const Integral operator-() const { return Integral(-val()); }
const Integral operator++() {
v_ = norm(v_ + 1);
return *this;
}
const Integral operator++(int) {
Integral ret = *this;
++(*this);
return ret;
}
const Integral operator--() {
v_ = norm(v_ - 1);
return *this;
}
const Integral operator--(int) {
Integral ret = *this;
--(*this);
return ret;
}
Integral power(long long b) const {
long long ret = 1 % MOD, a = v_;
for (; b; b >>= 1, a = a * a % MOD)
if (b & 1) ret = ret * a % MOD;
return ret;
}
Integral inv() const { return power(MOD - 2); }
};
template <int MOD>
std::string to_string(const Integral<MOD>& v) {
return std::string("Integral{v=") + std::to_string(v.val()) + "}";
}
template <int MOD>
struct Binomial {
std::vector<Integral<MOD>> factor, inv_factor;
explicit Binomial(int n = 0) : factor(n + 1), inv_factor(n + 1) {
factor[0] = 1;
for (int i = 1; i <= n; ++i) factor[i] = factor[i - 1] * i;
inv_factor[n] = factor[n].inv();
for (int i = n; i >= 1; --i) inv_factor[i - 1] = inv_factor[i] * i;
}
~Binomial() = default;
template <typename T>
Integral<MOD> operator()(T a, T b) const {
if (a < b || b < 0) return 0;
if (a < factor.size()) return factor[a] * inv_factor[b] * inv_factor[a - b];
}
};
const int MOD = 1e9 + 7;
using Mint = Integral<MOD>;
using Binom = Binomial<MOD>;
const int inv2 = (MOD + 1) / 2;
inline int get_bit(int v, int at) { return v >> at & 1; }
using Value = std::pair<Mint, Mint>;
Value operator+(const Value& lhs, const Value& rhs) {
return std::make_pair(lhs.first + rhs.first, lhs.second + rhs.second);
}
Value& operator+=(Value& lhs, const Value& rhs) { return lhs = lhs + rhs; }
const int kBit = 31;
Value calc_cache[kBit][2][2];
bool calc_vis[kBit][2][2];
Value calc(int at, int a1, int b1, bool sa1, bool sb1) {
if (at == -1) return std::make_pair(Mint(0), Mint(!sa1 && !sb1));
if (!sa1 && !sb1) {
int l = 0, r = (1 << (at + 1)) - 1;
Mint cnt = Mint(r - l + 1);
return std::make_pair((Mint(l + r) * (r - l + 1) * inv2) * cnt, cnt * cnt);
}
if (calc_vis[at][sa1][sb1]) return calc_cache[at][sa1][sb1];
calc_vis[at][sa1][sb1] = true;
Value& ret = calc_cache[at][sa1][sb1];
ret = std::make_pair(Mint(0), Mint(0));
for (int a = 0; a < 2; ++a) {
for (int b = 0; b < 2; ++b) {
if (sa1 && a > get_bit(a1, at)) continue;
if (sb1 && b > get_bit(b1, at)) continue;
Value val = calc(at - 1, a1, b1, sa1 && a == get_bit(a1, at),
sb1 && b == get_bit(b1, at));
ret += val;
if (a ^ b) ret.first += val.second * Mint(1 << at);
}
}
return ret;
}
Value solve_cache[kBit][2][2];
bool solve_vis[kBit][2][2];
Value solve(int at, int a1, int b1, int k, bool sa1, bool sb1) {
if (at == -1) return std::make_pair(Mint(0), Mint(0));
if (solve_vis[at][sa1][sb1]) return solve_cache[at][sa1][sb1];
solve_vis[at][sa1][sb1] = true;
Value& ret = solve_cache[at][sa1][sb1];
ret = std::make_pair(Mint(0), Mint(0));
for (int a = 0; a < 2; ++a) {
for (int b = 0; b < 2; ++b) {
if (sa1 && a > get_bit(a1, at)) continue;
if (sb1 && b > get_bit(b1, at)) continue;
if ((a ^ b) > get_bit(k, at)) continue;
if ((a ^ b) < get_bit(k, at)) {
ret += calc(at - 1, a1, b1, sa1 && a == get_bit(a1, at),
sb1 && b == get_bit(b1, at));
} else {
Value val = solve(at - 1, a1, b1, k, sa1 && a == get_bit(a1, at),
sb1 && b == get_bit(b1, at));
ret += val;
if (a ^ b) ret.first += val.second * Mint(1 << at);
}
}
}
return ret;
}
Mint work(int a1, int b1, int k) {
if (a1 <= 0 || b1 <= 0) return 0;
memset(calc_vis, 0, sizeof(calc_vis));
memset(solve_vis, 0, sizeof(solve_vis));
Value val = solve(kBit - 1, a1, b1, k, true, true);
return val.first + val.second;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::istream& reader = std::cin;
int q;
reader >> q;
while (q--) {
int a0, b0, a1, b1, k;
reader >> a0 >> b0 >> a1 >> b1 >> k;
Mint result = work(a1, b1, k) - work(a0 - 1, b1, k) - work(a1, b0 - 1, k) +
work(a0 - 1, b0 - 1, k);
printf("%d\n", result.val());
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
inline int in() {
int k = 0;
char ch = getchar();
bool p = 1;
while (ch < '-') ch = getchar();
if (ch == '-') ch = getchar(), p = 0;
while (ch > '-') k = k * 10 + ch - '0', ch = getchar();
return p ? k : -k;
}
const int N = 55, YL = 1e9 + 7;
inline int MO(const int &x) { return x >= YL ? x - YL : x; }
inline void add(int &x, int y) {
if ((x += y) >= YL) x -= YL;
}
inline void dec(int &x, int y) {
if ((x -= y) < 0) x += YL;
}
int sn[N], sm[N], sk[N], tot, pw[N];
std::pair<int, int> dp[N][2][2][2];
int o[N][2][2][2];
std::pair<int, int> dfs(int d, int a, int b, int c) {
if (d == -1) return std::make_pair(0, 1);
if (o[d][a][b][c]) return dp[d][a][b][c];
o[d][a][b][c] = 1;
std::pair<int, int> &res = dp[d][a][b][c];
res = std::make_pair(0, 0);
int ma = a ? 1 : sn[d], mb = b ? 1 : sm[d], mc = c ? 1 : sk[d], k;
for (int i = 0; i <= ma; ++i)
for (int j = 0; j <= mb; ++j)
if ((k = (i ^ j)) <= mc) {
std::pair<int, int> ret =
dfs(d - 1, i < ma || a, j < mb || b, k < mc || c);
add(res.second, ret.second);
add(res.first, ret.first);
if (!k) continue;
res.first = (1ll * ret.second * pw[d] + res.first) % YL;
}
return res;
}
int calc(int n, int m, int k) {
if (n < 0 || m < 0 || k < 0) return 0;
memset(o, 0, sizeof o);
tot = -1;
while (n || m || k)
sn[++tot] = n & 1, sm[tot] = m & 1, sk[tot] = k & 1, n >>= 1, m >>= 1,
k >>= 1;
return MO(dfs(tot, 0, 0, 0).first + dfs(tot, 0, 0, 0).second);
}
int main() {
for (int i = pw[0] = 1; i <= 31; ++i) pw[i] = MO(pw[i - 1] << 1);
for (int t = in(); t; --t) {
int x1 = in() - 1, y1 = in() - 1, x2 = in() - 1, y2 = in() - 1,
k = in() - 1, ans = 0;
add(ans, calc(x2, y2, k));
add(ans, calc(x1 - 1, y1 - 1, k));
dec(ans, calc(x2, y1 - 1, k));
dec(ans, calc(x1 - 1, y2, k));
printf("%d\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int q, f[50][2][2][2], g[50][2][2][2], a[50], b[50], c[50], al, bl, cl, ans;
void upd(int &x, int y) { x = (x + y) % 1000000007; }
int solve(int x, int y, int k) {
memset(f, 0, sizeof(f));
memset(g, 0, sizeof(g));
al = bl = cl = 0;
while (x) {
a[++al] = x % 2;
x /= 2;
}
for (int i = 1; i <= al / 2; i++) swap(a[i], a[al - i + 1]);
while (y) {
b[++bl] = y % 2;
y /= 2;
}
for (int i = 1; i <= bl / 2; i++) swap(b[i], b[bl - i + 1]);
while (k) {
c[++cl] = k % 2;
k /= 2;
}
for (int i = 1; i <= cl / 2; i++) swap(c[i], c[cl - i + 1]);
int n = max(al, max(bl, cl)), t = n - al;
for (int i = n; i > t; i--) a[i] = a[i - t];
for (int i = 1; i <= t; i++) a[i] = 0;
t = n - bl;
for (int i = n; i > t; i--) b[i] = b[i - t];
for (int i = 1; i <= t; i++) b[i] = 0;
t = n - cl;
for (int i = n; i > t; i--) c[i] = c[i - t];
for (int i = 1; i <= t; i++) c[i] = 0;
f[0][0][0][0] = 0;
g[0][0][0][0] = 1;
t = 1;
for (int i = 1; i < n; i++) t *= 2;
for (int i = 0; i < n; i++) {
for (int o1 = 0; o1 <= 1; o1++)
for (int o2 = 0; o2 <= 1; o2++)
for (int o3 = 0; o3 <= 1; o3++)
for (int j = 0; j <= 1; j++)
for (int k = 0; k <= 1; k++) {
int s1 = o1, s2 = o2, s3 = o3;
if ((o1 == 0 && j > a[i + 1]) || (o2 == 0 && k > b[i + 1]) ||
(o3 == 0 && (j ^ k) > c[i + 1]))
continue;
if (j < a[i + 1]) s1 = 1;
if (k < b[i + 1]) s2 = 1;
if ((j ^ k) < c[i + 1]) s3 = 1;
upd(g[i + 1][s1][s2][s3], g[i][o1][o2][o3]);
upd(f[i + 1][s1][s2][s3],
(f[i][o1][o2][o3] +
1ll * (j ^ k) * g[i][o1][o2][o3] * t % 1000000007) %
1000000007);
}
t /= 2;
}
return (f[n][1][1][1] + g[n][1][1][1]) % 1000000007;
}
int main() {
scanf("%d", &q);
while (q--) {
int x1, x2, y1, y2, k;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
ans = (solve(x2, y2, k) - solve(x2, y1 - 1, k) - solve(x1 - 1, y2, k) +
solve(x1 - 1, y1 - 1, k)) %
1000000007;
ans = (ans + 1000000007) % 1000000007;
printf("%d\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long MODD = 1e9 + 7;
int q;
long long a, b, u, v, k;
long long calc(long long u, long long v, long long k, long long o = 0) {
if (u > v) swap(u, v);
if (u < 1 || k <= 0) return 0;
long long p = 1, res = 0;
while (p * 2LL <= v) p *= 2LL;
long long tmp = min(p, k);
if (u <= p) {
res = (tmp * (tmp + 1) / 2 % MODD) * u % MODD;
res += (o * tmp % MODD) * u % MODD + calc(u, v - p, k - p, o + p);
res %= MODD;
} else {
res = (tmp * (tmp + 1) / 2 % MODD) * p % MODD;
res += (o * tmp % MODD) * p % MODD;
res += calc(u - p, v - p, k, o) + calc(p, v - p, k - p, o + p) +
calc(u - p, p, k - p, o + p);
}
return res;
}
int main() {
scanf("%d", &q);
while (q-- > 0) {
scanf("%lld %lld %lld %lld %lld", &a, &b, &u, &v, &k);
long long res = (calc(u, v, k) + calc(a - 1, b - 1, k)) % MODD;
res = (res - calc(a - 1, v, k) - calc(u, b - 1, k) + MODD * MODD) % MODD;
printf("%lld\n", res);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int c[31 + 1][2][2][2], s[31 + 1][2][2][2], T[2][2][2];
void Initialise() {
T[0][0][0] = 0;
T[0][0][1] = 1;
T[0][1][0] = 0;
T[0][1][1] = 0;
T[1][0][0] = 1;
T[1][0][1] = 1;
T[1][1][0] = 0;
T[1][1][1] = 1;
}
int Query(int n, int m, int k) {
if (n < 0 || m < 0) return 0;
for (int x = 0; x <= 31; x++)
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++) c[x][i][j][k] = s[x][i][j][k] = 0;
c[0][0][0][0] = 1;
for (int x = 0; x < 31; x++) {
int t1 = ((n & (1 << x)) > 0), t2 = ((m & (1 << x)) > 0),
t3 = ((k & (1 << x)) > 0);
for (int p1 = 0; p1 < 2; p1++) {
for (int p2 = 0; p2 < 2; p2++) {
for (int p3 = 0; p3 < 2; p3++) {
for (int c1 = 0; c1 < 2; c1++) {
for (int c2 = 0; c2 < 2; c2++) {
int c3 = (c1 ^ c2);
int n1 = T[p1][t1][c1], n2 = T[p2][t2][c2], n3 = T[p3][t3][c3];
c[x + 1][n1][n2][n3] =
(c[x + 1][n1][n2][n3] + c[x][p1][p2][p3]) % 1000000007;
s[x + 1][n1][n2][n3] =
(s[x + 1][n1][n2][n3] + s[x][p1][p2][p3]) % 1000000007;
if (c3)
s[x + 1][n1][n2][n3] =
(s[x + 1][n1][n2][n3] +
(1ll * (1 << x) * c[x][p1][p2][p3]) % 1000000007) %
1000000007;
}
}
}
}
}
}
return (s[31][0][0][0] + c[31][0][0][0]) % 1000000007;
}
int FindCars(int x1, int y1, int x2, int y2, int k) {
x1--, y1--, x2--, y2--, k--;
int ans = (Query(x2, y2, k) + Query(x1 - 1, y1 - 1, k)) % 1000000007 -
(Query(x1 - 1, y2, k) + Query(x2, y1 - 1, k)) % 1000000007;
ans = (1000000007 + ans) % 1000000007;
return ans;
}
int main() {
int q;
Initialise();
scanf("%d", &q);
while (q--) {
int x1, x2, y1, y2, k;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
printf("%d\n", FindCars(x1, y1, x2, y2, k));
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
inline long long sum(long long x, long long y, long long off, long long k) {
long long lo = off + 1;
long long hi = min(k, off + x);
if (hi < lo) return 0;
long long ret = (lo * (hi - lo + 1) + (hi - lo) * (hi - lo + 1) / 2) % mod;
(ret *= y) %= mod;
return ret % mod;
}
inline long long getSum(long long x, long long y, long long off, long long k) {
if (y > x) {
long long ret = sum(x, x, off, k);
ret += sum(x, y - x, off ^ x, k);
if (ret >= mod) ret -= mod;
return ret;
}
return sum(x, y, off, k);
}
long long compute(long long x, long long y, long long off, long long k) {
if (x < y) swap(x, y);
long long p = 1;
while (p * 2 <= x) p *= 2;
if (x == p) {
return getSum(p, y, off, k);
} else {
long long ret = getSum(p, y, off, k);
ret += compute(x - p, y, off ^ p, k);
if (ret >= mod) ret -= mod;
return ret;
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cout << setprecision(32);
int q;
cin >> q;
while (q--) {
long long x1, x2, y1, y2, k;
cin >> x1 >> y1 >> x2 >> y2 >> k;
long long ans = compute(x2, y2, 0, k);
if (x1 > 1) ans += mod - compute(x1 - 1, y2, 0, k);
if (y1 > 1) ans += mod - compute(x2, y1 - 1, 0, k);
if (x1 > 1 && y1 > 1) ans += compute(x1 - 1, y1 - 1, 0, k);
ans %= mod;
cout << ans << '\n';
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
long long p[32];
long long work(long long x, long long y, long long k, long long lazy) {
if (x <= 0 || y <= 0) return 0;
if (k <= 0) return 0;
if (x > y) swap(x, y);
int t = 0;
while (p[t] <= y) t++;
long long ans;
if (y == p[t - 1]) {
long long tmpa, tmpb;
if (p[t - 1] < k) {
tmpa = p[t - 1];
tmpa = tmpa * (tmpa + 1) / 2;
tmpa %= MOD;
tmpa *= x;
tmpa %= MOD;
tmpb = (lazy * y) % MOD;
tmpb *= x;
tmpb %= MOD;
ans = (tmpa + tmpb) % MOD;
} else {
tmpa = k;
tmpa = tmpa * (tmpa + 1) / 2;
tmpa %= MOD;
tmpa *= x;
tmpa %= MOD;
tmpb = (lazy * k) % MOD;
tmpb *= x;
tmpb %= MOD;
ans = (tmpa + tmpb) % MOD;
}
} else {
if (x < p[t - 1]) {
ans = work(x, p[t - 1], k, lazy);
ans = (ans + work(x, y - p[t - 1], k - p[t - 1], lazy + p[t - 1])) % MOD;
} else {
ans = work(p[t - 1], p[t - 1], k, lazy);
ans =
(ans + work(x - p[t - 1], p[t - 1], k - p[t - 1], lazy + p[t - 1])) %
MOD;
ans =
(ans + work(p[t - 1], y - p[t - 1], k - p[t - 1], lazy + p[t - 1])) %
MOD;
ans = (ans + work(x - p[t - 1], y - p[t - 1], k, lazy)) % MOD;
}
}
return ans;
}
int main() {
p[0] = 1;
for (int i = 1; i < 60; i++) p[i] = p[i - 1] * 2;
int q;
scanf("%d", &q);
int xa, xb, ya, yb, k;
long long result;
while (q--) {
scanf("%d%d%d%d%d", &xa, &ya, &xb, &yb, &k);
result = ((work((long long)xb, (long long)yb, (long long)k, 0) -
work((long long)(xa - 1), (long long)yb, (long long)k, 0) +
work((long long)(xa - 1), (long long)(ya - 1), (long long)k, 0) -
work((long long)xb, (long long)(ya - 1), k, 0)) %
MOD +
MOD) %
MOD;
printf("%lld\n", result);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.Arrays;
import java.io.IOException;
import java.io.UncheckedIOException;
import java.io.Closeable;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) throws Exception {
Thread thread = new Thread(null, new TaskAdapter(), "", 1 << 27);
thread.start();
thread.join();
}
static class TaskAdapter implements Runnable {
@Override
public void run() {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastInput in = new FastInput(inputStream);
FastOutput out = new FastOutput(outputStream);
CFindACar solver = new CFindACar();
solver.solve(1, in, out);
out.close();
}
}
static class CFindACar {
Modular mod = new Modular(1e9 + 7);
ArrayIndex ai = new ArrayIndex(31, 2, 2, 2, 2, 2);
int[][] dp = new int[ai.totalSize()][2];
int x1;
int x2;
int y1;
int y2;
int k;
int[] end = new int[]{1, 0};
int highest = 30;
public void solve(int testNumber, FastInput in, FastOutput out) {
int q = in.readInt();
for (int i = 0; i < q; i++) {
int x1 = in.readInt() - 1;
int y1 = in.readInt() - 1;
int x2 = in.readInt() - 1;
int y2 = in.readInt() - 1;
int k = in.readInt() - 1;
int ans = solve(x1, x2, y1, y2, k);
out.println(ans);
}
}
public int[] dp(int i, int xFloor, int xCeil, int yFloor, int yCeil, int kCeil) {
if (i < 0) {
return end;
}
int index = ai.indexOf(i, xFloor, xCeil, yFloor, yCeil, kCeil);
if (dp[index][0] == -1) {
int xL = Bits.bitAt(x1, i);
int xR = Bits.bitAt(x2, i);
int yL = Bits.bitAt(y1, i);
int yR = Bits.bitAt(y2, i);
int ki = Bits.bitAt(k, i);
dp[index][0] = dp[index][1] = 0;
for (int a = 0; a < 2; a++) {
for (int b = 0; b < 2; b++) {
int xor = a ^ b;
if (xFloor == 1 && a < xL
|| xCeil == 1 && a > xR
|| yFloor == 1 && b < yL
|| yCeil == 1 && b > yR
|| kCeil == 1 && xor > ki) {
continue;
}
int[] ans = dp(i - 1,
xFloor == 1 && a == xL ? 1 : 0,
xCeil == 1 && a == xR ? 1 : 0,
yFloor == 1 && b == yL ? 1 : 0,
yCeil == 1 && b == yR ? 1 : 0,
kCeil == 1 && ki == xor ? 1 : 0);
//cnt
dp[index][0] = mod.plus(dp[index][0], ans[0]);
dp[index][1] = mod.plus(dp[index][1], ans[1]);
dp[index][1] = mod.plus(dp[index][1], mod.mul(xor << i, ans[0]));
}
}
}
return dp[index];
}
public int solve(int x1, int x2, int y1, int y2, int k) {
SequenceUtils.deepFill(dp, -1);
this.x1 = x1;
this.x2 = x2;
this.y1 = y1;
this.y2 = y2;
this.k = k;
int[] ret = dp(highest, 1, 1, 1, 1, 1);
int ans = mod.plus(ret[1], ret[0]);
return ans;
}
}
static class Bits {
private Bits() {
}
public static int bitAt(int x, int i) {
return (x >>> i) & 1;
}
}
static class ArrayIndex {
int[] dimensions;
public ArrayIndex(int... dimensions) {
this.dimensions = dimensions;
}
public int totalSize() {
int ans = 1;
for (int x : dimensions) {
ans *= x;
}
return ans;
}
public int indexOf(int a, int b) {
return a * dimensions[1] + b;
}
public int indexOf(int a, int b, int c) {
return indexOf(a, b) * dimensions[2] + c;
}
public int indexOf(int a, int b, int c, int d) {
return indexOf(a, b, c) * dimensions[3] + d;
}
public int indexOf(int a, int b, int c, int d, int e) {
return indexOf(a, b, c, d) * dimensions[4] + e;
}
public int indexOf(int a, int b, int c, int d, int e, int f) {
return indexOf(a, b, c, d, e) * dimensions[5] + f;
}
}
static class FastOutput implements AutoCloseable, Closeable, Appendable {
private StringBuilder cache = new StringBuilder(10 << 20);
private final Writer os;
public FastOutput append(CharSequence csq) {
cache.append(csq);
return this;
}
public FastOutput append(CharSequence csq, int start, int end) {
cache.append(csq, start, end);
return this;
}
public FastOutput(Writer os) {
this.os = os;
}
public FastOutput(OutputStream os) {
this(new OutputStreamWriter(os));
}
public FastOutput append(char c) {
cache.append(c);
return this;
}
public FastOutput append(int c) {
cache.append(c);
return this;
}
public FastOutput println(int c) {
return append(c).println();
}
public FastOutput println() {
cache.append(System.lineSeparator());
return this;
}
public FastOutput flush() {
try {
os.append(cache);
os.flush();
cache.setLength(0);
} catch (IOException e) {
throw new UncheckedIOException(e);
}
return this;
}
public void close() {
flush();
try {
os.close();
} catch (IOException e) {
throw new UncheckedIOException(e);
}
}
public String toString() {
return cache.toString();
}
}
static class Modular {
int m;
public Modular(int m) {
this.m = m;
}
public Modular(long m) {
this.m = (int) m;
if (this.m != m) {
throw new IllegalArgumentException();
}
}
public Modular(double m) {
this.m = (int) m;
if (this.m != m) {
throw new IllegalArgumentException();
}
}
public int valueOf(int x) {
x %= m;
if (x < 0) {
x += m;
}
return x;
}
public int valueOf(long x) {
x %= m;
if (x < 0) {
x += m;
}
return (int) x;
}
public int mul(int x, int y) {
return valueOf((long) x * y);
}
public int plus(int x, int y) {
return valueOf(x + y);
}
public String toString() {
return "mod " + m;
}
}
static class FastInput {
private final InputStream is;
private byte[] buf = new byte[1 << 13];
private int bufLen;
private int bufOffset;
private int next;
public FastInput(InputStream is) {
this.is = is;
}
private int read() {
while (bufLen == bufOffset) {
bufOffset = 0;
try {
bufLen = is.read(buf);
} catch (IOException e) {
bufLen = -1;
}
if (bufLen == -1) {
return -1;
}
}
return buf[bufOffset++];
}
public void skipBlank() {
while (next >= 0 && next <= 32) {
next = read();
}
}
public int readInt() {
int sign = 1;
skipBlank();
if (next == '+' || next == '-') {
sign = next == '+' ? 1 : -1;
next = read();
}
int val = 0;
if (sign == 1) {
while (next >= '0' && next <= '9') {
val = val * 10 + next - '0';
next = read();
}
} else {
while (next >= '0' && next <= '9') {
val = val * 10 - next + '0';
next = read();
}
}
return val;
}
}
static class SequenceUtils {
public static void deepFill(Object array, int val) {
if (!array.getClass().isArray()) {
throw new IllegalArgumentException();
}
if (array instanceof int[]) {
int[] intArray = (int[]) array;
Arrays.fill(intArray, val);
} else {
Object[] objArray = (Object[]) array;
for (Object obj : objArray) {
deepFill(obj, val);
}
}
}
}
}
| JAVA |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 33;
void add(long long &a, long long b) {
(a += b) %= mod;
if (a < 0) a += mod;
}
int q;
int A[maxn], acnt;
int B[maxn], bcnt;
int C[maxn], ccnt;
bool vis[maxn][2][2][2];
struct node {
long long cnt, sum;
void clear() { cnt = sum = 0; }
} dp[maxn][2][2][2];
node dfs(int pos, int limit1, int limit2, int limit) {
if (pos == -1) return (node){1, 0};
if (vis[pos][limit1][limit2][limit]) return dp[pos][limit1][limit2][limit];
vis[pos][limit1][limit2][limit] = 1;
node &res = dp[pos][limit1][limit2][limit];
int up1 = limit1 ? A[pos] : 1, up2 = limit2 ? B[pos] : 1, sta = C[pos];
for (int a = 0; a <= up1; a++)
for (int b = 0; b <= up2; b++) {
int val = a ^ b;
if (!limit || val <= sta) {
node e = dfs(pos - 1, limit1 && a == up1, limit2 && b == up2,
limit && sta == val);
add(res.cnt, e.cnt);
add(res.sum, e.sum);
add(res.sum, 1ll * val * e.cnt % mod * ((1ll << pos) % mod) % mod);
}
}
return res;
}
long long solve(int n, int m, int k) {
if (n < 0 || m < 0) return 0;
memset(vis, 0, sizeof(vis));
for (int i = 0; i < maxn; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
for (int h = 0; h < 2; h++) dp[i][j][k][h].clear();
acnt = bcnt = ccnt = 0;
memset(A, 0, sizeof(A));
memset(B, 0, sizeof(B));
memset(C, 0, sizeof(C));
int a = n, b = m, c = k;
while (a) {
int k = a & 1;
a >>= 1;
A[acnt++] = k;
}
while (b) {
int k = b & 1;
b >>= 1;
B[bcnt++] = k;
}
while (c) {
int k = c & 1;
c >>= 1;
C[ccnt++] = k;
}
int pos = max(max(acnt - 1, bcnt - 1), ccnt - 1);
node ans = dfs(pos, 1, 1, 1);
return (ans.sum + ans.cnt) % mod;
}
int main() {
scanf("%d", &q);
for (int i = 1; i <= q; i++) {
int x1, x2, y1, y2, k;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
k--, x1--, x2--, y1--, y2--;
printf("%lld\n", ((solve(x2, y2, k) - solve(x1 - 1, y2, k) -
solve(x2, y1 - 1, k) + solve(x1 - 1, y1 - 1, k)) %
mod +
mod) %
mod);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
long long int k;
inline long long int sum(long long int l, long long int r) {
if (l > r)
return 0;
else
return (l + r) * (r - l + 1) / 2 % mod;
}
long long int solve(long long int x, long long int y, long long int h) {
if (x <= 0 || y <= 0) return 0;
if (x < y) swap(x, y);
long long int t = 1, ans = 0;
while (t * 2 <= x) t <<= 1;
if (t <= y) {
ans = sum(h + 1, min(t + h, k)) * t % mod;
ans = (ans + solve(x - t, y - t, h)) % mod;
ans = (ans + solve(x - t, t, h + t)) % mod;
ans = (ans + solve(t, y - t, h + t)) % mod;
} else {
ans = sum(h + 1, min(t + h, k)) * y % mod;
ans = (ans + solve(x - t, y, h + t)) % mod;
}
return ans;
}
int main() {
long long int x1, y1, x2, y2, q;
scanf("%lld", &q);
while (q--) {
scanf("%lld%lld%lld%lld%lld", &x1, &y1, &x2, &y2, &k);
long long int ans = solve(x2, y2, 0) + solve(x1 - 1, y1 - 1, 0) -
solve(x2, y1 - 1, 0) - solve(x1 - 1, y2, 0);
ans = (ans % mod + mod) % mod;
printf("%lld\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Div2_415E {
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
PrintWriter printer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
int numQ = Integer.parseInt(reader.readLine());
while (numQ-- > 0) {
StringTokenizer inputData = new StringTokenizer(reader.readLine());
long t = Long.parseLong(inputData.nextToken());
long l = Long.parseLong(inputData.nextToken());
long b = Long.parseLong(inputData.nextToken());
long r = Long.parseLong(inputData.nextToken());
T = Long.parseLong(inputData.nextToken());
// System.out.println(query(r, b));
// System.out.println(query(l - 1, b));
// System.out.println(query(r, t - 1));
// System.out.println(query(l - 1, b - 1));
long ans = (MOD * 2 + query(r, b) - query(l - 1, b) - query(r, t - 1) + query(l - 1, t - 1)) % MOD;
printer.println(ans);
// printer.println();
}
printer.close();
}
static long T;
static final long MOD = 1_000_000_007L;
static long query(long wid, long hei) {
return query(wid, hei, 0);
}
static long query(long wid, long hei, final long off) {
if (wid < hei) {
long temp = wid;
wid = hei;
hei = temp;
}
assert (hei <= wid);
assert (off >= 0);
if (off >= T || wid <= 0 || hei <= 0) {
return 0;
}
long blk = Long.highestOneBit(hei);
long ext = wid % blk;
long reg = wid - ext;
long ans;
// regOff indicates the item at position reg
long regOff = reg + off;
if (regOff < T) {
ans = (sum(1 + off, regOff) * blk % MOD + sum(regOff + 1, min(regOff + blk, T)) * ext) % MOD;
} else {
ans = sum(1 + off, T) * blk % MOD;
}
assert (0 <= ans && ans < MOD);
// computation of sub values
long rem = hei - blk;
assert (rem < blk);
if (rem == 0) {
return ans;
}
ext = wid % (blk << 1);
reg = wid - ext;
regOff = reg + off;
if (regOff < T) {
ans = (ans + sum(1 + off, regOff) * rem) % MOD;
assert (0 <= ans && ans < MOD);
if (ext >= blk) {
long skip = reg + blk;
if (skip + off < T) {
ans = (ans + sum(skip + 1 + off, min(skip + blk + off, T)) * rem) % MOD;
}
ext -= blk;
if (ext > 0) {
ans = (ans + query(ext, rem, reg + off)) % MOD;
}
} else if (ext > 0) {
ans = (ans + query(ext, rem, reg + blk + off)) % MOD;
}
} else {
ans = (ans + sum(1 + off, T) * rem) % MOD;
}
assert (0 <= ans && ans < MOD);
return ans;
}
static long sum(long start, long end) {
if (start > end) {
return 0;
}
return (((end - start + 1L) * (start + end)) >> 1) % MOD;
}
static long min(long a, long b) {
return a < b ? a : b;
}
}
| JAVA |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
struct par {
long long cuantos, suma;
} D[35][5][5][5], m;
long long mr, mc, k;
long long R[35], C[35], A[35];
bool Y[35][5][5][5];
long long P[35], modd = 1000000007;
par dp(long long idx, bool yr, bool yc, bool ya) {
if (Y[idx][yr][yc][ya]) return D[idx][yr][yc][ya];
if (idx == 32) {
m.suma = 0;
m.cuantos = 1;
return m;
}
Y[idx][yr][yc][ya] = true;
par ans;
ans.cuantos = ans.suma = 0;
for (int a = 0; a <= 1; a++) {
for (int b = 0; b <= 1; b++) {
if (a == 1 and !yr and R[idx] == 0) {
continue;
}
if (b == 1 and !yc and C[idx] == 0) {
continue;
}
if (a != b and !ya and A[idx] == 0) {
continue;
}
bool newr = yr or (R[idx] == 1 and a == 0);
bool newc = yc or (C[idx] == 1 and b == 0);
bool newa = ya or (A[idx] == 1 and a == b);
ans.cuantos =
(ans.cuantos + dp(idx + 1, newr, newc, newa).cuantos) % modd;
ans.suma = ans.suma + dp(idx + 1, newr, newc, newa).suma;
ans.suma = ans.suma % modd;
if (a != b)
ans.suma = ans.suma + P[idx] * dp(idx + 1, newr, newc, newa).cuantos;
ans.suma = ans.suma % modd;
}
}
D[idx][yr][yc][ya] = ans;
return ans;
}
void ponlo(long long *O, long long que) {
for (long long i = 0; i < 32; i++) {
O[i] = (que & (1 << (31 - i))) > 0;
}
}
long long saca(long long r, long long c, long long k) {
if (r < 0 or c < 0 or k < 0) return 0;
memset(Y, 0, sizeof(Y));
ponlo(R, r);
ponlo(C, c);
ponlo(A, k);
return (dp(0, 0, 0, 0).suma + dp(0, 0, 0, 0).cuantos) % modd;
}
long long sacar(int r1, int c1, int r2, int c2, int k) {
long long ayu = 0;
long long g = (ayu + saca(r2, c2, k) - saca(r1 - 1, c2, k) -
saca(r2, c1 - 1, k) + saca(r1 - 1, c1 - 1, k)) %
modd;
if (g < 0) g = g + modd;
return g;
}
int main() {
P[31] = 1;
for (int i = 30; i >= 0; i--) P[i] = P[i + 1] * 2;
ios::sync_with_stdio(false);
int ctos;
cin >> ctos;
while (ctos--) {
int r1, c1, r2, c2, k;
cin >> r1 >> c1 >> r2 >> c2 >> k;
--k;
--r1;
--r2;
--c1;
--c2;
cout << sacar(r1, c1, r2, c2, k) << "\n";
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int B = 27397, MOD = 1e9 + 7;
const int B1 = 33941, MOD1 = 1e9 + 9;
int q, r, s, k;
int memo_cnt[32][2][2][2], memo_sum[32][2][2][2];
inline int add(int a, int b) {
a += b;
while (a < 0) a += MOD;
return a % MOD;
}
inline int mul(const int &a, const int &b) { return (long long)a * b % MOD; }
inline void init() {
memset(memo_cnt, -1, sizeof memo_cnt);
memset(memo_sum, -1, sizeof memo_sum);
}
int get_cnt(int bit, bool flag_r, bool flag_s, bool flag_k) {
int &ref = memo_cnt[bit][flag_r][flag_s][flag_k];
if (ref != -1) return ref;
if (bit == 0) return ref = 1;
bool r_bit = (r & (1 << (bit - 1))) != 0;
bool s_bit = (s & (1 << (bit - 1))) != 0;
bool k_bit = (k & (1 << (bit - 1))) != 0;
ref = 0;
for (int i = 0; i < 2; ++i) {
if (flag_r && i > r_bit) continue;
for (int j = 0; j < 2; ++j) {
if (flag_s && j > s_bit) continue;
if (flag_k && (i ^ j) > k_bit) continue;
ref = add(ref,
get_cnt(bit - 1, flag_r && (i == r_bit), flag_s && (j == s_bit),
flag_k && ((i ^ j) == k_bit)));
}
}
return ref;
}
int get_sum(int bit, bool flag_r, bool flag_s, bool flag_k) {
int &ref = memo_sum[bit][flag_r][flag_s][flag_k];
if (ref != -1) return ref;
if (bit == 0) return ref = 1;
bool r_bit = (r & (1 << (bit - 1))) != 0;
bool s_bit = (s & (1 << (bit - 1))) != 0;
bool k_bit = (k & (1 << (bit - 1))) != 0;
ref = 0;
for (int i = 0; i < 2; ++i) {
if (flag_r && i > r_bit) continue;
for (int j = 0; j < 2; ++j) {
if (flag_s && j > s_bit) continue;
if (flag_k && (i ^ j) > k_bit) continue;
int cnt = get_cnt(bit - 1, flag_r && (i == r_bit), flag_s && (j == s_bit),
flag_k && ((i ^ j) == k_bit));
if (i ^ j) ref = add(ref, mul(cnt, 1 << (bit - 1)));
ref = add(ref,
get_sum(bit - 1, flag_r && (i == r_bit), flag_s && (j == s_bit),
flag_k && ((i ^ j) == k_bit)));
}
}
return ref;
}
int pref(int rr, int ss) {
r = rr;
s = ss;
if (r < 0 || s < 0) return 0;
init();
return get_sum(31, true, true, true);
}
int solve(int r1, int s1, int r2, int s2) {
return add(pref(r2, s2), add(pref(r1 - 1, s1 - 1),
add(-pref(r2, s1 - 1), -pref(r1 - 1, s2))));
}
int main(void) {
scanf("%d", &q);
for (int i = 0; i < q; ++i) {
int r1, s1, r2, s2;
scanf("%d%d%d%d%d", &r1, &s1, &r2, &s2, &k);
--r1;
--r2;
--s1;
--s2;
--k;
printf("%d\n", solve(r1, s1, r2, s2));
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long UNDEF = -1;
const long long INF = 1e18;
template <typename T>
inline bool chkmax(T& aa, T bb) {
return aa < bb ? aa = bb, true : false;
}
template <typename T>
inline bool chkmin(T& aa, T bb) {
return aa > bb ? aa = bb, true : false;
}
static char stdinBuffer[1024];
static char* stdinDataEnd = stdinBuffer + sizeof(stdinBuffer);
static const char* stdinPos = stdinDataEnd;
void readAhead(size_t amount) {
size_t remaining = stdinDataEnd - stdinPos;
if (remaining < amount) {
memmove(stdinBuffer, stdinPos, remaining);
size_t sz = fread(stdinBuffer + remaining, 1,
sizeof(stdinBuffer) - remaining, stdin);
stdinPos = stdinBuffer;
stdinDataEnd = stdinBuffer + remaining + sz;
if (stdinDataEnd != stdinBuffer + sizeof(stdinBuffer)) *stdinDataEnd = 0;
}
}
int readInt() {
readAhead(16);
int x = 0;
bool neg = false;
while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
if (*stdinPos == '-') {
++stdinPos;
neg = true;
}
while (*stdinPos >= '0' && *stdinPos <= '9') {
x *= 10;
x += *stdinPos - '0';
++stdinPos;
}
return neg ? -x : x;
}
const int me = 31;
long long k;
long long x, y;
pair<long long, long long> dp[me][2][2][2];
long long dphit[me][2][2][2];
const int LS = 0, EQ = 1, GR = 2;
int trans(int s, int b, int tb) {
if (s == LS) return LS;
if (b < tb)
return LS;
else if (b == tb)
return EQ;
else
return GR;
}
long long final = 0;
pair<long long, long long> f(int e, int xs, int ys, int ks) {
if (e == -1) {
if (xs + ys + ks == 0)
return make_pair(1ll, 1ll);
else
return make_pair(0ll, 0ll);
}
auto dpval = dp[e][xs][ys][ks];
if (dpval.first != -1) return dpval;
pair<long long, long long> ans = make_pair(0ll, 0ll);
for (int xb = 0; xb < 2; xb++) {
for (int yb = 0; yb < 2; yb++) {
{
int kb = xb ^ yb;
int nxs = trans(xs, xb, (x >> e) & 1);
int nys = trans(ys, yb, (y >> e) & 1);
int nks = trans(ks, kb, (k >> e) & 1);
if (nxs != GR && nys != GR && nks != GR) {
auto got = f(e - 1, nxs, nys, nks);
ans.first += got.first;
ans.second += got.second;
if (kb == 1) ans.second += (got.first << e) % 1000000007LL;
ans.first %= 1000000007LL;
ans.second %= 1000000007LL;
}
}
}
}
return dp[e][xs][ys][ks] = ans;
}
int bf(int x, int y, int k) {
int ans = 0;
for (int i = 0; i < x; i++)
for (int j = 0; j < y; j++) {
if ((i ^ j) < k) ans += (i ^ j) + 1;
}
return ans;
}
int query(int _x, int _y, int _k) {
k = _k;
x = _x;
y = _y;
if (x < 0 || y < 0) return 0;
memset(dp, -1, sizeof dp);
memset(dphit, 0, sizeof dphit);
auto ans = f(me - 1, 1, 1, 1);
return (ans.second) % 1000000007LL;
}
void tester() {
int lim = 4;
for (int i = 0; i < 1000; i++) {
int x = rand() % lim + 1, y = rand() % lim + 1, k = rand() % lim + 1;
if (bf(x, y, k) != query(x, y, k)) {
printf("x:%d y:%d k:%d. BF:%d QUE:%d\n", x, y, k, bf(x, y, k),
query(x, y, k));
}
}
printf("\n");
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int q = readInt();
for (int ii = 0; ii < q; ii++) {
int x1 = readInt(), y1 = readInt(), x2 = readInt(), y2 = readInt(),
_k = readInt();
--x1;
--y1;
int ans = query(x2, y2, _k) - query(x2, y1, _k) - query(x1, y2, _k) +
query(x1, y1, _k);
ans %= 1000000007LL;
if (ans < 0) ans += 1000000007LL;
printf("%d\n", ans);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
long long la, ra, lb, rb, K;
pair<long long, long long> dp[35][2][2][2];
pair<long long, long long> calc(int p, int f1, int f2, int pre, long long lim1,
long long lim2) {
if (p == -1) return pair<long long, long long>(0, 1);
if (dp[p][f1][f2][pre].first != -1) return dp[p][f1][f2][pre];
int e1 = f1 ? (lim1 >> p & 1) : 1, e2 = f2 ? (lim2 >> p & 1) : 1;
pair<long long, long long> tmp, ret(0, 0);
for (int i = 0; i <= e1; ++i) {
for (int j = 0; j <= e2; ++j) {
if (pre && (i ^ j) > (K >> p & 1)) continue;
tmp = calc(p - 1, f1 && i == e1, f2 && j == e2,
pre && (i ^ j) == (K >> p & 1), lim1, lim2);
ret.first =
(ret.first + tmp.second * (i ^ j) % MOD * (1ll << p) + tmp.first) %
MOD;
ret.second = (ret.second + tmp.second) % MOD;
}
}
dp[p][f1][f2][pre] = ret;
return ret;
}
int main() {
int Q;
scanf("%d", &Q);
while (Q--) {
scanf("%I64d%I64d%I64d%I64d%I64d", &la, &lb, &ra, &rb, &K);
--la, --lb, --ra, --rb, --K;
long long ans = 0, t1, t2;
memset(dp, -1, sizeof(dp));
tie(t1, t2) = calc(32, 1, 1, 1, ra, rb);
ans = (ans + t1 + t2) % MOD;
if (la != 0) {
memset(dp, -1, sizeof(dp));
tie(t1, t2) = calc(32, 1, 1, 1, la - 1, rb);
ans = (ans - t1 - t2 + MOD + MOD) % MOD;
}
if (lb != 0) {
memset(dp, -1, sizeof(dp));
tie(t1, t2) = calc(32, 1, 1, 1, ra, lb - 1);
ans = (ans - t1 - t2 + MOD + MOD) % MOD;
}
if (la != 0 && lb != 0) {
memset(dp, -1, sizeof(dp));
tie(t1, t2) = calc(32, 1, 1, 1, la - 1, lb - 1);
ans = (ans + t1 + t2) % MOD;
}
printf("%I64d\n", ans);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
int q, dp[31][8], sum[30][8], k;
void add(int &a, int b) {
a += b;
if (a >= M) a -= M;
}
void sub(int &a, int b) {
a -= b;
if (a < 0) a += M;
}
int find(int x, int y) {
if (x < 0 || y < 0) return 0;
int ret = 0;
memset(dp, 0, sizeof dp);
memset(sum, 0, sizeof sum);
dp[30][0] = 1;
for (int i = 30; i; i--)
for (int j = 0; j < 8; j++)
if (dp[i][j])
for (int at = 0; at < 4; at++) {
bool bitx = at & 1, bity = at & 2;
if (bitx && !(x & (1 << i)) && !(j & 4)) continue;
if (bity && !(y & (1 << i)) && !(j & 2)) continue;
if ((bitx ^ bity) && !(k & (1 << i)) && !(j & 1)) continue;
int to = j;
to |= 4 * (bitx < !!(x & (1 << i)));
to |= 2 * (bity < !!(y & (1 << i)));
to |= (bitx ^ bity) < !!(k & (1 << i));
add(dp[i - 1][to], dp[i][j]);
add(sum[i - 1][to], sum[i][j]);
add(sum[i - 1][to], ((bitx ^ bity) << i) * 1LL * dp[i][j] % M);
}
int i = 0;
for (int j = 0; j < 8; j++)
for (int at = 0; at < 4; at++) {
bool bitx = at & 1, bity = at & 2;
if (bitx && !(x & (1 << i)) && !(j & 4)) continue;
if (bity && !(y & (1 << i)) && !(j & 2)) continue;
if ((bitx ^ bity) && !(k & (1 << i)) && !(j & 1)) continue;
add(ret, (bitx ^ bity) * dp[i][j]);
add(ret, dp[i][j]);
add(ret, sum[i][j]);
}
return ret;
}
int main() {
scanf("%d", &q);
while (q--) {
int x, x1, y, y1;
scanf("%d%d%d%d%d", &x, &y, &x1, &y1, &k);
x--;
y--;
x1--;
y1--;
k--;
int ans = 0;
add(ans, find(x1, y1));
sub(ans, find(x - 1, y1));
sub(ans, find(x1, y - 1));
add(ans, find(x - 1, y - 1));
printf("%d\n", ans);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long ct = 0;
long long l(long long x) {
long long ans = 0, p = 1;
while (2 * p <= x) {
ans++;
p *= 2;
}
return ans;
}
long long sm(long long p) { return ((p * (p + 1)) / 2) % 1000000007; }
long long comp_less(long long x, long long y, long long k) {
if (x == 0 || y == 0 || k < 1) return 0;
long long m = max(l(x), l(y));
long long ans = 0, r;
if ((1LL << m) >= min(x, y)) {
if (x > y) swap(x, y);
r = min(1LL << m, k);
ans = (r * x) % 1000000007;
if (k > (1LL << m)) ans += comp_less(x, y - (1LL << m), k - (1LL << m));
ans %= 1000000007;
return ans;
}
m = min(l(x), l(y));
r = min(1LL << m, k);
ans = ((1LL << m) * r) % 1000000007;
if (k > (1LL << m))
ans += comp_less(1LL << m, y - (1LL << m), k - (1LL << m)) +
comp_less(1LL << m, x - (1LL << m), k - (1LL << m));
ans += comp_less(x - (1LL << m), y - (1LL << m), k);
ans %= 1000000007;
return ans;
}
long long comp(long long x, long long y, long long k) {
if (x == 0 || y == 0) return 0;
ct++;
long long m = max(l(x), l(y));
long long ans = 0, r, extray, extrax;
if ((1LL << m) >= min(x, y)) {
if (x > y) swap(x, y);
r = min(1LL << m, k);
ans = (sm(r) * x) % 1000000007;
extray = (comp_less(x, y, k) - comp_less(x, 1LL << m, k)) % 1000000007;
if (k > (1LL << m))
ans += (extray * (1LL << m)) % 1000000007 +
comp(x, y - (1LL << m), k - (1LL << m));
ans %= 1000000007;
return ans;
}
m = min(l(x), l(y));
r = min(1LL << m, k);
ans = ((1LL << m) * sm(r)) % 1000000007;
extray = (comp_less(1LL << m, y, k) - comp_less(1LL << m, 1LL << m, k)) %
1000000007;
extrax = (comp_less(x, 1LL << m, k) - comp_less(1LL << m, 1LL << m, k));
if (k > (1LL << m))
ans += (extray * (1LL << m)) % 1000000007 +
(extrax * (1LL << m)) % 1000000007 +
comp(1LL << m, y - (1LL << m), k - (1LL << m)) +
comp(1LL << m, x - (1LL << m), k - (1LL << m));
ans += comp(x - (1LL << m), y - (1LL << m), k);
ans %= 1000000007;
return ans;
}
int main() {
long long q, x1, y1, x2, y2, k;
cin >> q;
while (q--) {
cin >> x1 >> y1 >> x2 >> y2 >> k;
long long ans = comp(x2, y2, k) - comp(x1 - 1, y2, k) -
comp(x2, y1 - 1, k) + comp(x1 - 1, y1 - 1, k);
ans = (ans % 1000000007 + 1000000007) % 1000000007;
cout << ans << "\n";
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const int MOD = 1e9 + 7;
ll binpow(ll a, ll p, int mod = MOD) {
ll res = 1;
while (p) {
if (p & 1) {
(res *= a) %= mod;
}
p >>= 1;
(a *= a) %= mod;
}
return res;
}
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
const int M = 32;
ll dp[M][2][2][2];
ll cnt[M][2][2][2];
ll count(ll x, ll y, ll K) {
for (int i = 0; i < M; i++) {
for (int fx = 0; fx < 2; fx++)
for (int fy = 0; fy < 2; fy++)
for (int fk = 0; fk < 2; fk++) {
cnt[i][fx][fy][fk] = dp[i][fx][fy][fk] = 0;
}
}
cnt[0][1][1][1] = 1;
for (int pref = 1; pref < M; pref++) {
int bit = M - pref - 1;
int bx = x >> bit & 1;
int by = y >> bit & 1;
int bk = K >> bit & 1;
for (int fx = 0; fx < 2; fx++)
for (int fy = 0; fy < 2; fy++)
for (int fk = 0; fk < 2; fk++) {
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
int k = i ^ j;
if (fx && i > bx || fy && j > by || fk && k > bk) continue;
int ffx = fx && i == bx;
int ffy = fy && j == by;
int ffk = fk && k == bk;
(cnt[pref][ffx][ffy][ffk] += cnt[pref - 1][fx][fy][fk]) %= MOD;
(dp[pref][ffx][ffy][ffk] += cnt[pref - 1][fx][fy][fk] * k +
dp[pref - 1][fx][fy][fk] * 2) %= MOD;
}
}
}
}
ll res = cnt[M - 1][0][0][0] + dp[M - 1][0][0][0];
return res % MOD;
}
void solve() {
ll x1, y1, x2, y2, k;
cin >> x1 >> y1 >> x2 >> y2 >> k;
ll ans = count(x2, y2, k) - count(x1 - 1, y2, k) - count(x2, y1 - 1, k) +
count(x1 - 1, y1 - 1, k);
ans = ans % MOD + MOD;
cout << ans % MOD << endl;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
cin >> T;
for (int tc = 1; tc <= T; tc++) {
solve();
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
long long sum[35][2][2][2], cnt[35][2][2][2];
inline long long Calc(int _x, int _y, int _K) {
if (_x == -1 || _y == -1) return 0;
int x[35] = {0}, y[35] = {0}, K[35] = {0}, _p = 0;
_p = -1;
while (_x) x[++_p] = _x & 1, _x >>= 1;
_p = -1;
while (_y) y[++_p] = _y & 1, _y >>= 1;
_p = -1;
while (_K) K[++_p] = _K & 1, _K >>= 1;
memset(sum, 0, sizeof(sum));
memset(cnt, 0, sizeof(cnt));
cnt[31][0][0][0] = 1;
for (int i = 31; i; i--)
for (int t1 = 0; t1 < 2; t1++)
for (int t2 = 0; t2 < 2; t2++)
for (int t3 = 0; t3 < 2; t3++) {
for (int a = 0; a <= (t1 ? 1 : x[i - 1]); a++)
for (int b = 0; b <= (t2 ? 1 : y[i - 1]); b++) {
if ((a ^ b) > (t3 ? 1 : K[i - 1])) continue;
int nt1 = t1 | (a < x[i - 1]), nt2 = t2 | (b < y[i - 1]),
nt3 = t3 | ((a ^ b) < K[i - 1]);
(cnt[i - 1][nt1][nt2][nt3] += cnt[i][t1][t2][t3]) %= P;
(sum[i - 1][nt1][nt2][nt3] +=
sum[i][t1][t2][t3] +
cnt[i][t1][t2][t3] * ((a ^ b) << (i - 1)) % P) %= P;
}
}
long long Sum = 0, Cnt = 0;
for (int t1 = 0; t1 < 2; t1++)
for (int t2 = 0; t2 < 2; t2++)
for (int t3 = 0; t3 < 2; t3++)
(Sum += sum[0][t1][t2][t3]) %= P, (Cnt += cnt[0][t1][t2][t3]) %= P;
return (Sum + Cnt) % P;
}
int main() {
int Q, x1, x2, y1, y2, K;
scanf("%d", &(Q));
while (Q--) {
scanf("%d", &(x1));
scanf("%d", &(y1));
scanf("%d", &(x2));
scanf("%d", &(y2));
scanf("%d", &(K));
x1--;
y1--;
x2--;
y2--;
K--;
printf("%I64d\n", (Calc(x2, y2, K) + P - Calc(x2, y1 - 1, K) + P -
Calc(x1 - 1, y2, K) + Calc(x1 - 1, y1 - 1, K)) %
P);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int mxn = 10005;
int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
int x[35], y[35], k[35], len;
long long cnt[33][2][2][2];
long long f[33][2][2][2];
long long calc(int X, int Y, int K) {
if (X < Y) swap(X, Y);
if (!X || !Y) return 0;
K--;
X--;
Y--;
len = 31;
int i, j;
for (i = 1; i <= len; i++) {
k[i] = K & 1;
K >>= 1;
}
for (i = 1; i <= len; i++) {
y[i] = Y & 1;
Y >>= 1;
}
for (i = 1; i <= len; i++) {
x[i] = X & 1;
X >>= 1;
}
memset(cnt, 0, sizeof cnt);
memset(f, 0, sizeof f);
cnt[32][0][0][0] = 1;
for (i = 31; i; i--) {
for (int c1 = 0; c1 < 2; c1++)
for (int c2 = 0; c2 < 2; c2++)
for (int nk = 0; nk < 2; nk++)
for (int a = 0; a <= (c1 ? 1 : x[i]); a++) {
for (int b = 0; b <= (c2 ? 1 : y[i]); b++) {
if ((a ^ b) > (nk ? 1 : k[i])) continue;
int t1 = c1 | (a < x[i]), t2 = c2 | (b < y[i]),
t3 = nk | ((a ^ b) < k[i]);
(cnt[i][t1][t2][t3] += cnt[i + 1][c1][c2][nk]) %= mod;
(f[i][t1][t2][t3] +=
f[i + 1][c1][c2][nk] +
(((a ^ b) << (i - 1)) % mod * cnt[i + 1][c1][c2][nk]) % mod) %=
mod;
}
}
}
long long smm = 0, g = 0;
for (int c1 = 0; c1 < 2; c1++)
for (int c2 = 0; c2 < 2; c2++)
for (int nk = 0; nk < 2; nk++) {
smm += f[1][c1][c2][nk];
if (smm >= mod) smm -= mod;
g += cnt[1][c1][c2][nk];
if (g >= mod) g -= mod;
}
return (smm + g) % mod;
}
int main() {
int i, j, x1, x2, y1, y2, k;
int Q = read();
while (Q--) {
x1 = read();
y1 = read();
x2 = read();
y2 = read();
k = read();
long long ans = ((long long)calc(x2, y2, k) - calc(x1 - 1, y2, k) -
calc(x2, y1 - 1, k) + calc(x1 - 1, y1 - 1, k)) %
mod;
ans = (ans + mod) % mod;
printf("%I64d\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline long long add(long long x, long long y) {
return x + y >= ((long long)1e9 + 7) ? x + y - ((long long)1e9 + 7) : x + y;
}
inline long long mns(long long x, long long y) {
return x - y < 0 ? x - y + ((long long)1e9 + 7) : x - y;
}
struct data {
long long sum, cnt;
data() {}
data(long long ss, long long cc) { sum = ss, cnt = cc; }
};
int vis[35][2][2][2];
data f[35][2][2][2];
int A[35], B[35], C[35];
data dfs(int len, int p1, int p2, int p3) {
if (!len) {
return data(0, 1);
}
if (vis[len][p1][p2][p3]) {
return f[len][p1][p2][p3];
}
data ans = data(0, 0);
int r1 = p1 ? A[len] : 1, r2 = p2 ? B[len] : 1, r3 = p3 ? C[len] : 1;
for (int i = 0; i <= r1; i++) {
for (int j = 0; j <= r2; j++) {
if ((i ^ j) <= r3) {
data tmp = dfs(len - 1, (p1 & (i == A[len])), (p2 & (j == B[len])),
(p3 & ((i ^ j) == C[len])));
ans.cnt = add(ans.cnt, tmp.cnt);
ans.sum = add(ans.sum, add(tmp.sum, 1LL * (i ^ j) * (1LL << (len - 1)) %
((long long)1e9 + 7) * tmp.cnt %
((long long)1e9 + 7)));
}
}
}
vis[len][p1][p2][p3] = 1;
f[len][p1][p2][p3] = ans;
return ans;
}
int get(int a[], long long x) {
int cnt = 0;
while (x) {
a[++cnt] = (x & 1);
x /= 2;
}
return cnt;
}
long long cal(long long x, long long y, long long K) {
if (x < 0 || y < 0 || K < 0) {
return 0;
}
memset(A, 0, sizeof(A));
memset(B, 0, sizeof(B));
memset(C, 0, sizeof(C));
memset(vis, 0, sizeof(vis));
int cnta = get(A, x), cntb = get(B, y), cntc = get(C, K);
data ans = dfs(max(cnta, max(cntb, cntc)), 1, 1, 1);
return add(ans.sum, ans.cnt);
}
int main() {
int Task;
{
Task = 0;
char ch;
int f = 0;
do {
ch = getchar();
if (ch == '-') {
f = 1;
}
} while (ch < '0' || ch > '9');
while ('0' <= ch && ch <= '9') {
Task = (Task << 1) + (Task << 3) + ch - 48;
ch = getchar();
}
if (f) Task = -Task;
}
long long X1, X2, Y1, Y2, K;
while (Task--) {
{
X1 = 0;
char ch;
int f = 0;
do {
ch = getchar();
if (ch == '-') {
f = 1;
}
} while (ch < '0' || ch > '9');
while ('0' <= ch && ch <= '9') {
X1 = (X1 << 1) + (X1 << 3) + ch - 48;
ch = getchar();
}
if (f) X1 = -X1;
}
{
Y1 = 0;
char ch;
int f = 0;
do {
ch = getchar();
if (ch == '-') {
f = 1;
}
} while (ch < '0' || ch > '9');
while ('0' <= ch && ch <= '9') {
Y1 = (Y1 << 1) + (Y1 << 3) + ch - 48;
ch = getchar();
}
if (f) Y1 = -Y1;
}
{
X2 = 0;
char ch;
int f = 0;
do {
ch = getchar();
if (ch == '-') {
f = 1;
}
} while (ch < '0' || ch > '9');
while ('0' <= ch && ch <= '9') {
X2 = (X2 << 1) + (X2 << 3) + ch - 48;
ch = getchar();
}
if (f) X2 = -X2;
}
{
Y2 = 0;
char ch;
int f = 0;
do {
ch = getchar();
if (ch == '-') {
f = 1;
}
} while (ch < '0' || ch > '9');
while ('0' <= ch && ch <= '9') {
Y2 = (Y2 << 1) + (Y2 << 3) + ch - 48;
ch = getchar();
}
if (f) Y2 = -Y2;
}
{
K = 0;
char ch;
int f = 0;
do {
ch = getchar();
if (ch == '-') {
f = 1;
}
} while (ch < '0' || ch > '9');
while ('0' <= ch && ch <= '9') {
K = (K << 1) + (K << 3) + ch - 48;
ch = getchar();
}
if (f) K = -K;
}
X1--, X2--, Y1--, Y2--, K--;
long long t1 = cal(X1 - 1, Y1 - 1, K), t2 = cal(X1 - 1, Y2, K),
t3 = cal(X2, Y1 - 1, K), t4 = cal(X2, Y2, K);
printf("%lld\n", mns(add(t1, t4), add(t2, t3)));
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
constexpr int64_t MOD = (int64_t)1e9 + 7LL;
struct quad {
int64_t x1, y1;
int64_t x2, y2;
bool contains(const quad& other) {
return (x1 <= other.x1 && other.x2 <= x2) ||
(y1 <= other.y1 && other.y2 <= y2);
}
bool intersects(const quad& other) {
return !(x2 < other.x1 || x1 > other.x2 || y2 < other.y1 || y1 > other.y2);
}
};
int64_t contribute(int64_t L, int64_t R, int64_t k, int64_t n) {
if (k < L) return 0;
int64_t A = L + min(k, R);
int64_t B = min(k, R) - L + 1;
if (A % 2LL)
B /= 2LL;
else
A /= 2LL;
A %= MOD;
B %= MOD;
return (((A * B) % MOD) * n) % MOD;
}
int64_t dfs(quad space, int64_t L, int64_t R, quad target, int64_t k) {
if (!target.intersects(space)) {
return 0;
}
if (target.contains(space)) {
int64_t n = min(min(target.x2, space.x2) - max(target.x1, space.x1) + 1,
min(target.y2, space.y2) - max(target.y1, space.y1) + 1);
return contribute(L, R, k, n % MOD);
}
int64_t ans = 0;
int64_t M = (L + R) / 2LL;
int64_t m = (R - L) / 2LL;
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < 2; ++j) {
quad next_space;
next_space.x1 = space.x1 + (i ? m + 1 : 0);
next_space.x2 = (i ? space.x2 : space.x1 + m);
next_space.y1 = space.y1 + (j ? m + 1 : 0);
next_space.y2 = (j ? space.y2 : space.y1 + m);
if (i ^ j) {
ans += dfs(next_space, M + 1, R, target, k);
ans %= MOD;
} else {
ans += dfs(next_space, L, M, target, k);
ans %= MOD;
}
}
}
return ans;
}
void solve(int x1, int x2, int y1, int y2, int64_t k) {
int64_t L = 1;
int64_t R = 1LL << 31;
quad space;
space.x1 = space.y1 = L;
space.x2 = space.y2 = R;
quad target;
target.x1 = x1;
target.x2 = x2;
target.y1 = y1;
target.y2 = y2;
cout << dfs(space, L, R, target, k) << endl;
}
int main() {
int q;
scanf("%d", &q);
for (int i = 0; i < q; ++i) {
int64_t x1, x2, y1, y2, k;
cin >> x1 >> y1 >> x2 >> y2 >> k;
solve(x1, x2, y1, y2, k);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int UP(int x) {
int l = 0, r = 32;
int ans = 0;
while (l <= r) {
int mid = (l + r) >> 1;
if (!(x >> mid)) {
ans = mid;
r = mid - 1;
} else
l = mid + 1;
}
return ans;
}
int qury(int x, int y, int k, int va) {
if (k <= 0) return 0;
if (x == 0 || y == 0) return 0;
int up1 = UP(x) - 1;
int up2 = UP(y) - 1;
int ret = 0;
int n = min(k, max(1 << up1, 1 << up2));
ret = ((1LL * n * (n + 1) / 2) % mod * min(1 << up1, 1 << up2)) % mod;
ret = (ret + (1LL * n * min(1 << up1, 1 << up2) % mod * va % mod)) % mod;
if (up1 > up2) {
ret =
(ret + qury(x - (1 << up1), y, k - (1 << up1), va + (1 << up1))) % mod;
ret = (ret + (1LL * n * (n + 1) / 2) % mod * (y - (1 << up2)) % mod) % mod;
ret = (ret + (1LL * n * (y - (1 << up2)) % mod * va % mod)) % mod;
} else if (up1 < up2) {
ret = (ret + (1LL * n * (n + 1) / 2) % mod * (x - (1 << up1)) % mod) % mod;
ret = (ret + (1LL * n * (x - (1 << up1)) % mod * va % mod)) % mod;
ret =
(ret + qury(x, y - (1 << up2), k - (1 << up2), va + (1 << up2))) % mod;
} else {
ret = (ret +
qury(1 << up1, y - (1 << up2), k - (1 << up2), va + (1 << up2))) %
mod;
ret = (ret +
qury(x - (1 << up1), 1 << up2, k - (1 << up1), va + (1 << up1))) %
mod;
ret = (ret + qury(x - (1 << up1), y - (1 << up2), k, va)) % mod;
}
return ret;
}
int main() {
int x1, y1, x2, y2;
int k;
int n;
scanf("%d", &n);
while (n--) {
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
int q1 = qury(x1 - 1, y1 - 1, k, 0);
int q2 = qury(x2, y1 - 1, k, 0);
int q3 = qury(x1 - 1, y2, k, 0);
int q4 = qury(x2, y2, k, 0);
int ans = (((q4 + q1) % mod - (q2 + q3) % mod) % mod + mod) % mod;
printf("%d\n", ans);
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long k;
int numx[35], numy[35], numk[35];
long long sum[35][2][2][2], num[35][2][2][2];
long long ask(long long x, long long y) {
long long ans = 0;
long long z = k;
if (x < 0 || y < 0) return 0;
memset(sum, 0, sizeof(sum));
memset(num, 0, sizeof(num));
for (int i = 0; i < 32; i++) {
numx[i] = x % 2;
numy[i] = y % 2;
numk[i] = z % 2;
x >>= 1;
y >>= 1;
z >>= 1;
}
num[32][1][1][1] = 1;
sum[32][1][1][1] = 0;
int im, jm;
for (int lens = 32; lens >= 0; lens--)
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int kk = 0; kk < 2; kk++)
if (num[lens][i][j][kk]) {
if (lens == 0) {
ans = (ans + num[lens][i][j][kk] + sum[lens][i][j][kk]) %
1000000007;
continue;
}
im = jm = 1;
if (i) im = numx[lens - 1];
if (j) jm = numy[lens - 1];
for (int nowi = 0; nowi <= im; nowi++)
for (int nowj = 0; nowj <= jm; nowj++) {
if (!kk || (nowi ^ nowj) <= numk[lens - 1]) {
num[lens - 1][nowi == im && i][nowj == jm && j]
[(nowi ^ nowj == numk[lens - 1]) && kk] =
(num[lens - 1][nowi == im && i][nowj == jm && j]
[(nowi ^ nowj == numk[lens - 1]) && kk] +
num[lens][i][j][kk]) %
1000000007;
sum[lens - 1][nowi == im && i][nowj == jm && j]
[(nowi ^ nowj == numk[lens - 1]) && kk] =
(sum[lens - 1][nowi == im && i][nowj == jm && j]
[(nowi ^ nowj == numk[lens - 1]) && kk] +
(sum[lens][i][j][kk] << 1) +
(nowi ^ nowj) * num[lens][i][j][kk]) %
1000000007;
}
}
}
return ans;
}
int main() {
int q;
long long x1, y1, x2, y2;
scanf("%d", &q);
while (q--) {
scanf("%lld%lld%lld%lld%lld", &x1, &y1, &x2, &y2, &k);
x1--;
y1--;
x2--;
y2--;
k--;
printf("%lld\n", (ask(x2, y2) - ask(x2, y1 - 1) - ask(x1 - 1, y2) +
ask(x1 - 1, y1 - 1) + 1000000007 * 2) %
1000000007);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
/**
* @author Don Li
*/
public class FindACar2 {
int N = 31;
int MOD = (int) 1e9 + 7;
void solve() {
int q = in.nextInt();
while (q-- > 0) {
int x1 = in.nextInt() - 1, y1 = in.nextInt() - 1, x2 = in.nextInt() - 1, y2 = in.nextInt() - 1, k = in.nextInt() - 1;
long ans = 0;
ans = add(ans, calc(x2, y2, k));
ans = sub(ans, calc(x1 - 1, y2, k));
ans = sub(ans, calc(x2, y1 - 1, k));
ans = add(ans, calc(x1 - 1, y1 - 1, k));
out.println(ans);
}
}
int calc(int x, int y, int k) {
if (x < 0 || y < 0) return 0;
int[][][][] cnt = new int[N + 1][2][2][2];
int[][][][] sum = new int[N + 1][2][2][2];
int[] a = new int[N], b = new int[N], c = new int[N];
for (int i = 0; i < N; i++) {
a[i] = (x >> i) & 1;
b[i] = (y >> i) & 1;
c[i] = (k >> i) & 1;
}
reverse(a);
reverse(b);
reverse(c);
cnt[0][1][1][1] = 1;
for (int i = 0; i < N; i++) {
for (int p = 0; p < 2; p++) for (int q = 0; q < 2; q++) for (int r = 0; r < 2; r++) {
for (int u = 0; u < 2; u++) for (int v = 0; v < 2; v++) {
int w = u ^ v;
if (p == 1 && u == 1 && a[i] == 0) continue;
if (q == 1 && v == 1 && b[i] == 0) continue;
if (r == 1 && w == 1 && c[i] == 0) continue;
int np = p == 1 && a[i] == u ? 1 : 0;
int nq = q == 1 && b[i] == v ? 1 : 0;
int nr = r == 1 && c[i] == w ? 1 : 0;
cnt[i + 1][np][nq][nr] = add(cnt[i + 1][np][nq][nr], cnt[i][p][q][r]);
sum[i + 1][np][nq][nr] = add(sum[i + 1][np][nq][nr], sum[i][p][q][r]);
sum[i + 1][np][nq][nr] = add(sum[i + 1][np][nq][nr], mul(w << (30 - i), cnt[i][p][q][r]));
}
}
}
long res = 0;
for (int p = 0; p < 2; p++) for (int q = 0; q < 2; q++) for (int r = 0; r < 2; r++) {
res = add(res, sum[N][p][q][r]);
res = add(res, cnt[N][p][q][r]);
}
return (int) res;
}
void reverse(int[] a) {
int l = 0, r = a.length - 1;
while (l < r) {
int tmp = a[l];
a[l] = a[r];
a[r] = tmp;
l++;
r--;
}
}
int add(long a, long b) {
return (int) ((a + b) % MOD);
}
int sub(long a, long b) {
return (int) ((a - b + MOD) % MOD);
}
int mul(long a, long b) {
return (int) (a * b % MOD);
}
public static void main(String[] args) {
in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
out = new PrintWriter(System.out);
new FindACar2().solve();
out.close();
}
static FastScanner in;
static PrintWriter out;
static class FastScanner {
BufferedReader in;
StringTokenizer st;
public FastScanner(BufferedReader in) {
this.in = in;
}
public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(in.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(nextToken());
}
public long nextLong() {
return Long.parseLong(nextToken());
}
public double nextDouble() {
return Double.parseDouble(nextToken());
}
}
}
| JAVA |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline long long read() {
long long x = 0, f = 1;
char c = getchar();
while (!isdigit(c)) {
if (c == '-') f = -1;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return x * f;
}
long long T;
long long a, b, c, d, k;
const long long mod = 1e9 + 7;
long long f[2][2][2][2], g[2][2][2][2];
bool flag;
long long sol(long long x, long long y, long long z) {
if (x < 0 || y < 0 || z < 0) return 0;
memset(f, 0, sizeof(f));
memset(g, 0, sizeof(g));
f[flag][1][1][1] = 1;
for (long long i = 31; i >= 1; --i) {
flag ^= 1;
long long s1 = ((x >> i - 1) & 1), s2 = ((y >> i - 1) & 1),
s3 = ((z >> i - 1) & 1);
for (long long _ = 0; _ <= 1; ++_) {
for (long long __ = 0; __ <= 1; ++__) {
for (long long ___ = 0; ___ <= 1; ++___) {
for (long long X = 0; X <= (s1 | (!_)); ++X) {
for (long long Y = 0; Y <= (s2 | (!__)); ++Y) {
if ((X ^ Y) > s3 && ___) continue;
f[flag][_ & (X == s1)][__ & (Y == s2)][___ & ((X ^ Y) == s3)] +=
f[flag ^ 1][_][__][___];
if (f[flag][_ & (X == s1)][__ & (Y == s2)]
[___ & ((X ^ Y) == s3)] > mod)
f[flag][_ & (X == s1)][__ & (Y == s2)][___ & ((X ^ Y) == s3)] -=
mod;
g[flag][_ & (X == s1)][__ & (Y == s2)][___ & ((X ^ Y) == s3)] =
(g[flag][_ & (X == s1)][__ & (Y == s2)]
[___ & ((X ^ Y) == s3)] +
g[flag ^ 1][_][__][___] +
1ll * f[flag ^ 1][_][__][___] * (((X ^ Y) & 1) << i - 1) %
mod) %
mod;
}
}
f[flag ^ 1][_][__][___] = g[flag ^ 1][_][__][___] = 0;
}
}
}
}
long long res = 0;
for (long long _ = 0; _ <= 1; ++_)
for (long long __ = 0; __ <= 1; ++__)
for (long long ___ = 0; ___ <= 1; ++___) {
res += f[flag][_][__][___] + g[flag][_][__][___];
if (res > mod) res -= mod;
if (res > mod) res -= mod;
}
return res;
}
signed main() {
T = read();
while (T--) {
long long ans = 0;
a = read(), b = read(), c = read(), d = read(), k = read();
printf("%d\n",
(sol(c - 1, d - 1, k - 1) - sol(c - 1, b - 2, k - 1) -
sol(a - 2, d - 1, k - 1) + sol(a - 2, b - 2, k - 1) + mod + mod) %
mod);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e6 + 5;
const long long mod = 1e9 + 7;
const long long base = 3e18;
pair<long long, long long> dp[33][2][2][2];
long long a[33];
long long b[33];
long long c[33];
pair<long long, long long> f(long long nw, long long x, long long y,
long long k) {
if (nw == -1) return make_pair(1, 0);
if (dp[nw][x][y][k].first != -1) return dp[nw][x][y][k];
pair<long long, long long> h = make_pair(0, 0);
for (int p = 0; p <= (x ? a[nw] : 1); p++) {
for (int t = 0; t <= (y ? b[nw] : 1); t++) {
long long z = p ^ t;
if (k && c[nw] < z) continue;
h.first +=
f(nw - 1, x & (p == a[nw]), y & (t == b[nw]), k & (z == c[nw])).first;
h.second +=
f(nw - 1, x & (p == a[nw]), y & (t == b[nw]), k & (z == c[nw]))
.second;
if (z) {
h.second += (1ll << nw) * f(nw - 1, x & (p == a[nw]), y & (t == b[nw]),
k & (z == c[nw]))
.first;
}
}
}
h.first %= mod;
h.second %= mod;
return dp[nw][x][y][k] = h;
}
long long get(long long x, long long y, long long k) {
if (x < 0 || y < 0) return 0;
memset(dp, -1, sizeof(dp));
long long len = 0;
while (x || y || k) {
a[len] = x % 2;
b[len] = y % 2;
c[len] = k % 2;
len++;
x /= 2;
y /= 2;
k /= 2;
}
len--;
auto p = f(len, 1, 1, 1);
return (p.first + p.second) % mod;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
if (fopen("t.inp", "r")) {
freopen("test.inp", "r", stdin);
freopen("test.out", "w", stdout);
}
long long t;
cin >> t;
while (t--) {
long long x, y, x1, y1, k;
cin >> x >> y >> x1 >> y1 >> k;
k--;
x--;
y--;
x1--;
y1--;
cout << ((get(x1, y1, k) - get(x1, y - 1, k) - get(x - 1, y1, k) +
get(x - 1, y - 1, k)) %
mod +
mod) %
mod
<< "\n";
}
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long sol(long long x, long long y, long long off, long long k) {
if (x > y) swap(x, y);
long long mod = 1000000007;
long long res = 0;
if (x == 0 && y == 0) {
if (off <= k)
return off;
else
return 0;
}
if (x < 0 || k < 0) return 0;
if (off > k) return 0;
long long c = 1;
while (c <= y) c *= 2;
c /= 2;
if (x < c) {
long long res2 = 0;
if (k >= (c + off - 1)) {
res2 = (c * (c - 1) / 2) % mod * (x + 1) % mod;
res2 = (res2 + off * (c * (x + 1) % mod) % mod) % mod;
} else if (k >= off) {
res2 = (k * (k + 1) / 2 - (off - 1) * off / 2) % mod * (x + 1) % mod;
}
long long res3 = sol(x, y - c, c + off, k);
res = (res2 + res3) % mod;
} else {
long long res2 = 0;
if (k >= (c + off - 1)) {
res2 = (c * (c - 1) / 2) % mod * c % mod;
res2 = (res2 + off * (c * c % mod) % mod) % mod;
} else if (k >= off) {
res2 = (k * (k + 1) / 2 - (off - 1) * off / 2) % mod * c % mod;
}
long long len = (x - c + y - c + 2);
long long res3 = 0;
if (k >= (c + c + off - 1)) {
res3 = (c * (c - 1) / 2) % mod * len % mod;
res3 = (res3 + (c + off) * (c * len % mod) % mod) % mod;
} else if (k >= (c + off)) {
res3 = (k * (k + 1) / 2 - (off + c - 1) * (off + c) / 2) % mod * len;
}
long long res4 = sol(x - c, y - c, off, k);
res = (res4 + res2 + res3) % mod;
}
return res;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int q;
cin >> q;
long long mod = 1000000007;
for (int i = 0; i < q; i++) {
long long x0, y0, x1, y1, k;
cin >> x0 >> y0 >> x1 >> y1 >> k;
x0--, y0--, x1--, y1--;
long long res = sol(x1, y1, 1, k) - sol(x1, y0 - 1, 1, k) -
sol(x0 - 1, y1, 1, k) + sol(x0 - 1, y0 - 1, 1, k);
res %= mod;
res = (res + mod) % mod;
cout << res << "\n";
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long solve(long long u, long long v, long long k, long long o = 0) {
if (u > v) swap(u, v);
if (u < 1 || k < 1) return 0;
long long p = 1, res = 0;
while (p * 2 <= v) p *= 2;
long long tmp = min(p, k);
if (u <= p) {
res = ((tmp * (tmp + 1) / 2 % 1000000007LL) * u) % 1000000007LL;
res = (res + (o * tmp % 1000000007LL) * u % 1000000007LL +
solve(u, v - p, k - p, o + p)) %
1000000007LL;
} else {
res = ((tmp * (tmp + 1) / 2 % 1000000007LL) * p) % 1000000007LL;
res = (res + o * tmp % 1000000007LL * p % 1000000007LL) % 1000000007LL;
res = (res + solve(u - p, v - p, k, o) + solve(p, v - p, k - p, o + p) +
solve(u - p, p, k - p, o + p)) %
1000000007LL;
}
return res;
}
int main() {
long long q, a, b, u, v, k;
ios::sync_with_stdio(0);
cin >> q;
while (q--) {
cin >> a >> b >> u >> v >> k;
long long res = (solve(u, v, k) + solve(a - 1, b - 1, k)) % 1000000007LL;
res = (res - solve(a - 1, v, k) - solve(u, b - 1, k) +
1000000007LL * 1000000007LL) %
1000000007LL;
cout << res << endl;
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long calc1(long long x, long long y, long long k) {
if (k < 1 || y == 0 || x == 0) return 0;
if (x > y) swap(x, y);
int u = 63 - __builtin_clzll(y), v = 63 - __builtin_clzll(x);
long long c = (1LL << u), d = (1LL << v);
if (c > d) {
if (k <= c) return k * x % 1000000007;
return c * x % 1000000007 + calc1(x, y - c, k - c);
} else {
long long kk = min(k, c), ans = kk * c % 1000000007;
ans += calc1(x - c, c, k - c);
ans += calc1(y - c, c, k - c);
ans += calc1(x - c, y - c, k);
return ans %= 1000000007;
}
}
long long inv2 = 500000004LL;
long long calc(long long x, long long y, long long k) {
if (k < 1 || y == 0 || x == 0) return 0;
if (x > y) swap(x, y);
int u = 63 - __builtin_clzll(y), v = 63 - __builtin_clzll(x);
long long c = (1LL << u), d = (1LL << v);
if (c > d) {
long long f = min(c, k);
return ((f + 1) * f / 2 % 1000000007 * x % 1000000007 +
calc(x, y - c, k - c) + c * calc1(x, y - c, k - c)) %
1000000007;
} else {
long long f = min(k, c),
ans = (f + 1LL) * f / 2 % 1000000007 * c % 1000000007;
ans += calc(d, y - c, k - c) + c * calc1(d, y - c, k - c) % 1000000007;
ans += calc(c, x - d, k - d) + c * calc1(c, x - d, k - d) % 1000000007;
ans += calc(x - c, y - c, k);
return ans % 1000000007;
}
}
int main() {
int Q;
long long x1, y1, x2, y2, k;
for (scanf("%d", &Q); Q--;) {
scanf("%I64d %I64d %I64d %I64d %I64d", &x1, &y1, &x2, &y2, &k);
long long ans = calc(x2, y2, k);
ans -= calc(x1 - 1, y2, k);
ans -= calc(x2, y1 - 1, k);
ans += calc(x1 - 1, y1 - 1, k);
ans %= 1000000007;
if (ans < 0) ans += 1000000007;
printf("%I64d\n", ans);
}
return 0;
}
| CPP |
809_C. Find a car | After a wonderful evening in the restaurant the time to go home came. Leha as a true gentlemen suggested Noora to give her a lift. Certainly the girl agreed with pleasure. Suddenly one problem appeared: Leha cannot find his car on a huge parking near the restaurant. So he decided to turn to the watchman for help.
Formally the parking can be represented as a matrix 109 × 109. There is exactly one car in every cell of the matrix. All cars have their own machine numbers represented as a positive integer. Let's index the columns of the matrix by integers from 1 to 109 from left to right and the rows by integers from 1 to 109 from top to bottom. By coincidence it turned out, that for every cell (x, y) the number of the car, which stands in this cell, is equal to the minimum positive integer, which can't be found in the cells (i, y) and (x, j), 1 ≤ i < x, 1 ≤ j < y.
<image> The upper left fragment 5 × 5 of the parking
Leha wants to ask the watchman q requests, which can help him to find his car. Every request is represented as five integers x1, y1, x2, y2, k. The watchman have to consider all cells (x, y) of the matrix, such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2, and if the number of the car in cell (x, y) does not exceed k, increase the answer to the request by the number of the car in cell (x, y). For each request Leha asks the watchman to tell him the resulting sum. Due to the fact that the sum can turn out to be quite large, hacker asks to calculate it modulo 109 + 7.
However the requests seem to be impracticable for the watchman. Help the watchman to answer all Leha's requests.
Input
The first line contains one integer q (1 ≤ q ≤ 104) — the number of Leha's requests.
The next q lines contain five integers x1, y1, x2, y2, k (1 ≤ x1 ≤ x2 ≤ 109, 1 ≤ y1 ≤ y2 ≤ 109, 1 ≤ k ≤ 2·109) — parameters of Leha's requests.
Output
Print exactly q lines — in the first line print the answer to the first request, in the second — the answer to the second request and so on.
Example
Input
4
1 1 1 1 1
3 2 5 4 5
1 1 5 5 10000
1 4 2 5 2
Output
1
13
93
0
Note
Let's analyze all the requests. In each case the requested submatrix is highlighted in blue.
In the first request (k = 1) Leha asks only about the upper left parking cell. In this cell the car's number is 1. Consequentally the answer is 1.
<image>
In the second request (k = 5) suitable numbers are 4, 1, 2, 3, 2, 1. Consequentally the answer is 4 + 1 + 2 + 3 + 2 + 1 = 13.
<image>
In the third request (k = 10000) Leha asks about the upper left frament 5 × 5 of the parking. Since k is big enough, the answer is equal to 93.
<image>
In the last request (k = 2) none of the cur's numbers are suitable, so the answer is 0.
<image> | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int Imx = 2147483647;
const int mod = 1000000007;
const long long Lbg = 2e18;
inline long long getnum() {
register long long r = 0;
register bool ng = 0;
register char c;
c = getchar();
while (c != '-' && (c < '0' || c > '9')) c = getchar();
if (c == '-') ng = 1, c = getchar();
while (c >= '0' && c <= '9') r = r * 10 + c - '0', c = getchar();
if (ng) r = -r;
return r;
}
template <class T>
inline void putnum(T x) {
if (x < 0) putchar('-'), x = -x;
register short a[20] = {}, sz = 0;
while (x > 0) a[sz++] = x % 10, x /= 10;
if (sz == 0) putchar('0');
for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
}
inline void putsp() { putchar(' '); }
inline void putendl() { putchar('\n'); }
inline char mygetchar() {
register char c = getchar();
while (c == ' ' || c == '\n') c = getchar();
return c;
}
int q;
long long pw2[111], sum[31];
long long calc(int t, int qx, int qy, int k, int bs = 0) {
if (qx <= 0 || qy <= 0 || k <= 0) return 0;
if (t == 0) return bs + 1;
if (qx > qy) return calc(t, qy, qx, k, bs);
int l = 1 << t, hl = l >> 1;
if (qx == l && qy == l) {
if (k >= l)
return (sum[t] + 1ll * bs * pw2[t * 2]) % mod;
else {
return (calc(t - 1, hl, hl, k, bs) * 2 +
calc(t - 1, hl, hl, k - hl, bs + hl) * 2) %
mod;
}
}
int xu = min(hl, qx), xd = max(0, qx - hl);
int yl = min(hl, qy), yr = max(0, qy - hl);
if (qy == l && k >= l) {
long long ret = 0;
return (calc(t - 1, xu, hl, k, bs) * 2 + calc(t - 1, xd, hl, k, bs) * 2 +
1ll * hl * qx % mod * hl) %
mod;
} else {
long long ret = 0;
ret += calc(t - 1, xd, yr, k, bs);
ret += calc(t - 1, xu, yl, k, bs);
ret += calc(t - 1, xd, yl, k - hl, bs + hl);
ret += calc(t - 1, xu, yr, k - hl, bs + hl);
return ret % mod;
}
}
int main() {
pw2[0] = 1;
for (int i = 1; i <= 100; i++) pw2[i] = pw2[i - 1] * 2 % mod;
sum[0] = 1;
for (int i = 1; i <= 30; i++)
sum[i] = (sum[i - 1] * 4 + pw2[3 * (i - 1) + 1]) % mod;
q = getnum();
while (q--) {
int X1, Y1, X2, Y2, k;
X1 = getnum(), Y1 = getnum(), X2 = getnum();
Y2 = getnum(), k = getnum();
long long ans1 = +calc(30, X2, Y2, k);
long long ans2 = -calc(30, X1 - 1, Y2, k);
long long ans3 = -calc(30, X2, Y1 - 1, k);
long long ans4 = +calc(30, X1 - 1, Y1 - 1, k);
putnum(((ans1 + ans2 + ans3 + ans4) % mod + mod) % mod), putendl();
}
return 0;
}
| CPP |
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