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p02802 AtCoder Beginner Contest 151 - Welcome to AtCoder | Takahashi participated in a contest on AtCoder.
The contest had N problems.
Takahashi made M submissions during the contest.
The i-th submission was made for the p_i-th problem and received the verdict S_i (`AC` or `WA`).
The number of Takahashi's correct answers is the number of problems on which he received an `AC` once or more.
The number of Takahashi's penalties is the sum of the following count for the problems on which he received an `AC` once or more: the number of `WA`s received before receiving an `AC` for the first time on that problem.
Find the numbers of Takahashi's correct answers and penalties.
Constraints
* N, M, and p_i are integers.
* 1 \leq N \leq 10^5
* 0 \leq M \leq 10^5
* 1 \leq p_i \leq N
* S_i is `AC` or `WA`.
Input
Input is given from Standard Input in the following format:
N M
p_1 S_1
:
p_M S_M
Output
Print the number of Takahashi's correct answers and the number of Takahashi's penalties.
Examples
Input
2 5
1 WA
1 AC
2 WA
2 AC
2 WA
Output
2 2
Input
100000 3
7777 AC
7777 AC
7777 AC
Output
1 0
Input
6 0
Output
0 0 | 6 | 0 | import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
sc.nextLine();
Set<Integer> acList = new HashSet<>();
Map<Integer, Integer> waMap = new HashMap<>();
for (int i = 0; i < m; i++){
int q = sc.nextInt();
String r = sc.next();
if(!acList.contains(q)) {
if (r.equals("AC")) {
acList.add(q);
} else {
waMap.put(q, waMap.getOrDefault(q, 0) + 1);
}
}
sc.nextLine();
}
System.out.println(acList.size() + " " + acList.stream().mapToInt(q -> waMap.getOrDefault(q, 0)).sum());
}
}
| JAVA |
p02802 AtCoder Beginner Contest 151 - Welcome to AtCoder | Takahashi participated in a contest on AtCoder.
The contest had N problems.
Takahashi made M submissions during the contest.
The i-th submission was made for the p_i-th problem and received the verdict S_i (`AC` or `WA`).
The number of Takahashi's correct answers is the number of problems on which he received an `AC` once or more.
The number of Takahashi's penalties is the sum of the following count for the problems on which he received an `AC` once or more: the number of `WA`s received before receiving an `AC` for the first time on that problem.
Find the numbers of Takahashi's correct answers and penalties.
Constraints
* N, M, and p_i are integers.
* 1 \leq N \leq 10^5
* 0 \leq M \leq 10^5
* 1 \leq p_i \leq N
* S_i is `AC` or `WA`.
Input
Input is given from Standard Input in the following format:
N M
p_1 S_1
:
p_M S_M
Output
Print the number of Takahashi's correct answers and the number of Takahashi's penalties.
Examples
Input
2 5
1 WA
1 AC
2 WA
2 AC
2 WA
Output
2 2
Input
100000 3
7777 AC
7777 AC
7777 AC
Output
1 0
Input
6 0
Output
0 0 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int n, m, ans, pan;
int vis[maxn], cnt[maxn];
int main() {
scanf("%d%d", &n, &m);
while(m--) {
int p;
char s[10];
scanf("%d%s", &p, s);
if(s[0] == 'A') {
if(!vis[p]) {
vis[p] = 1;
++ans;
pan += cnt[p];
}
}
if(s[0] == 'W') {
++cnt[p];
}
}
printf("%d %d\n", ans, pan);
return 0;
}
| CPP |
p02802 AtCoder Beginner Contest 151 - Welcome to AtCoder | Takahashi participated in a contest on AtCoder.
The contest had N problems.
Takahashi made M submissions during the contest.
The i-th submission was made for the p_i-th problem and received the verdict S_i (`AC` or `WA`).
The number of Takahashi's correct answers is the number of problems on which he received an `AC` once or more.
The number of Takahashi's penalties is the sum of the following count for the problems on which he received an `AC` once or more: the number of `WA`s received before receiving an `AC` for the first time on that problem.
Find the numbers of Takahashi's correct answers and penalties.
Constraints
* N, M, and p_i are integers.
* 1 \leq N \leq 10^5
* 0 \leq M \leq 10^5
* 1 \leq p_i \leq N
* S_i is `AC` or `WA`.
Input
Input is given from Standard Input in the following format:
N M
p_1 S_1
:
p_M S_M
Output
Print the number of Takahashi's correct answers and the number of Takahashi's penalties.
Examples
Input
2 5
1 WA
1 AC
2 WA
2 AC
2 WA
Output
2 2
Input
100000 3
7777 AC
7777 AC
7777 AC
Output
1 0
Input
6 0
Output
0 0 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
long long a,b,c,d,e,n,m,pas1,pas2,g[100009];
bool bo[100009];
string s;
int main(){
ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
cin>>n>>m;
for(b=1; b<=m; b++){
cin>>c>>s;
if(s=="AC"){
if(bo[c]==0){
bo[c]=1;
pas1++;
pas2+=g[c];
}
}else{
g[c]++;
}
}
cout<<pas1<<" "<<pas2;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
#define ll long long
using namespace std;
ll f[64][2][2][2][2];ll l,r;ll ans1=0;int l1[64],r1[64];ll mod=1e9+7;
ll dfs(int pos,int tl,int tr,int ts,int th){
if(pos<0) return 1;ll &ans=f[pos][tl][tr][ts][th];
if(ans) return ans;
for(int i=0;i<=1;i++){
for(int j=0;j<=1;j++){int fs=ts,fl=tl,fr=tr,fh=th;
if(!th&&i)if(i!=j)continue;
if((j&i)!=j) continue;
if(!tl&&l1[pos]>j) continue;
if(!tr&&r1[pos]<i) continue;
if(!ts&&j>i) continue;
if(j<i) fs=1;if(j>l1[pos])fl=1;if(i<r1[pos])fr=1;if(j) fh=1;
ans+=dfs(pos-1,fl,fr,fs,fh);
}
}ans%=mod;
return ans;
}
int main(){
cin>>l>>r;
for(int i=63;i>=0;i--)
{
l1[i]=(bool)((1ll<<i)&l);r1[i]=(bool((1ll<<i)&r));
}
cout<<dfs(63,0,0,0,0);
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import java.io.*;
import java.util.*;
class Solver {
private static final long M = 1_000_000_007L;
private final long l, r;
private long[][][][] dptable;
Solver(long l, long r) {
this.l = l;
this.r = r;
}
private int getIndexFromBoolean(boolean b) {
return b ? 1 : 0;
}
private long getBit(long v, int bitPosition) {
return (v >> bitPosition) & 1L;
}
private long solve(int bitPosition, boolean xAlreadyHigherThanL, boolean yAlreadyLowerThanR, boolean mostSignificantBitAlreadyFlipped) {
if (bitPosition < 0) {
return 1;
}
long cacheValue = dptable[bitPosition][getIndexFromBoolean(xAlreadyHigherThanL)][getIndexFromBoolean(yAlreadyLowerThanR)][getIndexFromBoolean(mostSignificantBitAlreadyFlipped)];
if (cacheValue >= 0) {
return cacheValue;
}
long output = 0L;
// x: 0, y: 0
if (xAlreadyHigherThanL || getBit(l, bitPosition) == 0L) {
output += solve(
bitPosition - 1,
xAlreadyHigherThanL,
yAlreadyLowerThanR || getBit(r, bitPosition) == 1L,
mostSignificantBitAlreadyFlipped);
}
// x: 1, y: 1
if (yAlreadyLowerThanR || getBit(r, bitPosition) == 1L) {
output += solve(
bitPosition - 1,
xAlreadyHigherThanL || getBit(l, bitPosition) == 0L,
yAlreadyLowerThanR,
true);
}
// x: 0, y: 1
if ((xAlreadyHigherThanL || getBit(l, bitPosition) == 0L)
&& (yAlreadyLowerThanR || getBit(r, bitPosition) == 1L)
&& mostSignificantBitAlreadyFlipped) {
output += solve(
bitPosition - 1,
xAlreadyHigherThanL,
yAlreadyLowerThanR,
true);
}
output %= M;
dptable[bitPosition][getIndexFromBoolean(xAlreadyHigherThanL)][getIndexFromBoolean(yAlreadyLowerThanR)][getIndexFromBoolean(mostSignificantBitAlreadyFlipped)] = output;
return output;
}
public long solve() {
dptable = new long[60][2][2][2];
for (long[][][] d1 : dptable) {
for (long[][] d2 : d1) {
for (long[] d3 : d2) {
Arrays.fill(d3, -1);
}
}
}
return solve(59, false, false, false);
}
}
public class Main {
private static void execute(ContestReader reader, PrintWriter out) {
long l = reader.nextLong();
long r = reader.nextLong();
out.println(new Solver(l, r).solve());
}
public static void main(String[] args) {
ContestReader reader = new ContestReader(System.in);
PrintWriter out = new PrintWriter(System.out);
execute(reader, out);
out.flush();
}
}
class ContestReader {
private BufferedReader reader;
private StringTokenizer tokenizer;
ContestReader(InputStream in) {
reader = new BufferedReader(new InputStreamReader(in));
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new java.util.StringTokenizer(reader.readLine());
} catch (Exception e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; i++) {
array[i] = nextInt();
}
return array;
}
public long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; i++) {
array[i] = nextLong();
}
return array;
}
}
class Algorithm {
private static void swap(Object[] list, int a, int b) {
Object tmp = list[a];
list[a] = list[b];
list[b] = tmp;
}
public static <T extends Comparable<? super T>> boolean nextPermutation(T[] ts) {
int rightMostAscendingOrderIndex = ts.length - 2;
while (rightMostAscendingOrderIndex >= 0 &&
ts[rightMostAscendingOrderIndex].compareTo(ts[rightMostAscendingOrderIndex + 1]) >= 0) {
rightMostAscendingOrderIndex--;
}
if (rightMostAscendingOrderIndex < 0) {
return false;
}
int rightMostGreatorIndex = ts.length - 1;
while (ts[rightMostAscendingOrderIndex].compareTo(ts[rightMostGreatorIndex]) >= 0) {
rightMostGreatorIndex--;
}
swap(ts, rightMostAscendingOrderIndex, rightMostGreatorIndex);
for (int i = 0; i < (ts.length - rightMostAscendingOrderIndex - 1) / 2; i++) {
swap(ts, rightMostAscendingOrderIndex + 1 + i, ts.length - 1 - i);
}
return true;
}
public static void shuffle(int[] array) {
Random random = new Random();
int n = array.length;
for (int i = 0; i < n; i++) {
int randomIndex = i + random.nextInt(n - i);
int temp = array[i];
array[i] = array[randomIndex];
array[randomIndex] = temp;
}
}
public static void shuffle(long[] array) {
Random random = new Random();
int n = array.length;
for (int i = 0; i < n; i++) {
int randomIndex = i + random.nextInt(n - i);
long temp = array[i];
array[i] = array[randomIndex];
array[randomIndex] = temp;
}
}
public static void sort(int[] array) {
shuffle(array);
Arrays.sort(array);
}
public static void sort(long[] array) {
shuffle(array);
Arrays.sort(array);
}
}
| JAVA |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
#define ll long long
using namespace std;
const int P=1e9+7;
ll L,R;
int num[71],len,mi[71];
int dp[71][2][2][2];
int dfs(int x,bool fl1,bool fl2,bool fl3){
if(x==0)return fl1;
if(dp[x][fl1][fl2][fl3]!=-1)return dp[x][fl1][fl2][fl3];
int up=fl2?1:num[x],res=0;
if(fl3||mi[x]==0)res=(res+dfs(x-1,fl1,num[x]|fl2,fl3))%P;
if(up){
res=(res+dfs(x-1,1,fl2,fl3|(mi[x]^1)))%P;
if(fl1&&(fl3||mi[x]==0))res=(res+dfs(x-1,1,fl2,fl3))%P;
}
dp[x][fl1][fl2][fl3]=res;
return res;
}
int Solve(ll x,ll y){
while(x){
num[++len]=x&1;
x>>=1;
}
int tot=0;
while(y){
mi[++tot]=y&1;
y>>=1;
}
return dfs(len,0,0,0);
}
int main(){
memset(dp,-1,sizeof(dp));
scanf("%lld%lld",&L,&R);
printf("%d\n",Solve(R,L));
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | def i1():
return int(input())
def i2():
return [int(i) for i in input().split()]
[l,r]=i2()
z=10**9+7
dp=[[[[0 for l in range(2)]for k in range(2)]for j in range(2)]for i in range(61)]
dp[60][0][0][0]=1
for n in range(60)[::-1]:
lb=l>>n&1
rb=r>>n&1
for x in range(2):
for y in range(2):
for i in range(2):
for j in range(2):
for k in range(2):
ni=i+0
nj=j+0
nk=k+0
if x>y:
continue
if k==0 and x!=y:
continue
if x&y:
nk=1
if i==0 and x<lb:
continue
if x>lb:
ni=1
if j==0 and y>rb:
continue
if y<rb:
nj=1
dp[n][ni][nj][nk]+=dp[n+1][i][j][k]
dp[n][ni][nj][nk]%=z
print((dp[0][1][1][1]+dp[0][0][1][1]+dp[0][1][0][1]+dp[0][0][0][1])%z) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007ll;
const int N_DIGITS = 60;
ll L[N_DIGITS], R[N_DIGITS];
ll Dp[1 + N_DIGITS][2][2][2]; // これまでで L<x, y<R, x,y>0 が 不成立/成立
const bool Bs[2] = {false, true};
void go(ll &tgt, ll src) { (tgt += src) %= MOD; }
int main() {
ll l, r;
cin >> l >> r;
for (int i = N_DIGITS - 1; i >= 0; --i) {
L[i] = l % 2;
R[i] = r % 2;
l /= 2;
r /= 2;
}
Dp[0][0][0][0] = 1;
for (int i = 0; i < N_DIGITS; ++i) {
for (bool overL : Bs) {
for (bool underR : Bs) {
for (bool some : Bs) {
ll src = Dp[i][overL][underR][some];
// 0,0
if (overL || L[i] == 0) {
go(Dp[i + 1][overL][underR || R[i] == 1][some], src);
}
// 0,1
if ((overL || L[i] == 0) && (underR || R[i] == 1) && some) {
go(Dp[i + 1][overL][underR][true], src);
}
// 1,1
if (underR || R[i] == 1) {
go(Dp[i + 1][overL || L[i] == 0][underR][true], src);
}
}
}
}
}
ll res = 0;
for (bool overL : Bs) {
for (bool underR : Bs) {
for (bool some : Bs) {
(res += Dp[N_DIGITS][overL][underR][some]) %= MOD;
}
}
}
cout << res << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | /*input
*/
#include <bits/stdc++.h>
#define up(i,a,b) for(int (i) = (a);(i)<=(b);++(i))
#define down(i,b,a) for(int (i) = (b);i>=(a);--i)
#define debug(x) cerr << (x) << '\n';
#define bits(x,i) ((x >> i) & 1)
#define mid ((l+r)/2)
#define pr pair<int,int>
using namespace std;
const int logA = 60;
const int mod = 1e9 + 7;
long long dp[logA][2][2][2];
long long l,r;
long long solve(int pos,int isLow, int isHigh,int firstBit){
if (pos < 0) return 1;
long long& res = dp[pos][isLow][isHigh][firstBit];
if (res != -1) return res;
res = 0;
int low = (isLow ? 0 : bits(l, pos)), high = (isHigh ? 1 : bits(r, pos));
for(int choice = low; choice <= high;++choice){
res +=
solve(pos - 1,isLow | (choice != low),isHigh | (choice != high), firstBit | choice == 1);
if (choice == 1 && low == 0 && firstBit == 1)
res += solve(pos-1, isLow, isHigh,1);
res %= mod;
}
return res;
}
int main(){
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> l >> r;
memset(dp, -1, sizeof(dp));
cout << solve(59, 0,0,0) << '\n';
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;
ll P = 1000000007;
ll L, R;
ll memo[64][2][2][2] = {0};
ll f(int k, int m, int l, int r) {
if (k < 0) return m == 1;
if (memo[k][m][l][r] >= 0) return memo[k][m][l][r];
ll ret = 0;
for (int li = 0; li < 2; li++)
for (int ri = 0; ri < 2; ri++) {
if (li == 0 && l && (L & (1LL << k)) > 0) continue;
if (ri == 1 && r && (R & (1LL << k)) == 0) continue;
if (li > ri) continue;
if (m == 0 && li != ri) continue;
ret += f(k - 1, m || (li == 1 && ri == 1), l && ((L >> k) & 1) == li, r && ((R >> k) & 1) == ri);
}
return memo[k][m][l][r] = ret % P;
}
int main() {
cin.tie(nullptr);
ios::sync_with_stdio(false);
cin >> L >> R;
for (int i = 0; i < 64; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
for (int l = 0; l < 2; l++)
memo[i][j][k][l] = -1;
cout << f(61, 0, 1, 1) << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
using ll = long long;
using namespace std;
ll f[61][2][2][2];
const int M=1000000007;
int main() {
ll L, R;
cin >> L >> R;
/*
i 从左往右前 i 位 ，
j 是否>L 1是， 0表示==L
k 是否<R 1是， 0表示==R
x L的第i位
y R的第i位
//l 0 表示 小的数==大的数 1表示 小的数<大的数
m 0 表示 第一个1 填了没有。
d 表示 小的数填d
e 表示 大的数填e */
f[60][0][0][0]=1;
for (int i=59; i>=0; i--)
rep(j,2) rep(k,2) rep(m,2) {
int x=(L>>i)&1;
int y=(R>>i)&1;
int low=j?0:x;
int high=k?1:y;
for (int d=low;d<=high; d++)
for (int e=d;e<=high; e++) {
if (!m && (d^e)) continue;
ll& t=f[i][j||d>low][k||e<high][m||d];
t = (t+f[i+1][j][k][m])%M;
}
}
ll ans=0;
rep(j,2) rep(k,2) ans=(ans+f[0][j][k][1])%M;
cout<<ans<<endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (n); ++i)
typedef long long ll;
typedef pair<ll, ll> P;
const int MAX = 1e5 + 10;
const int INF = 1e9;
const int MOD = 1e9 + 7;
ll L, R;
ll dp[61][2][2][2];
int main(){
cin >> L >> R;
dp[60][0][0][0] = 1;
for(int i = 59; i >= 0; --i){
int r = R >> i & 1;
int l = L >> i & 1;
rep(j, 2)rep(k, 2)rep(s, 2){
rep(x, 2)rep(y, 2){
if(!y && x) continue;
int nj, nk, ns;
nj = j, nk = k, ns = s;
// s:既に最上位ビットを通過したか
if(!s && y && !x) continue;
if(x && y) ns = 1;
// j:L <= x を既に満たしているか
if(!j && l && !x) continue;
if(!l && x) nj = 1;
// k:y <= R を既に満たしているか
if(!k && !r && y) continue;
if(r && !y) nk = 1;
dp[i][nj][nk][ns] += dp[i + 1][j][k][s];
dp[i][nj][nk][ns] %= MOD;
}
}
}
ll ans = 0;
rep(j, 2)rep(k, 2)rep(s, 2){
ans += dp[0][j][k][s];
ans %= MOD;
}
cout << ans << endl;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
long long int dp[61][2][2][2];// 10-18 < 2^60
int main(){
long long int l, r;
cin >> l >> r;
dp[60][0][0][0] = 1;
for(int i=59; i>=0; i--){// digit DP
for(int j=0; j<2; j++){
for(int k=0; k<2; k++){
for(int s=0; s<2; s++){
for(int nx=0; nx<2; nx++){
for(int ny=0; ny<2; ny++){
if(nx == 1 && ny == 0) continue;
int nj = j, nk = k, ns = s;
// s : MSB
if(s == 0 && nx != ny) continue;
if(nx == 1 && ny == 1) ns = 1;
// j : L <= x
if(j == 0 && nx == 0 && ((l>>i) & 1)) continue;// x < L
if(nx == 1 && !((l>>i) & 1)) nj = 1;
// k : y <= R
if(k == 0 && ny == 1 && !((r>>i) & 1)) continue;// y > R
if(ny == 0 && ((r>>i) & 1)) nk = 1;
dp[i][nj][nk][ns] += dp[i+1][j][k][s];
dp[i][nj][nk][ns] %= MOD;
}
}
}
}
}
}
long long int ans = 0;
for(int j=0; j<2; j++){
for(int k=0; k<2; k++){
for(int s=0; s<2; s++){
ans += dp[0][j][k][s];
ans %= MOD;
}
}
}
cout << ans << endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | mod = 10**9+7
def count(L, R):
res = 0
for j in range(70):
res = (res + subcount(j, L, R)) % mod
return res
def subcount(j, L, R):
if j >= L.bit_length():
return 0
dp = [[0]*4 for _ in range(70)]
cnt = 0
if R.bit_length() != j+1:
cnt += 1
if L.bit_length() != j+1:
cnt += 2
dp[j][cnt] = 1
for i in range(j, 0, -1):
yf = R & (1<<(i-1))
xf = L & (1<<(i-1))
d0 = dp[i][0]
if yf and xf:
dp[i-1][0] += d0
dp[i-1][2] += d0
dp[i-1][3] += d0
elif yf:
dp[i-1][0] += d0
dp[i-1][1] += d0
elif xf:
dp[i-1][2] += d0
else:
dp[i-1][0] += d0
d1 = dp[i][1]
if xf:
dp[i-1][1] += d1
dp[i-1][3] += 2*d1
else:
dp[i-1][1] += 2*d1
d2 = dp[i][2]
if yf:
dp[i-1][2] += 2*d2
dp[i-1][3] += d2
else:
dp[i-1][2] += d2
d3 = dp[i][3]
dp[i-1][3] += 3*d3
dp[i-1][0] %= mod
dp[i-1][1] %= mod
dp[i-1][2] %= mod
dp[i-1][3] %= mod
return sum(dp[0][i] for i in range(4))
L, R = map(int, input().split())
print((count(R, R) - count(L-1, R))%mod)
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
ll L,R;
ll f[64][2][2][2][2];
int l[100],r[100];
ll dfs(int x,int fl,int fr,int s,int ht)
{
if(x<0)return 1;
// x：第x位 （二进制转十进制下权值是1的为第0位）
// fl： y之前是否有填过小于 R的某一位的数字 例如：R:11000 y:10??? fl即为1 表示后面数字可以往大了填
// fr： x之前是否有填过大于 L的某一位的数字 类似fl
// s： 之前x是否有某一位 < y的某一位 如果有，后面的位数就没有相对大小限制了 (确保 x<=y)
//ht：确保 x，y 最高位一样都为1 （满足推导的要求）
ll &ans=f[x][fl][fr][s][ht];
if(ans) return ans;
for(int i=0;i<2;i++)//y
{
for(int j=0;j<2;j++)//x
{
int tl=fl,tr=fr,ts=s,tt=ht;
if(!tt&&i&&i!=j)continue;
if((j&i)!=j)continue;
if(!tl&&j<l[x])continue;
if(!tr&&i>r[x])continue;
if(!ts&&j>i)continue;
if(j>l[x])tl=1;if(i<r[x])tr=1;if(j<i)ts=1;if(j)tt=1;
ans+=dfs(x-1,tl,tr,ts,tt);
}
}
ans%=mod;
return ans;
}
int main()
{
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
cin>>L>>R;
for(int i=63;i>=0;i--)
{
l[i]=(bool)((1ll<<i)&L);r[i]=(bool)((1ll<<i)&R);
}
cout<<dfs(63,0,0,0,0)<<endl;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }
const int MOD = TEN(9) + 7;
inline void add(int &x, int y) { x += y; if (x >= MOD) x -= MOD; }
const int maxn = 62;
int dp[maxn][2][2][2]; //L<x, y<R, highest (x <= y can be ignored)
int main() {
ll L, R; cin >> L >> R;
--L; ++R;
int now = 0;
dp[0][0][0][0] = 1;
for (int i = 60; i >= 0; --i) {
int p = (L >> i) & 1;
int q = (R >> i) & 1;
rep(a, 2) rep(b, 2) rep(c, 2) {
if (dp[now][a][b][c] == 0) continue;
rep(r, 2) rep(s, 2) { //for x, y
int na = a, nb = b, nc = c;
if (p == 1 && r == 0 && !a) continue;
if (p == 0 && r == 1) na = 1;
if (s == 1 && q == 0 && !b) continue;
if (s == 0 && q == 1) nb = 1;
if (!c && r + s == 1) continue;
if (r || s) nc = 1;
if (r > s) continue;
add(dp[now + 1][na][nb][nc], dp[now][a][b][c]);
}
}
now++;
}
cout << dp[now][1][1][1] << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define int long long
#define inf 1000000007
#define LINF 100000000000000007LL
#define ll long long
using namespace std;
int dp[62][2][2]; // bit, r=y? l=x? allzero?
signed main(){
int l,r;
cin>>l>>r;
bool fl = false, fr = false, ffl = false, ffr = false;
for(int i=60;i>=0;i--){
int br = ( r >> i ) % 2;
int bl = ( l >> i ) % 2;
if( !fl && bl ){ fl = true; ffl = true; }
if( !fr && br ){ fr = true; ffr = true; }
if( fr && ( !fl || ffl ) ){
dp[i][ffr][ffl]++;
}
if( br == 0 && bl == 0){
dp[i][0][0] += 3*dp[i+1][0][0] + dp[i+1][0][1];
dp[i][1][0] += dp[i+1][1][0];
dp[i][0][1] += 2*dp[i+1][0][1];
dp[i][1][1] += dp[i+1][1][1];
}
if( br == 1 && bl == 0){
dp[i][0][0] += 3*dp[i+1][0][0] + dp[i+1][0][1] + dp[i+1][1][0];
dp[i][1][0] += 2*dp[i+1][1][0] + dp[i+1][1][1];
dp[i][0][1] += 2*dp[i+1][0][1] + dp[i+1][1][1];
dp[i][1][1] += dp[i+1][1][1];
}
if( br == 0 && bl == 1){
dp[i][0][0] += 3*dp[i+1][0][0];
dp[i][1][0] += dp[i+1][1][0];
dp[i][0][1] += dp[i+1][0][1];
}
if( br == 1 && bl == 1){
dp[i][0][0] += 3*dp[i+1][0][0] + dp[i+1][1][0];
dp[i][1][0] += 2*dp[i+1][1][0];
dp[i][0][1] += dp[i+1][0][1];
dp[i][1][1] += dp[i+1][1][1];
}
dp[i][0][0] %=inf; dp[i][0][1]%=inf; dp[i][1][0] %=inf; dp[i][1][1]%=inf;
ffl = false; ffr = false;
}
int ans = dp[0][0][0]+dp[0][1][0]+dp[0][0][1]+dp[0][1][1];
cout<<ans%inf<<endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define next Next
#define int long long
const int mod=1e9+7;
int dp[65][2][2][2];
/*char buf[1<<21],*p1=buf,*p2=buf;
inline int gc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}*/
#define gc getchar
inline int read()
{
int ret=0,f=0;char c=gc();
while(!isdigit(c)){if(c=='-')f=1;c=gc();}
while(isdigit(c)){ret=ret*10+c-48;c=gc();}
if(f)return -ret;return ret;
}
int solve(int l,int r,int len,int lim1,int lim2,int zero)
{
if(len<0)return 1;
int &ans=dp[len][lim1][lim2][zero];
if(ans!=-1)return ans;
ans=0;
int up2=lim2?((r>>len)&1):1;
int up1=lim1?((l>>len)&1):0;
for(int i=up1;i<=1;i++)
for(int j=0;j<=up2;j++)
{
if (i>j)continue;
if(zero&&i!=j)continue;
(ans+=solve(l,r,len-1,lim1&&i==up1,lim2&&j==up2,zero&&j==0))%=mod;
}
return ans;
}
signed main()
{
memset(dp,-1,sizeof(dp));
int L=read(),R=read();
cout<<solve(L,R,60,1,1,1);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
long long L, R;
vector<bool> l, r;
const long long mod = 1000000007;
vector<vector<vector<vector<long long>>>> memo(60,
vector<vector<vector<long long>>>(2,
vector<vector<long long>>(2,
vector<long long>(2, -1))));
long long f(int pos, bool flagX, bool flagY, bool flagZ) {
if (pos == -1) return 1;
if (memo.at(pos).at(flagX).at(flagY).at(flagZ) != -1)
return memo.at(pos).at(flagX).at(flagY).at(flagZ);
long long ret = 0;
if (flagX || !l.at(pos)) {
if (r.at(pos))
ret += f(pos - 1, flagX, 1, flagZ);
else
ret += f(pos - 1, flagX, flagY, flagZ);
}
if ((flagX || !l.at(pos)) && (flagY || r.at(pos)) && flagZ) {
ret += f(pos - 1, flagX, flagY, flagZ);
}
if (flagY || r.at(pos)) {
if (l.at(pos))
ret += f(pos - 1, flagX, flagY, 1);
else
ret += f(pos - 1, 1, flagY, 1);
}
ret %= mod;
memo.at(pos).at(flagX).at(flagY).at(flagZ) = ret;
return ret;
}
int main() {
cin >> L >> R;
for (int i = 59; i >= 0; --i) {
if (L & ((long long)1 << i))
l.push_back(1);
else
l.push_back(0);
if (R & ((long long)1 << i))
r.push_back(1);
else
r.push_back(0);
}
reverse(l.begin(), l.end());
reverse(r.begin(), r.end());
long long ans = f(59, 0, 0, 0);
cout << ans << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 |
import bisect
import collections
import itertools
def getint(): return int(input())
def getints(): return list(map(int, input().split()))
def getint2d(rows): return [getints() for _ in range(rows)]
def getgrid(rows): return [input() for _ in range(rows)]
def array1d(n, value): return [value for _ in range(n)]
def array2d(n, m, value): return [array1d(m, value) for _ in range(n)]
min_val,max_val=getints()
def get_key(pos, is_small, is_large, found):
return ((pos * 2 + is_small) * 2 + is_large) * 2 + found
mod = 10**9 + 7
cache = [-1] * get_key(64,1,1,1)
def solve(pos, is_small, is_large, found):
if pos < 0:
return 1
key = get_key(pos,is_small,is_large,found)
res = cache[key]
if res >= 0:
return res
res = 0
for x in [0,1]:
for y in [0,1]:
if x == 1 and y == 0: continue
if not is_small and x == 0 and (min_val >> pos & 1) == 1:
continue
if not is_large and y == 1 and (max_val >> pos & 1) == 0:
continue
if not found and x != y:
continue
new_is_small = True if x > (min_val >> pos & 1) else is_small
new_is_large = True if y < (max_val >> pos & 1) else is_large
new_found = True if x == y and x == 1 else found
res += solve(pos-1, new_is_small, new_is_large, new_found)
res = res % mod
cache[key] = res
return res
print(solve(63, False, False, False)) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#include "bits/stdc++.h"
using namespace std;
#define pb push_back
#define F first
#define S second
#define f(i,a,b) for(int i = a; i < b; i++)
using ll = long long;
using db = long double;
using ii = pair<int, int>;
const int N = 1e6 + 5, LG = 19, MOD = 1e9 + 7;
const int SQ =225;
const long double EPS = 1e-7;
ll XL, XR;
ll L, R;
int dp[66][2][2];
int solve(int i, bool f2, bool f3){
if(i < 0)
return 1;
int &ret = dp[i][f2][f3];
if(~ret)
return ret;
ret = 0;
int val = (XR >> i & 1);
int val2 = (XL >> i & 1);
f(a,0,2)
f(b,a,2){
if(a <= val2 || f3){
if(b <= val || f2){
ret += solve(i - 1, f2 | (b < val), f3 | (a < val2));
if(ret >= MOD)ret -= MOD;
}
}
}
return ret;
}
int solve(ll x1,ll x2){
memset(dp, -1, sizeof dp);
XL = x1;
XR = x2;
bool ok = false;
bool ok2 = false;
int ret = 0;
for(int i = 59; i >= 0; --i){
if(XL >> i & 1 || ok){
ret += solve(i-1,ok2,ok);
if(ret >= MOD)ret -= MOD;
}
ok |= (XL >> i & 1);
ok2 |= (XR >> i & 1);
}
return ret;
}
int32_t main(){
#ifdef ONLINE_JUDGE
ios_base::sync_with_stdio(0);
cin.tie(0);
#endif
// cout << solve(R,R) << '\n';
cin >> L >> R;
cout << (solve(R,R) - solve(L-1,R) + MOD) % MOD << '\n';
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | def f_coincidence(MOD=10 ** 9 + 7):
# 参考: https://atcoder.jp/contests/abc138/submissions/7013926
from itertools import product
L, R = [int(i) for i in input().split()]
dp = [[[[0 for s in range(2)] for k in range(2)]
for j in range(2)] for i in range(61)]
dp[60][0][0][0] = 1
for i in range(59, -1, -1):
lb = (L >> i) & 1 # 'b' はビット値であることを示す(以下も同じ)
rb = (R >> i) & 1
r2 = range(2)
for j, k, s in product(r2, r2, r2):
pre = dp[i + 1][j][k][s]
# ビットごとに値を試す
for xb, yb in product(r2, r2):
# editorial の通り、このビットの組は元の条件を満たさない
if xb == 1 and yb == 0:
continue
# editorial の通り、これ以降は x <= y が満たされる
nj, nk, ns = j, k, s
# 今注目している桁でMSBが立つか？
if s == 0 and xb != yb:
continue # どちらかが立っていない
if s == 0 and xb == 1 and yb == 1:
ns = 1 # ここで初めてMSBが立った
# L <= x を満たすか？
if j == 0 and xb == 0 and lb == 1:
# jが0のまま xb<lb となれば、L<=xになりようがない
# jが1なら、xbとlbがどうなっていようがL<=xである
continue
if j == 0 and xb == 1 and lb == 0:
nj = 1 # この桁で L<=x が確定する
# y <= R を満たすか？
if k == 0 and yb == 1 and rb == 0:
continue # L<=x の場合分けと同じこと
if k == 0 and yb == 0 and rb == 1:
nk = 1
# 反映
dp[i][nj][nk][ns] += pre
dp[i][nj][nk][ns] %= MOD
ans = sum([dp[0][j][k][s] for j in range(2) for k in range(2)
for s in range(2)]) % MOD
return ans
print(f_coincidence()) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | def examA():
N = DI()/dec(7)
ans = N
print(N)
return
def examB():
ans = 0
print(ans)
return
def examC():
ans = 0
print(ans)
return
def examD():
ans = 0
print(ans)
return
def examE():
ans = 0
print(ans)
return
def examF():
def bitdp(l, r):
n = r.bit_length()
dp = defaultdict(int)
dp[0, 0, 0, 0] = 1
for i, less, greater, start in itertools.product(range(n), (0, 1), (0, 1), (0, 1)):
R_ = 1 if less else (r>>(n-i-1))&1
L_ = 0 if greater else (l>>(n-i-1))&1
for y in range(R_ + 1):
for x in range(L_, y + 1):
less_ = less or y < R_
greater_ = greater or L_ < x
start_ = start or (y==1 and x==1)
if not start and (y==1 and x==0):
# xorがあまりを確実に上回る
continue
dp[i + 1, less_, greater_, start_] += dp[i, less, greater, start]
dp[i + 1, less_, greater_, start_] %= mod
#print(dp[i + 1, less_, greater_, start_],i+1,less_,greater,start_,start)
#for i in range(n):
# print(dp[i,0,0,0],dp[i,1,0,0],dp[i,0,1,0],dp[i,1,1,0])
res = sum(dp[n, less, greater, 1] for less, greater in itertools.product((0, 1), (0, 1)))
return res
L, R = LI()
ans = bitdp(L,R) % mod
print(ans)
return
from decimal import getcontext,Decimal as dec
import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = dec("0.000000000001")
alphabet = [chr(ord('a') + i) for i in range(26)]
alphabet_convert = {chr(ord('a') + i): i for i in range(26)}
getcontext().prec = 28
sys.setrecursionlimit(10**7)
if __name__ == '__main__':
examF()
"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
""" | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | from functools import lru_cache
import sys
sys.setrecursionlimit(1000000)
P = 10**9+7
@lru_cache(maxsize=None)
def subcalc(l, r):
if l < 0 or r < 0:
print("ERROR")
print("l, r =", l, r)
print(1//0)
if l > r: return 0
if r == 0: return 1
aa, bb = l.bit_length(), r.bit_length()
if aa == bb:
return subcalc(l-(1<<aa-1), r-(1<<bb-1))
if (r & (r+1) == 0) and (l == 0):
return pow(3, r.bit_length(), P)
t = (subcalc(l, r-(1<<bb-1)) + subcalc(l, (1<<bb-1)-1) + subcalc(0, r-(1<<bb-1))) % P
# print("subcalc", l, r, t)
return t
@lru_cache(maxsize=None)
def calc(L, R):
if L < 0 or R < 0:
print("ERROR")
print("l, r =", l, r)
print(1//0)
if L > R: return 0
a = L.bit_length()
b = R.bit_length()
if b > a:
t = (calc(L, (1<<b-1)-1) + calc(1<<b-1, R)) % P
# print("calc", L, R, t)
return t
a = 1 << L.bit_length() - 1 if L else 0
t = subcalc(L-a, R-a)
# print("calc", L, R, t)
return t
L, R = map(int, input().split())
print(calc(L, R)) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define REP(i,n) for(ll (i) = (0);(i) < (n);++i)
#define MOD 1000000007
typedef long long ll;
ll dp[63][1<<3];
void mod_add(ll &a, ll b){
a += b;
a %= MOD;
}
int main(){
ll l, r;cin >> l >> r;
dp[0][0] = 1;
REP(i, 62){
REP(j, (1LL<<3)){
REP(x, 2)REP(y, 2){
if(x == 1 && y == 0)continue;
int nj = j;
if(!(nj & 1)){
int tmp = ((l & (1LL << (61-i))) != 0);
if(x < tmp)continue;
else if(x > tmp)nj |= 1;
}
if(!(nj & 2)){
int tmp = ((r & (1LL << (61-i))) != 0);
if(y > tmp)continue;
else if(y < tmp)nj |= 2;
}
if(!(nj & 4)){
if(x + y != 0 && x + y != 2)continue;
if(x + y == 2)nj |= 4;
}
mod_add(dp[i+1][nj], dp[i][j]);
}
}
}
ll ans = 0;
REP(i, 8)mod_add(ans, dp[62][i]);
cout << ans << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | def p_f():
from itertools import product
L, R = map(int, input().split())
mod = 10 ** 9 + 7
# dp[i][j][k][l]: iは桁, jはL<=xか, kはy<=Rか, lは桁数が同じか
dp = [[[[0] * 2 for _ in range(2)] for _ in range(2)] for _ in range(61)]
dp[60][0][0][0] = 1
for i in reversed(range(60)):
# LとRのiビット目を取り出す
lb = L >> i & 1
rb = R >> i & 1
for j, k, l in product(range(2), repeat=3):
# 前の状態
pre = dp[i + 1][j][k][l]
for x, y in product(range(2), repeat=2):
nj, nk, nl = j, k, l
if x and (not y): # xが1でyが0
continue
if (not l) and x != y:
continue
if x and y:
nl = 1
# j: L <= x
if (not j) and (not x) and lb:
continue
if x and (not lb):
nj = 1
# k: y <= R
if (not k) and y and (not rb):
continue
if (not y) and rb:
nk = 1
dp[i][nj][nk][nl] += pre
ans = 0
for j, k, l in product(range(2), repeat=3):
ans += dp[0][j][k][l]
ans %= mod
print(ans)
if __name__ == '__main__':
p_f()
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import sys
sys.setrecursionlimit(2147483647)
INF=float("inf")
input=lambda :sys.stdin.buffer.readline().rstrip()
class ModInt(object):
__MOD=10**9+7
def __init__(self,x):
self.__x=x%self.__MOD
def __repr__(self):
return str(self.__x)
def __add__(self,other):
return ModInt(self.__x+other.__x) if isinstance(other,ModInt) else ModInt(self.__x+other)
def __sub__(self,other):
return ModInt(self.__x-other.__x) if isinstance(other,ModInt) else ModInt(self.__x-other)
def __mul__(self,other):
return ModInt(self.__x*other.__x) if isinstance(other,ModInt) else ModInt(self.__x*other)
def __truediv__(self,other):
return ModInt(self.__x*pow(other.__x,self.__MOD-2,self.__MOD)) if isinstance(other, ModInt) else ModInt(self.__x*pow(other, self.__MOD-2,self.__MOD))
def __pow__(self,other):
return ModInt(pow(self.__x,other.__x,self.__MOD)) if isinstance(other,ModInt) else ModInt(pow(self.__x,other,self.__MOD))
__radd__=__add__
def __rsub__(self,other):
return ModInt(other.__x-self.__x) if isinstance(other,ModInt) else ModInt(other-self.__x)
__rmul__=__mul__
def __rtruediv__(self,other):
return ModInt(other.__x*pow(self.__x,self.__MOD-2,self.__MOD)) if isinstance(other,ModInt) else ModInt(other*pow(self.__x,self.__MOD-2,self.__MOD))
def __rpow__(self,other):
return ModInt(pow(other.__x,self.__x,self.__MOD)) if isinstance(other,ModInt) else ModInt(pow(other,self.__x,self.__MOD))
def resolve():
from itertools import product
L,R=map(int,input().split())
D=R.bit_length()
dp=[[[[ModInt(0)]*2 for _ in range(2)] for _ in range(2)] for _ in range(D+1)]
dp[D][0][0][0]=ModInt(1)
for d in range(D-1,-1,-1):
lb=L>>d&1; rb=R>>d&1
for i,j,m,x,y in product([0,1],repeat=5):
ni,nj,nm=i,j,m
if(x>y): continue
# i:L<=X
if(i==0 and lb>x): continue
if(lb<x): ni=1
# j:Y<=R
if(j==0 and y>rb): continue
if(y<rb): nj=1
# m:MSB
if(m==0 and x!=y): continue
if(x==1 and y==1): nm=1
dp[d][ni][nj][nm]+=dp[d+1][i][j][m];
print(sum(dp[0][i][j][m] for i,j,m in product([0,1],repeat=3)))
resolve() | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using P = pair<int, int>;
using PL = pair<lint, lint>;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
#define ALL(a) (a).begin(),(a).end()
constexpr int MOD = 1000000007;
constexpr int INF = 2147483647;
void yes(bool expr) {cout << (expr ? "Yes" : "No") << "\n";}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
lint L, R;
cin >> L >> R;
lint ans = 0;
REP(x, 60) {
vector<vector<vector<lint>>> dp(61, vector<vector<lint>>(2, vector<lint>(2)));
dp[60][0][0] = 1;
IREP(k, 60) {
lint l = L>>k&1;
lint r = R>>k&1;
REP(i, 2) REP(j, 2) {
//00
if(k != x) {
if(l == 0 || i == 1) {
if(r == 0) dp[k][i][j] = (dp[k][i][j] + dp[k+1][i][j]) % MOD;
else dp[k][i][min(1, j+1)] = (dp[k][i][min(1, j+1)] + dp[k+1][i][j]) % MOD;
}
}
if(k > x) continue;
//10
if(k != x) {
if((l == 0 || i == 1) && (r == 1 || j == 1)) {
dp[k][i][j] = (dp[k][i][j] + dp[k+1][i][j]) % MOD;
}
}
//11
if(r == 1 || j == 1) {
if(l == 1) dp[k][i][j] = (dp[k][i][j] + dp[k+1][i][j]) % MOD;
else dp[k][min(1, i+1)][j] = (dp[k][min(1, i+1)][j] + dp[k+1][i][j]) % MOD;
}
}
}
REP(i, 2) REP(j, 2) ans = (ans + dp[0][i][j]) % MOD;
}
cout << ans << endl;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #!/usr/bin/env python
from collections import deque, defaultdict
import itertools as ite
import sys
import math
import decimal
sys.setrecursionlimit(1000000)
INF = 10 ** 18
MOD = 10 ** 9 + 7
L_, R_ = map(int, raw_input().split())
ans = 0
for MSB in range(70):
S = 2 ** MSB
if S * 2 <= L_ or R_ < S:
continue
L, R = L_, R_
if L <= S:
L = S
if 2 * S - 1 <= R:
R = 2 * S - 1
n1 = nL = nR = 0
nLR = 1
SL = SR = 1
for i in range(MSB)[::-1]:
flag1 = ((1 << i) & L != 0)
flag2 = ((1 << i) & R != 0)
SL *= 2
SR *= 2
SL += flag1
SR += flag2
n1 *= 3
if flag1 and not flag2:
nLR = 0
if SR - SL >= 2:
n1 += SR / 2 - SL / 2 - 1
if not flag1:
n1 += nL
if flag2:
n1 += nR
if not flag1:
nL *= 2
nL += 1
if flag2:
nR *= 2
nR += 1
if not flag1 and flag2:
nL += nLR
nR += nLR
if L == R:
ans += 1
else:
ans += n1 + (SR - SL + 1) + nL + nR + nLR
print ans % MOD
| PYTHON |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | MOD = 10**9 + 7
class mint:
def __init__(self, i):
self.i = i
def __add__(self, m):
t = self.i + (m.i if isinstance(m, mint) else m)
if t > MOD:
t -= MOD
return mint(t)
def __radd__(self, m):
t = self.i + (m.i if isinstance(m, mint) else m)
if t > MOD:
t -= MOD
return mint(t)
def __mul__(self, m):
return mint(self.i * (m.i if isinstance(m, mint) else m) % MOD)
def __sub__(self, m):
t = self.i - m.i
if t < 0:
t += MOD
return mint(t)
def __pow__(self, m):
i = self.i
res = 1
while(m > 0):
if m & 1:
res = res * i % MOD
i = i * i % MOD
m >>= 1
return mint(res)
def __truediv__(self, m):
return mint(self.i * (m ** (MOD - 2)).i % MOD)
def __repr__(self):
return repr(self.i)
L, R = map(int, input().split())
dp = [[mint(0) for _ in range(4)] for _ in range(61)]
for d in range(60, 0, -1):
l = L >> d - 1 & 1
r = R >> d - 1 & 1
if (L >> d - 1) == 0:
if (R >> d - 1) > 1:
dp[d-1][3] += 1
elif (R >> d - 1) == 1:
dp[d-1][2] += 1
elif (L >> d - 1) == 1:
if (R >> d - 1) > 1:
dp[d-1][1] += 1
else:
dp[d-1][0] += 1
# x bound, y bound
if l == r:
dp[d-1][0] += dp[d][0]
elif l < r:
dp[d-1][0] += dp[d][0]
dp[d-1][1] += dp[d][0]
dp[d-1][2] += dp[d][0]
# x bound, y free
if l == 0:
dp[d-1][1] += dp[d][1] * 2
dp[d-1][3] += dp[d][1]
else:
dp[d-1][1] += dp[d][1]
# x free, y bound
if r == 1:
dp[d-1][2] += dp[d][2] * 2
dp[d-1][3] += dp[d][2]
else:
dp[d-1][2] += dp[d][2]
# x free, y free
dp[d-1][3] += dp[d][3] * 3
print(sum(dp[0])) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll mod = 1e9+7;
ll esq, dir;
ll dp[65][5][5][5];
ll solve(int pos, int biggerMin, int smallerMax, int k){
// cout << pos << " " << biggerMin << " " << smallerMax << endl;
if(pos < 0) return 1;
if(dp[pos][biggerMin][smallerMax][k] != -1) return dp[pos][biggerMin][smallerMax][k];
int a = (dir >> pos)&1;
int b = (esq >> pos)&1;
// cout << a << " " << b << endl;
ll res = 0;
if( b == 0 || biggerMin) // placing (0,0) in both numbers
res += solve(pos-1, biggerMin, smallerMax || (a == 1), k);
if( (b == 0 || biggerMin) && (a == 1 || smallerMax) && k)
res += solve(pos-1, biggerMin, smallerMax, k);
if( a == 1 || smallerMax)
res += solve(pos-1, biggerMin || (b == 0), smallerMax, 1);
res %= mod;
return dp[pos][biggerMin][smallerMax][k] = res;
}
int main(){
memset(dp, -1, sizeof dp);
cin >> esq >> dir;
cout << solve(60, 0, 0, 0) << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L,R = map(int,input().split())
mod = 10**9+7
def f(L,R):
if L>R : return 0
R = bin(R)[2:]
N = len(R)
ret = f(L,int("0"+"1"*(N-1),2))
L = bin(L)[2:]
if len(L) != N : L = "1"+"0"*(N-1)
for i in range(N):
if R[i] == "0" : continue
R2 = R[:i] + "0" + "?"*(N-i-1)
if i==0: R2 = R
for j in range(N):
if j==0 : L2 = L
elif L[j] == "1" : continue
else : L2 = L[:j] + "1" + "?"*(N-j-1)
tmp = 1
for r,l in zip(R2,L2):
if r=="0" and l=="1" : tmp *= 0 ; break
elif r=="?" and l=="?" : tmp = tmp*3%mod
elif r=="?" and l=="0" : tmp = tmp*2%mod
elif r=="1" and l=="?" : tmp = tmp*2%mod
ret += tmp
ret %= mod
return ret
print(f(L,R)) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int LGN = 60;
const int Z = 1e9+7;
void addm(int &a, int b) {
a = a + b >= Z ? a + b - Z : a + b;
}
int main() {
ios::sync_with_stdio(false);
ll l, r;
cin >> l >> r;
int dp[LGN+1][2][2][2];
fill(&dp[0][0][0][0], &dp[LGN+1][0][0][0], 0);
fill(&dp[0][0][0][0], &dp[1][0][0][0], 1);
for (int i = 1; i <= LGN; i++) {
bool bl = l >> (i - 1) & 1, br = r >> (i - 1) & 1;
for (int fl = 0; fl <= 1; fl++) {
for (int fr = 0; fr <= 1; fr++) {
for (int sb = 0; sb <= 1; sb++) {
for (int cl = 0; cl <= 1; cl++) {
for (int cr = 0; cr <= 1; cr++) {
if (cl <= cr && !(sb && (cl < cr)) && !(fl && (cl < bl)) && !(fr && (cr > br))) {
addm(dp[i][fl][fr][sb], dp[i-1][fl&&cl==bl][fr&&cr==br][sb&&cl!=1]);
}
}
}
}
}
}
}
cout << dp[LGN][1][1][1] << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
vector<ll> lbits, rbits;
ll dp[60][2][2][2];
void get_bits(ll n, vector<ll>& out){
out.clear();
for(int i = 0; n > 0; ++i){
out.push_back(n % 2);
n /= 2;
}
}
ll calc(ll pos, bool x_gt_l, bool y_lt_r, bool has_top){
if(pos < 0){
return 1;
}
if(dp[pos][x_gt_l][y_lt_r][has_top] != -1){
return dp[pos][x_gt_l][y_lt_r][has_top];
}
ll ret = 0;
// x = 0, y = 0
if((x_gt_l || lbits.at(pos) == 0)){
ret += calc(pos - 1, x_gt_l, y_lt_r || rbits.at(pos) == 1, has_top);
}
// x = 0, y = 1
if(has_top &&
(x_gt_l || lbits.at(pos) == 0) &&
(y_lt_r || rbits.at(pos) == 1)){
ret += calc(pos - 1, x_gt_l, y_lt_r, has_top);
}
// x = 1, y = 1
if(y_lt_r || rbits.at(pos) == 1){
ret += calc(pos - 1, x_gt_l || lbits.at(pos) == 0, y_lt_r, true);
}
ret %= MOD;
dp[pos][x_gt_l][y_lt_r][has_top] = ret;
return ret;
}
int main(){
ll l, r;
cin >> l >> r;
get_bits(l, lbits);
get_bits(r, rbits);
lbits.resize(60, 0);
rbits.resize(60, 0);
fill(
&dp[0][0][0][0],
&dp[0][0][0][0] + sizeof(dp) / sizeof(dp[0][0][0][0]),
-1
);
ll ans = calc(59, 0, 0, 0);
cout << ans << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
long mod = 1_000_000_007;
long[][][][] dp;
public static void main(String[] args) throws IOException {
Main mainObj = new Main();
mainObj.solve();
}
public void solve() throws IOException {
FastScanner fs = new FastScanner();
long l = fs.nextLong();
long r = fs.nextLong();
dp = new long[61][2][2][2];
dp[60][0][0][0] = 1;
for(int i = 59; i >= 0 ; i--) {
for(int j = 0; j < 2; j++) {
for(int k = 0; k < 2; k++) {
for(int s = 0; s < 2; s++) {
long pre = dp[i+1][j][k][s];
long lb = (l >> i) & 1;
long rb = (r >> i) & 1;
for(int x = 0; x < 2; x++) {
for(int y = 0; y < 2; y++) {
if(x == 1 && y == 0) {
continue;
}
int nj = j;
int nk = k;
int ns = s;
if(s == 0 && x != y) {
continue;
}
if(x == 1 && y == 1) {
ns = 1;
}
if(j == 0 && lb == 1 && x == 0) {
continue;
}
if(lb == 0 && x == 1) {
nj = 1;
}
if(k == 0 && rb == 0 && y == 1) {
continue;
}
if(rb == 1 && y == 0) {
nk = 1;
}
dp[i][nj][nk][ns] = (dp[i][nj][nk][ns] + pre) % mod;
}
}
}
}
}
}
long ans = 0;
for(int j = 0; j < 2; j++) {
for(int k = 0; k < 2; k++) {
for(int s = 0; s < 2; s++) {
ans = (ans + dp[0][j][k][s]) % mod;
}
}
}
System.out.println(ans);
}
public class FastScanner {
BufferedReader reader;
private StringTokenizer st;
public FastScanner() {
st = null;
reader = new BufferedReader(new InputStreamReader(System.in));
}
public String next() throws IOException {
if (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(reader.readLine());
}
return st.nextToken();
}
public String nextLine() throws IOException {
st = null;
String readLine = null;
readLine = reader.readLine();
return readLine;
}
public int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(next());
}
public long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(next());
}
public int[] nextIntArr(int n) throws NumberFormatException, IOException {
int[] retArr = new int[n];
for (int i = 0; i < n; i++) {
retArr[i] = nextInt();
}
return retArr;
}
public long[] nextLongArr(int n) throws NumberFormatException, IOException {
long[] retArr = new long[n];
for (int i = 0; i < n; i++) {
retArr[i] = nextLong();
}
return retArr;
}
public void close() throws IOException {
reader.close();
}
}
}
| JAVA |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using i64 = long long;
constexpr i64 mod = 1000000007;
std::string l, r;
i64 dp[61][2][2][2];
// l <= x, y <= r, MSB is fixed
i64 rec(int p, int a, int b, int c) {
if (!p) return 1;
if (dp[p][a][b][c] >= 0) return dp[p][a][b][c];
i64 ret = 0;
if (a || l[60 - p] == '0') {
ret = (ret + rec(p - 1, a, r[60 - p] == '1' || b, c)) % mod;
}
if (b || r[60 - p] == '1') {
ret = (ret + rec(p - 1, l[60 - p] == '0' || a, b, 1)) % mod;
}
if (c && (l[60 - p] == '0' || a) && (r[60 - p] == '1' || b)) {
ret = (ret + rec(p - 1, a, b, c)) % mod;
}
return dp[p][a][b][c] = ret;
}
int main() {
i64 ll, rr;
std::cin >> ll >> rr;
for (i64 i = 59; i >= 0; --i) l += '0' + ((ll >> i) & 1);
for (i64 i = 59; i >= 0; --i) r += '0' + ((rr >> i) & 1);
for (auto &w : dp) for (auto &x : w) for (auto &y : x) for (auto &z : y) z = -1;
std::cout << rec(60, 0, 0, 0) << std::endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <complex>
#include <string>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <limits>
#include <climits>
#include <cfloat>
#include <functional>
#include <iterator>
#include <memory>
#include <regex>
using namespace std;
const int MOD = 1000000007;
int main()
{
long long l, r;
cin >> l >> r;
vector<int> dp(8, 0);
dp[0] = 1;
for(int i=0; i<63; ++i){
int l2 = (l >> i) & 1;
int r2 = (r >> i) & 1;
vector<int> nextDp(8, 0);
for(int j=0; j<8; ++j){
for(int a=0; a<2; ++a){
for(int b=0; b<2; ++b){
if(a < b)
continue;
int k = j;
if(a > r2)
k |= 1;
else if(a < r2)
k &= ~1;
if(b < l2)
k |= 2;
else if(b > l2)
k &= ~2;
if(a == 1 && b == 1)
k |= 4;
else if(a != b)
k &= ~4;
nextDp[k] += dp[j];
nextDp[k] %= MOD;
}
}
}
dp = move(nextDp);
}
cout << dp[4] << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L,R = map(int,input().split())
mod = 10**9+7
def f(L,R):
if L>R : return 0
R = bin(R)[2:]
N = len(R)
ret = f(L,int("0"+"1"*(N-1),2))
L = bin(L)[2:]
if len(L) != N : L = "1"+"0"*(N-1)
for i in range(N):
if R[i] == "0" : continue
R2 = R[:i] + "0" + "?"*(N-i-1)
if i==0: R2 = R
for j in range(N):
if L[j] == "1" and j!=0 : continue
L2 = L[:j] + "1" + "?"*(N-j-1)
if j==0 : L2 = L
tmp = 1
for r,l in zip(R2,L2):
if r=="0" and l=="1" : tmp *= 0 ; break
elif r=="?" and l=="?" : tmp = tmp*3%mod
elif r=="?" and l=="0" : tmp = tmp*2%mod
elif r=="1" and l=="?" : tmp = tmp*2%mod
ret += tmp
ret %= mod
return ret
print(f(L,R)) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import sys
stdin = sys.stdin
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip() # ignore trailing spaces
# 最高次は等しくないといけないので、
# y % x = y-xになる
# yがxを包含していればok
mod = 1000000007
def ca(x, y):
ct = 0
for i in range(x, y+1):
for j in range(i, y+1):
if j % i == (j^i):
ct += 1
return ct
ans = 0
l, r = na()
dp = [0] * 4
for i in range(60, -1, -1):
ndp = [0] * 4
base = 1<<i
if base <= r and l < base*2:
ptn = 0
if base+1 <= l:
ptn += 1
if r < base*2-1:
ptn += 2
ndp[ptn] += 1
if l>>i != r>>i:
if (l>>i&1) == 0 and (r>>i&1) == 1:
ndp[1] += dp[3]
ndp[2] += dp[3]
if ((l>>i)&(r>>i)) == (l>>i):
ndp[3] += dp[3]
if (l >> i & 1) == 0:
ndp[0] += dp[1]
ndp[1] += dp[1]
ndp[1] += dp[1]
if (r >> i & 1) == 1:
ndp[0] += dp[2]
ndp[2] += dp[2]
ndp[2] += dp[2]
ndp[0] += dp[0] * 3
for k in range(4):
ndp[k] %= mod
dp = ndp
print(sum(dp) % mod)
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define BIT_COUNT 70
int add(int x,int y) {
int ret = x+y;
if(ret>=mod) {
ret -= mod;
}
if(ret<0) {
ret += mod;
}
return ret;
}
int dp[BIT_COUNT][2][2][2];
ll L,R;
int solveDp(int pos,int mbit,int LeqX,int ReqY) {
if(pos<0) {
return mbit==1;
}
int &ret = dp[pos][mbit][LeqX][ReqY];
if(ret!=-1) {
return ret;
}
ret = 0;
for(int a=0;a<2;++a) {
for(int b=0;b<2;++b) {
if(a==1 && b==0) continue;
if(LeqX && a==0 && (L&(1LL<<pos))>0 ) continue;
if(ReqY && b==1 && (R&(1LL<<pos) )==0)continue;
if(mbit==0 && (a!=b)) continue;
int nmbit = mbit | (a==1 && b==1);
int nLeqX = LeqX && ( ((L>>pos)&1) == a);
int nReqY = ReqY && ( ((R>>pos)&1) ==b);
ret = add(ret, solveDp(pos-1,nmbit,nLeqX,nReqY));
}
}
return ret;
}
void solve() {
scanf("%lld %lld",&L,&R);
memset(dp,-1,sizeof(dp));
cout<<solveDp(62,0,1,1)<<endl;
}
int main() {
solve();
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod=1e9+7;
ll l,r;
ll f[64][2][2][2][2];
int L[100],R[100];
ll dfs(int pos,int fl,int fr,int s,int ht)
{
if(pos<0)return 1;
ll &ans=f[pos][fl][fr][s][ht];
if(ans)return ans;
for(int i=0;i<2;i++)//y
{
for(int j=0;j<2;j++)//x
{
int tl=fl,tr=fr,ts=s,tt=ht;
if(!tt&&i)if(i!=j)continue;
if((i&j)!=j)continue;
if(!tl&&j<L[pos])continue;
if(!tr&&i>R[pos])continue;
if(!ts&&j>i)continue;
if(j>L[pos])tl=1;if(i<R[pos])tr=1;if(j<i)ts=1;if(j)tt=1;
ans+=dfs(pos-1,tl,tr,ts,tt);
}
}
ans%=mod;
return ans;
}
int main()
{
scanf("%lld%lld",&l,&r);
for(int i=63;i>=0;i--)
{
L[i]=(int)((l>>i)&1),R[i]=(int)((r>>i)&1);
}
printf("%lld\n",dfs(63,0,0,0,0));
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | def count(lb, rb):
assert lb[0] == '1'
assert rb[0] == '1'
assert len(lb) == len(rb)
dp = [1, 0, 0, 0]
for lc, rc in zip(lb[1:], rb[1:]):
ndp = [dp[0], 0, 0, 0]
if rc == '0':
ndp[1] += dp[1]
if lc == '1':
ndp[0] = 0
else:
ndp[1] += dp[1] * 2
ndp[2] += dp[1]
if lc == '0':
ndp[1] += dp[0]
ndp[3] += dp[0]
if lc == '0':
ndp[2] += dp[3]
ndp[3] += dp[3] * 2
else:
ndp[3] += dp[3]
ndp[2] += dp[2] * 3
dp = ndp
return sum(dp)
l, r = map(int, input().split())
lb = bin(l)[2:]
rb = bin(r)[2:]
ld = len(lb)
rd = len(rb)
ans = 0
MOD = 10 ** 9 + 7
for d in range(ld, rd + 1):
tlb = lb if d == ld else '1' + '0' * (d - 1)
trb = rb if d == rd else '1' * d
ans = (ans + count(tlb, trb)) % MOD
print(ans)
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | from itertools import product
MOD = 10**9 + 7
L, R = map(int, input().split())
binR = bin(R+1)[2:]
maxD = len(binR)
binL = bin(L-1)[2:].zfill(maxD)
dp = [[[[0]*(2) for k in range(2)] for j in range(2)] for i in range(maxD+1)]
dp[0][0][0][0] = 1
for d, (Ld, Rd) in enumerate(zip(binL, binR)):
Ld, Rd = int(Ld), int(Rd)
for isLleX, isYleR, isNum in product(range(2), repeat=3):
for x, y in [(0,0), (0,1), (1,1)]:
if not isNum and (x, y) == (0, 1): continue
if not isLleX and x < Ld: continue
if not isYleR and y > Rd: continue
isLleX2 = isLleX or x > Ld
isYleR2 = isYleR or y < Rd
isNum2 = isNum or y == 1
dp[d+1][isLleX2][isYleR2][isNum2] += dp[d][isLleX][isYleR][isNum]
dp[d+1][isLleX2][isYleR2][isNum2] %= MOD
print(dp[-1][1][1][1])
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L, R = map(int, raw_input().split())
"""
R ????????1??????
y 0001????0??????
^d ^rd
x 0001???????1??
^ld
L ???????????0???
"""
ans = 0
for d in range(70):
if d<len(bin(L))-3:
LD = []
elif d==len(bin(L))-3:
LD = [i for i in range(-1,d) if i==-1 or (L>>i&1)==0]
else:
LD = [d]
if d<len(bin(R))-3:
RD = [d]
elif d==len(bin(R))-3:
RD = [i for i in range(-1,d) if i==-1 or (R>>i&1)==1]
else:
RD = []
for ld in LD:
for rd in RD:
a = 1
for i in range(d):
if i<ld:
xc = [0, 1]
elif i==ld:
xc = [1]
else:
xc = [L>>i&1]
if i<rd:
yc = [0, 1]
elif i==rd:
yc = [0]
else:
yc = [R>>i&1]
c = 0
for x in xc:
for y in yc:
if y>=x:
c += 1
a *= c
ans += a
print ans%(10**9+7)
| PYTHON |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | l, r = input().split()
l = int(l)
r = int(r)
dp = [[[-1 for k in range(2)] for j in range(2)] for i in range(64)]
def bbin(num):
res = []
while num:
res.append(num % 2)
num //= 2
return res
numr = bbin(r)
numl = bbin(l)
while len(numl) < len(numr):
numl.append(0)
def rec(pos, brr, brl):
global numl, numr, dp
if pos < 0:
return 1
if dp[pos][brr][brl] != -1:
return dp[pos][brr][brl]
res = 0
if brr:
if brl:
res = 3 * rec(pos - 1, brr, brl)
else:
if numl[pos] == 1:
res = rec(pos - 1, brr, brl)
else:
res += (2 * rec(pos - 1, brr, 0))
res += rec(pos - 1, brr, 1)
else:
if numr[pos] == 1:
if brl:
res += (2 * rec(pos - 1, 0, brl))
res += rec(pos - 1, 1, brl)
else:
if numl[pos] == 1:
res += rec(pos - 1, 0, brl)
else:
res += rec(pos - 1, 0, 0)
res += rec(pos - 1, 0, 1)
res += rec(pos - 1, 1, 0)
else:
if brl:
res += rec(pos - 1, 0, brl)
elif numl[pos] == 0:
res += rec(pos - 1, 0, 0)
dp[pos][brr][brl] = res % (10**9 + 7)
return dp[pos][brr][brl]
c = 0
for i in range(len(numl) - 1, -1, -1):
if numl[i] == 1:
c += rec(i - 1, 1 if i < len(numl) - 1 else 0, 0)
break
else:
c += rec(i - 1, 1 if i < len(numl) - 1 else 0, 1)
print(c % (10**9 + 7)) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); ++i)
const int MOD = int(1e9)+7;
struct mint {
ll x;
mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
mint& operator+=(const mint a) { if ((x += a.x) >= MOD) x -= MOD; return *this; }
mint operator+(const mint a) const { return mint(*this) += a; }
};
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
const int M = 60;
mint dp[M+1][2][2][2];
int main() {
ll L, R; cin >> L >> R;
dp[0][0][0][0] = 1;
rep(i, M) {
int Li = L>>(M-1-i)&1; // L の上から i 桁目
int Ri = R>>(M-1-i)&1; // R の上から i 桁目
rep(j, 2) rep(l, 2) rep(r, 2) rep(xi, 2) rep(yi, 2) {
int j2 = j, l2 = l, r2 = r; // 遷移前の状態をコピー
if (xi > yi) continue; // (x の i 桁目) ≦ (y の i 桁目) でなければ無視
if (!j && (xi^yi)) continue; // (1, 1) が現れる前に (0, 1) か (1, 0) が現れたら無視
if (xi&yi) j2 = 1; // (1, 1) が現れたら j2 を更新
if (!l && (xi < Li)) continue;
if (xi > Li) l2 = 1;
if (!r && (yi > Ri)) continue;
if (yi < Ri) r2 = 1;
dp[i+1][j2][l2][r2] += dp[i][j][l][r]; // 遷移式
}
}
mint ans = 0;
rep(l, 2) rep(r, 2) ans += dp[M][1][l][r];
cout << ans << '\n';
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#define mkp make_pair
#define rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
const ll MOD=1e9+7;
ll L,R;
ll dp[61][2][2][2];
void add(ll &a,ll b){
a=(a+b)%MOD;
}
void mul(ll &a,ll b){
a=a*b%MOD;
}
int main(){
cin>>L>>R;
dp[60][0][0][0]=1;
for(int i=59;i>=0;i--){
int lb=(L>>i)&1;
int rb=(R>>i)&1;
rep(j,2)rep(k,2)rep(s,2){
rep(y,2)rep(x,2){
if(y==0&&x==1) continue;
int nj=j,nk=k,ns=s;
ll cost=1;
if(lb&&j==0&&x==0) continue;
if(rb==0&&k==0&&y==1) continue;
if(s==0&&(y==1&&x==0)) continue;
if(y==1&&x==1){
if(s==0) ns=1;
if(lb==0&&j==0) nj=1;
}
if(y==1&&x==0){
}
if(y==0&&x==0){
if(rb&&k==0) nk=1;
}
ll val=cost;
mul(val,dp[i+1][j][k][s]);
add(dp[i][nj][nk][ns],val);
}
}
}
ll ans=0;
rep(j,2)rep(k,2) add(ans,dp[0][j][k][1]);
cout<<ans<<endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)
const ll mod=1e9+7;
ll dp[65][2][2][2];
int main()
{
ll l,r;
scanf("%lld%lld",&l,&r);
int n=64;
vector<ll> bl(n),br(n);
for(int i=0;i<n;i++)
{
bl[i]=l%2;
br[i]=r%2;
l/=2;
r/=2;
}
reverse(begin(bl),end(bl));
reverse(begin(br),end(br));
dp[0][0][0][0]=1;
rep(i,n)rep(fl,2)rep(fr,2)rep(g,2)
{
if(dp[i][fl][fr][g]==0)continue;
rep(bx,2)rep(by,2)
{
if(bx==1&&by==0)continue;
if(!g&&bx+by==1)continue;
if(!fl&&bl[i]&&!bx)continue;
if(!fr&&!br[i]&&by)continue;
dp[i+1][fl | (bx > bl[i])][fr | (by < br[i])][g | bx | by] += dp[i][fl][fr][g]%=mod;
}
}
ll ans=0;
rep(fl,2)rep(fr,2)rep(g,2)ans=(ans+dp[n][fl][fr][g])%mod;
printf("%lld\n",ans);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include "bits/stdc++.h"
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;
const ll INF = 1LL << 60;
ll dp[61][2][2]; // 何ケタ目か,L=x or L<x,R=y or y<R
int main() {
ll L, R;
cin >> L >> R;
ll ans = 0;
for (int n = 0; n <= 60; n++) {
ll p = 1LL << n;
ll mask = ~(p - 1);
if ((L&mask) > p || p > (R&mask)) continue;
memset(dp, 0, sizeof(dp));
dp[n][(L&mask) < p][p < (R&mask)] = 1;
for (int i = n - 1; i >= 0; i--) {
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
// 0 0
if (j || ((L >> i) & 1) == 0) {
(dp[i][j][k | (((R >> i) & 1) == 1)] += dp[i + 1][j][k]) %= MOD;
}
// 1 0
if ((j || ((L >> i) & 1) == 0) && (k || ((R >> i) & 1) == 1)) {
(dp[i][j][k] += dp[i + 1][j][k]) %= MOD;
}
// 1 1
if (k || ((R >> i) & 1) == 1) {
(dp[i][j | (((L >> i) & 1) == 0)][k] += dp[i + 1][j][k]) %= MOD;
}
}
}
}
for (int i = 0; i < 2; i++) for (int j = 0; j < 2; j++) (ans += dp[0][i][j]) %= MOD;
}
cout << ans << endl;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll l, r;
const int MOD = 1000000007;
bool seen[2][2][2][100];
int cache[2][2][2][100];
//vector<pair<ll, ll>> sols[2][2][2][100];
int dp(bool top, bool bottom, bool same, int bit) {
if (bit < 0) {
//if (!same) sols[top][bottom][same][(bit + 100)%100].emplace_back(0, 0);
return !same;
}
if (seen[top][bottom][same][bit]) return cache[top][bottom][same][bit];
seen[top][bottom][same][bit] = true;
int &ans = cache[top][bottom][same][bit];
//auto &v = sols[top][bottom][same][bit];
ans = 0;
const ll mask = 1ll<<bit;
// try 1
if (!top || (mask & r)) {
ans += dp(top, bottom && (l & mask), false, bit-1);
ans %= MOD;
//for (auto p : sols[top][bottom && (l & mask)][0][(bit+99)%100])
// v.emplace_back(p.first | mask, p.second);
}
// try 0
if (!bottom || !(mask & l)) {
ans += dp(top && !(r & mask), bottom, same, bit-1);
ans %= MOD;
//for (auto p : sols[top && !(r & mask)][bottom][same][(bit+99)%100]) {
// v.emplace_back(p.first, p.second);
//}
if (!same && (!top || (mask & r))) {
ans += dp(top, bottom, same, bit-1);
//for (auto p : sols[top][bottom][same][(bit+99)%100]) {
// v.emplace_back(p.first, p.second);
//}
}
ans %= MOD;
}
return ans;
}
int main() {
cin >> l >> r;
cout << dp(true, true, true, 62) << '\n';
//for (auto p : sols[1][1][1][62]) {
// ll x = p.first, y = p.first + p.second, z = p.second;
// cout << x << ' ' << y << ' ' << z << '\n';
// assert(l <= x && x <= y && y <= r);
// assert((y % x) == (y ^ x));
//}
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define FOR(i, a, n) for(ll i = (ll)a; i < (ll)n; i++)
#define FORR(i, n) for(ll i = (ll)n - 1LL; i >= 0LL; i--)
#define rep(i, n) FOR(i, 0, n)
#define ALL(x) (x).begin(), (x).end()
using namespace std;
using ll = long long;
template <typename T> using V = vector<T>;
constexpr int Mod = 998244353;
constexpr int mod = 1e9 + 7;
constexpr ll inf = 1LL << 60;
template <typename T> constexpr bool chmax(T &a, const T &b) {
if(a >= b) return false;
a = b;
return true;
}
template <typename T> constexpr bool chmin(T &a, const T &b) {
if(a <= b) return false;
a = b;
return true;
}
/*-------------------------------------------*/
ll L, R;
ll dp[2][2][2][65];
ll f(int d, int S, int l, int r) {
if(d == -1) return S;
if(dp[S][l][r][d] >= 0) return dp[S][l][r][d];
ll ret = 0;
rep(a, 2) rep(b, 2) {
if(!a && l && (L >> d & 1)) continue;
if(b && r && !(R >> d & 1)) continue;
if(a > b) continue;
if(!S && a != b) continue;
int Nl = l && (L >> d & 1) == a;
int Nr = r && (R >> d & 1) == b;
int NS = S || a == 1;
ret += f(d - 1, NS, Nl, Nr);
}
return dp[S][l][r][d] = ret % mod;
}
int main() {
cin.tie(0);
ios::sync_with_stdio(0);
cin >> L >> R;
memset(dp, -1, sizeof(dp));
cout << f(61, 0, 1, 1) << endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | /*
如何动态维护第k大元素？
*/
#include<bits/stdc++.h>
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const ll MAXN=2e5+5;
const ll INF=1e18;
const ll MOD=1e9+7;
ll l,r,numx[60],numy[60],cntx,cnty;
int dp[60][2][2][2];
ll add(ll x,ll y)
{
ll ret=x+y;
if(ret>=MOD)
{
return ret-MOD;
}
return ret;
}
void getnum()
{
while(l)
{
numx[cntx++]=l%2;
l/=2;
}
while(r)
{
numy[cnty++]=r%2;
r/=2;
}
}
ll dfs(ll step,ll fx,ll fy,ll fz)
{
if(step==-1)
{
return 1;
}
if(dp[step][fx][fy][fz]!=-1)
{
return dp[step][fx][fy][fz];
}
ll ret=0;
//0 0
if(fx||numx[step]==0)
{
ret=add(ret,dfs(step-1,fx,numy[step]==1?1:fy,fz));
}
//0 1
if((fx||numx[step]==0)&&(fy||numy[step]==1)&&(fz))
{
ret=add(ret,dfs(step-1,fx,fy,fz));
}
if(fy||numy[step]==1)
{
ret=add(ret,dfs(step-1,numx[step]==0?1:fx,fy,1));
}
dp[step][fx][fy][fz]=ret;
return ret;
}
void solve()
{
memset(dp,-1,sizeof(dp));
getnum();
printf("%lld\n",dfs(59,0,0,0));
}
int main()
{
while(~scanf("%lld %lld",&l,&r))
{
solve();
}
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
using namespace std;
const int M = 1000000007;
long long modexp(int x, long long e, int m) {
long long ans = 1, p = x % m;
while (e > 0) {
if (e % 2 != 0) ans = (ans * p) % m;
p = (p * p) % m;
e >>= 1;
}
return ans;
}
long long L, R;
long long calc(long long lx, long long ly, int i, int j) {
long long rx = lx + (1LL<<i), ry = ly + (1LL<<j);
if (ry <= lx || rx <= L || ry <= L || R <= lx || R <= ly) return 0;
if (L <= lx && rx <= R)
if (L <= ly && ry <= R) {
if (i >= j) {
if (((lx >> i) & (ly >> i)) == (lx >> i)) {
int p = __builtin_popcountll((ly>>j)&((1LL<<(i-j))-1));
return modexp(3, j, M) * modexp(2, p, M) % M;
} else
return 0;
} else {
if (((lx >> j) & (ly >> j)) == (lx >> j)) {
int p = __builtin_popcountll((lx>>i)&((1LL<<(j-i))-1));
return modexp(3, i, M) * modexp(2, j-i-p, M) % M;
} else
return 0;
}
} else {
long long my = ly + (1LL<<(j-1));
return (calc(lx, ly, i, j-1) + calc(lx, my, i, j-1)) % M;
}
else {
long long mx = lx + (1LL<<(i-1));
return (calc(lx, ly, i-1, j) + calc(mx, ly, i-1, j)) % M;
}
}
int main() {
cin >> L >> R;
R++;
long long ans = 0;
for (int k = 0; k < 60; k++)
ans += calc(1LL<<k, 1LL<<k, k, k);
cout << ans % M << endl;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <vector>
#include <map>
#include <algorithm>
#include <queue>
#include <string>
#define REP(i,a,b) for (int i = a; i < (b); i++)
#define sz(t) int(t.size())
#define INF 1000000000
const int DIV = 1000000000 + 7;
using namespace std;
typedef long long ll;
int dp[62][2][2][2];
int main(void) {
ll l,r;
cin >> l >> r;
dp[60][0][0][0] = 1;
for (int i = 59; i >= 0; i--) {
int lb = l >> i & 1;
int rb = r >> i & 1;
// cout << lb << " " << rb << endl;
// s means most significant bit is the same or unknown
// j means x is greater than l
// k means y is less than r
REP(j, 0, 2) REP(k, 0, 2) REP(s, 0, 2) {
int pre = dp[i + 1][j][k][s];
REP(x, 0, 2) REP(y, 0, 2) {
int ni = i, nj = j, nk = k, ns = s;
if (x && !y) continue;
if (!s && x != y) continue;
if (!s && x&y) ns = 1;
if (!j && lb && !x) continue;
if (!j && !lb && x) nj = 1;
if (!k && !rb && y) continue;
if (!k && rb && !y) nk = 1;
dp[ni][nj][nk][ns] += pre;
dp[ni][nj][nk][ns] %= DIV;
}
}
}
ll ans = 0;
REP(j, 0, 2) REP(k, 0, 2) REP(s, 0, 2) {
ans += dp[0][j][k][s];
ans %= DIV;
}
cout << ans << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | from functools import *
MOD=10**9+7
@lru_cache(maxsize=None)
def solve(L,R):
if L > R: return 0
if L == 1: return (1 + solve(2, R)) % MOD
return (solve(L//2,(R-1)//2) + solve((L+1)//2,R//2) + solve((L+1)//2,(R-1)//2)) % MOD
print(solve(*map(int,input().split()))) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); ++i)
const int MOD = int(1e9)+7;
struct mint {
ll x;
mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
mint& operator+=(const mint a) { if ((x += a.x) >= MOD) x -= MOD; return *this; }
mint operator+(const mint a) const { return mint(*this) += a; }
};
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
const int M = 60;
mint dp[M+1][2][2][2];
int main() {
ll L, R; cin >> L >> R;
dp[0][0][0][0] = 1;
rep(i, M) {
int Li = L>>(M-1-i)&1;
int Ri = R>>(M-1-i)&1;
rep(j, 2) rep(l, 2) rep(r, 2) rep(xi, 2) rep(yi, 2) {
int j2 = j, l2 = l, r2 = r;
if (xi > yi) continue;
if (!j && (xi^yi)) continue;
if (xi&yi) j2 = 1;
if (!l && (xi < Li)) continue;
if (xi > Li) l2 = 1;
if (!r && (yi > Ri)) continue;
if (yi < Ri) r2 = 1;
dp[i+1][j2][l2][r2] += dp[i][j][l][r];
}
}
mint ans = 0;
rep(l, 2) rep(r, 2) ans += dp[M][1][l][r];
cout << ans << '\n';
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MO=1e9+7;
const int MB=65;
// x<=y x^y==y%x
// highbit() Most Significant Bit
// highbit(x) != highbit(y)
// x^y>x && x%y<x no solution
// so highbit(x) == highbit(y)
// so floor(y/x) == 1
// x^y==y%x -> x^y == y-x
// for y-x
// i-th bit
// 1-1 0-0 = 0 -> xor same
// 1-0 = 1 -> xor same
// 0-1 change higher bit! impossible!
void add(int &x,int y)
{
x+=y;
if(x>=MO) x-=MO;
}
int dp[MB][2][2][2];
// dp[i-th digit][upperbound][lowerbound][leading zero]
LL L,R;
int getdp(int d,int llim,int rlim,int lead)
{
if(d==-1)
return 1;
int &ans=dp[d][llim][rlim][lead];
if(ans!=-1) return ans;
ans=0;
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
{
if(i==0&&j==1) continue;
if(lead&&(i!=j)) continue;
int l=((L>>d)&1);
int r=((R>>d)&1);
if(llim&&j<l) continue;
if(rlim&&i>r) continue;
add(ans,getdp(d-1,llim&&l==j,rlim&&r==i,lead&&!i&&!j));
}
return ans;
}
int main()
{
cin>>L>>R;
memset(dp,-1,sizeof(dp));
cout<<getdp(61,1,1,1)<<endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define ll long long
namespace io{
const int l=1<<20;
char buf[l],*s,*t;
inline char gc(){
if(s==t){
t=(s=buf)+fread(buf,1,l,stdin);
return s==t?EOF:*s++;
}
return *s++;
}
char c;
template<class IT>inline void gi(IT &x){
x=0;c=gc();while(c<'0'||c>'9')c=gc();
while('0'<=c&&c<='9'){x=(x<<1)+(x<<3)+(c^'0');c=gc();}
}
};
using io::gi;
template<class IT>inline void cmin(IT &a,IT b){if(b<a)a=b;}
template<class IT>inline void cmax(IT &a,IT b){if(a<b)a=b;}
const int p=1000000007;
const int N=200005;
int dp[64][2][2];
inline void upd(int &a,int b){a=a+b<p?a+b:a+b-p;}
inline int solve(ll l,ll r){
int s=0,t=0,n,i,a,b,c,d,x,y,ans=0;
while((1ll<<s)<=l)++s;
while((1ll<<t)<=r)++t;
// printf("l=%lld r=%lld s=%d t=%d\n",l,r,s,t);
for(n=1;n<=s;++n){
memset(dp,0,sizeof(dp));
dp[n-1][n!=s][n!=t]=1;
for(i=n-2;~i;--i){
x=l>>i&1ll;
y=r>>i&1ll;
for(a=0;a<2;++a)for(b=0;b<2;++b)if(dp[i+1][a][b]){
for(c=0;c<2;++c)for(d=c;d<2;++d){
if((a||c<=x)&&(b||d<=y))upd(dp[i][a|(c<x)][b|(d<y)],dp[i+1][a][b]);
}
}
}
// printf("%d\n",n);
// for(i=n-1;~i;--i)printf("%d %d %d %d\n",dp[i][0][0],dp[i][0][1],dp[i][1][0],dp[i][1][1]);
for(a=0;a<2;++a)for(b=0;b<2;++b)upd(ans,dp[0][a][b]);
}
// printf("%d\n",ans);
return ans;
}
int main(){
ll l,r;
scanf("%lld%lld",&l,&r);
printf("%d",(solve(r,r)-solve(l-1ll,r)+p)%p);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxd = 64;
const int mod = 1e9+7;
int a[maxd],b[maxd];
long long dp[maxd][2][2][2][2],n,m;
long long dfs(int p,int l,int r,int f,int t)
{
//位数，是否填过小于m的值，是否填过大于n的值，是否填过a<b,最高位是否都为1
if(p < 0) return 1;
long long &ans = dp[p][l][r][f][t];
// 记忆化搜索
if(ans) return ans;
for(int i=0;i<2;i++) // y
for(int j=0;j<2;j++) // x
{
int nl = l,nr = r,nf = f,nt = t;
if(!nt && i && i!=j) continue;
if((j&i)!=j) continue;
if(!nl && j < a[p]) continue; // 如果值小于n了
if(!nr && i > b[p]) continue; // 如果值大于m了
if(!nf && j > i) continue;
if(j > a[p]) nl = 1;
if(i < b[p]) nr = 1;
if(j) nt = 1;
if(i > j) nf = 1;
ans += dfs(p-1,nl,nr,nf,nt);
ans %= mod;
}
return ans;
}
int main()
{
// freopen("a.in","r",stdin);
// freopen("k.out","w",stdout);
scanf("%lld %lld",&n,&m);
for(int i=0;i<=63;i++) a[i] = (n>>i)&1 , b[i] = (m>>i)&1;
long long ans = dfs(63,0,0,0,0);
printf("%lld",ans);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L, R = map(int, input().split())
ans = 0
mod = 10 ** 9 + 7
for d in range(62): # 最上位 bit
# dp[n] := n 桁目まで見た時に条件を満たすものが何個あるか
# dp_small[n] := n 桁目まで見た時に、l と x が n 桁目まで一致していて条件を満たすものが何個あるか
l = max(1<<d, L)
r = min((2<<d)-1, R)
if l > r:
continue
dp_small = [0] * (d+2)
dp_ok = [0] * (d+2)
dp_large = [0] * (d+2)
dp_small_large = [0] * (d+2)
dp_small_large[0] = 1
for n in range(1, d+2):
i = d-n+1
dp_ok[n] = dp_ok[n-1] * 3 \
+ (dp_small[n-1] if l>>i&1==0 else 0) \
+ (dp_large[n-1] if r>>i&1 else 0)
dp_small[n] = dp_small[n-1] * (1 if l>>i&1 else 2) \
+ (dp_small_large[n-1] if l>>i&1 < r>>i&1 else 0)
dp_large[n] = dp_large[n-1] * (2 if r>>i&1 else 1) \
+ (dp_small_large[n-1] if l>>i&1 < r>>i&1 else 0)
dp_small_large[n] = dp_small_large[n-1] if l>>i&1 <= r>>i&1 else 0
a = dp_ok[d+1] + dp_large[d+1] + dp_small[d+1] + dp_small_large[d+1]
ans += a
# print(f"d={d},a={a},l={l},r={r}")
# print(f"dp_ok={dp_ok}")
# print(f"dp_small={dp_small}")
# print(f"dp_large={dp_large}")
# print(f"dp_small_large={dp_small_large}")
# print()
print(ans % mod)
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L, R = map(int, input().split())
dp = [[[[0]*2 for _ in range(2)] for _ in range(2)] for _ in range(65)]
dp[0][0][0][0] = 1
MOD = 10**9+7
for i in range(64):
Li = (L>>(63-i))&1
Ri = (R>>(63-i))&1
for j in range(2):
for k in range(2):
for l in range(2):
for x in (range(2) if j else range(Li, 2)):
for y in range(2 if k else Ri+1):
if x==1 and y==0:
continue
if l==0 and x==0 and y==1:
continue
dp[i+1][j|(x>Li)][k|(y<Ri)][l|(x==1 and y==1)] += dp[i][j][k][l]
dp[i+1][j|(x>Li)][k|(y<Ri)][l|(x==1 and y==1)] %= MOD
ans = 0
for i in range(2):
for j in range(2):
for k in range(2):
ans += dp[-1][i][j][k]
ans %= MOD
print(ans) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
ll L,R; ll f[64][2][2][2][2];
int l[100],r[100];
ll dfs(int x,int fl,int fr,int s,int ht)
{ if(x<0)return 1;
ll &ans=f[x][fl][fr][s][ht];
if(ans)return ans;
for(int i=0;i<2;i++)
{ for(int j=0;j<2;j++){
int tl=fl,tr=fr,ts=s,tt=ht;
if(!tt&&i)if(i!=j)continue;
if((j&i)!=j)continue;
if(!tl&&j<l[x])continue;
if(!tr&&i>r[x])continue;
if(j>l[x])tl=1;if(i<r[x])tr=1;if(j<i)ts=1;if(j)tt=1;
ans+=dfs(x-1,tl,tr,ts,tt);
}
}
ans%=mod; return ans;
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
cin>>L>>R;
for(int i=63;i>=0;i--) {
l[i]=(bool)((1ll<<i)&L);r[i]=(bool((1ll<<i)&R));
}
cout<<dfs(63,0,0,0,0);
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | from functools import lru_cache
P = 10**9+7
@lru_cache(maxsize=None)
def subcalc(l, r):
if l > r: return 0
if r == 0: return 1
aa, bb = l.bit_length(), r.bit_length()
if aa == bb:
return subcalc(l-(1<<aa-1), r-(1<<bb-1))
if (r & (r+1) == 0) and (l == 0):
return pow(3, r.bit_length(), P)
return (subcalc(l, r-(1<<bb-1)) + subcalc(l, (1<<bb-1)-1) + subcalc(0, r-(1<<bb-1))) % P
@lru_cache(maxsize=None)
def calc(L, R):
if L > R: return 0
a = L.bit_length()
b = R.bit_length()
if b > a:
return (calc(L, (1<<b-1)-1) + calc(1<<b-1, R)) % P
a = 1 << L.bit_length() - 1 if L else 0
return subcalc(L-a, R-a)
L, R = map(int, input().split())
print(calc(L, R))
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int LGN = 60;
const int Z = 1e9+7;
void addm(int &a, int b) {
a = a + b >= Z ? a + b - Z : a + b;
}
int main() {
ios::sync_with_stdio(false);
ll l, r;
cin >> l >> r;
int dp[LGN+1][2][2][2];
fill(&dp[0][0][0][0], &dp[LGN+1][0][0][0], 0);
fill(&dp[0][0][0][0], &dp[1][0][0][0], 1);
for (int i = 1; i <= LGN; i++) {
bool bl = l >> (i - 1) & 1, br = r >> (i - 1) & 1;
for (int fl = 0; fl < 2; fl++) {
for (int fr = 0; fr < 2; fr++) {
for (int sb = 0; sb < 2; sb++) {
for (int cl = 0; cl < 2; cl++) {
for (int cr = 0; cr < 2; cr++) {
if (cl <= cr && !(sb && (cl < cr)) && !(fl && (cl < bl)) && !(fr && (cr > br))) {
addm(dp[i][fl][fr][sb], dp[i-1][fl&&cl==bl][fr&&cr==br][sb&&cl!=1]);
}
}
}
}
}
}
}
cout << dp[LGN][1][1][1] << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | /*Author: Satyajeet Singh, Delhi Technological University*/
import java.io.*;
import java.util.*;
import java.text.*;
import java.lang.*;
import java.math.*;
public class Main{
/*********************************************Constants******************************************/
static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static long mod=(long)1e9+7;
static long mod1=998244353;
static boolean sieve[];
static ArrayList<Integer> primes;
static ArrayList<Long> factorial;
static HashSet<Pair> graph[];
static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
/****************************************Solutions Begins***************************************/
static ArrayList<Long> list,ls;
static long dp[][][][][][];
static long dfs(int i,int x,int y,int f,int e,int f2){
if(i==list.size()){
if(f<=1&&e==0&&f2<=1){
return 1;
}
else{
return 0;
}
}
if(dp[i][x][y][f][e][f2]!=-1){
return dp[i][x][y][f][e][f2];
}
long ans=0;
//0,0
int aa=0;
if(ls.get(i)==1)aa=2;
else aa=f2;
if(list.get(i)==1){
ans=(ans+dfs(i+1,0,0,0,e,aa))%mod;
}
else{
ans=(ans+dfs(i+1,0,0,f,e,aa))%mod;
}
//1,0
if(list.get(i)==1){
ans=(ans+dfs(i+1,1,0,f,1,aa))%mod;
}
else{
ans=(ans+dfs(i+1,1,0,2,1,aa))%mod;
}
//1,1
if(ls.get(i)==1)aa=f2;
else aa=0;
if(list.get(i)==1){
ans=(ans+dfs(i+1,1,1,f,0,aa))%mod;
}
else{
ans=(ans+dfs(i+1,1,1,2,0,aa))%mod;
}
dp[i][x][y][f][e][f2]=ans;
return ans;
}
static long fn(long a,long b){
list=new ArrayList<>();
long k=a;
while(k!=0){
list.add(k%2);
k/=2;
}
ls=new ArrayList<>();
k=b;
while(k!=0){
ls.add(k%2);
k/=2;
}
while(ls.size()!=list.size())ls.add(0l);
// debug(list);
dp=new long[list.size()][2][2][3][2][3];
for(long arr[][][][][]:dp)
for(long ar1[][][][]:arr)
for(long ar2[][][]:ar1)
for(long aa[][]:ar2)
for(long hh[]:aa)
Arrays.fill(hh,-1);
long ans=dfs(0,0,0,1,0,1);
return ans;
}
public static void main (String[] args) throws Exception {
String st[]=br.readLine().split(" ");
long a=pl(st[0]);
long b=pl(st[1]);
// debug(b,a);
// debug(fn(4));
//debug(fn(b),fn(a-1));
long ans=fn(b,a);
out.println(ans);
/****************************************Solutions Ends**************************************************/
out.flush();
out.close();
}
/****************************************Template Begins************************************************/
static String[] nl() throws Exception{
return br.readLine().split(" ");
}
static String[] nls() throws Exception{
return br.readLine().split("");
}
static int pi(String str) {
return Integer.parseInt(str);
}
static long pl(String str){
return Long.parseLong(str);
}
static double pd(String str){
return Double.parseDouble(str);
}
/***************************************Precision Printing**********************************************/
static void printPrecision(double d){
DecimalFormat ft = new DecimalFormat("0.00000000000000000");
out.println(ft.format(d));
}
/**************************************Bit Manipulation**************************************************/
static void printMask(long mask){
System.out.println(Long.toBinaryString(mask));
}
static int countBit(int mask){
int ans=0;
while(mask!=0){
if(mask%2==1){
ans++;
}
mask/=2;
}
return ans;
}
/******************************************Graph*********************************************************/
static void Makegraph(int n){
graph=new HashSet[n];
for(int i=0;i<n;i++){
graph[i]=new HashSet<>();
}
}
static void addEdge(int a,int b,int c){
graph[a].add(new Pair(b,c));
}
/*********************************************PAIR********************************************************/
static class PairComp implements Comparator<Pair>{
public int compare(Pair p1,Pair p2){
if(p1.u!=p2.u){
return p1.u-p2.u;
}
else{
return p1.u-p2.u;
}
}
}
static class Pair implements Comparable<Pair> {
int u;
int v;
int index=-1;
public Pair(int u, int v) {
this.u = u;
this.v = v;
}
public int hashCode() {
int hu = (int) (u ^ (u >>> 32));
int hv = (int) (v ^ (v >>> 32));
return 31 * hu + hv;
}
public boolean equals(Object o) {
Pair other = (Pair) o;
return u == other.u && v == other.v;
}
public int compareTo(Pair other) {
if(index!=other.index)
return Long.compare(index, other.index);
return Long.compare(v, other.v)!=0?Long.compare(v, other.v):Long.compare(u, other.u);
}
public String toString() {
return "[u=" + u + ", v=" + v + "]";
}
}
/******************************************Long Pair*************************************************/
static class PairCompL implements Comparator<Pairl>{
public int compare(Pairl p1,Pairl p2){
long aa=p2.v-p1.v;
if(aa<0){
return -1;
}
else if(aa>0){
return 1;
}
else{
return 0;
}
}
}
static class Pairl implements Comparable<Pairl> {
long u;
long v;
int index=-1;
public Pairl(long u, long v) {
this.u = u;
this.v = v;
}
public int hashCode() {
int hu = (int) (u ^ (u >>> 32));
int hv = (int) (v ^ (v >>> 32));
return 31 * hu + hv;
}
public boolean equals(Object o) {
Pair other = (Pair) o;
return u == other.u && v == other.v;
}
public int compareTo(Pairl other) {
if(index!=other.index)
return Long.compare(index, other.index);
return Long.compare(v, other.v)!=0?Long.compare(v, other.v):Long.compare(u, other.u);
}
public String toString() {
return "[u=" + u + ", v=" + v + "]";
}
}
/*****************************************DEBUG***********************************************************/
public static void debug(Object... o) {
if(!oj)
System.out.println(Arrays.deepToString(o));
}
/************************************MODULAR EXPONENTIATION***********************************************/
static long modulo(long a,long b,long c) {
long x=1;
long y=a;
while(b > 0){
if(b%2 == 1){
x=(x*y)%c;
}
y = (y*y)%c; // squaring the base
b /= 2;
}
return x%c;
}
/********************************************GCD**********************************************************/
static long gcd(long x, long y)
{
if(x==0)
return y;
if(y==0)
return x;
long r=0, a, b;
a = (x > y) ? x : y; // a is greater number
b = (x < y) ? x : y; // b is smaller number
r = b;
while(a % b != 0)
{
r = a % b;
a = b;
b = r;
}
return r;
}
/******************************************SIEVE**********************************************************/
static void sieveMake(int n){
sieve=new boolean[n];
Arrays.fill(sieve,true);
sieve[0]=false;
sieve[1]=false;
for(int i=2;i*i<n;i++){
if(sieve[i]){
for(int j=i*i;j<n;j+=i){
sieve[j]=false;
}
}
}
primes=new ArrayList<Integer>();
for(int i=0;i<n;i++){
if(sieve[i]){
primes.add(i);
}
}
}
/********************************************End***********************************************************/
} | JAVA |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | MOD = 10**9 + 7
l, r = map(int, input().split())
def func(x, y):
if y == 0:
return 1
dp = [[0 for _ in range(6)] for _ in range(61)]
dp[60][0] = 1
for i in range(59, -1, -1):
if (y>>i) & 1 == 0 and (x>>i) & 1 == 0:
dp[i][0] = dp[i+1][0]
dp[i][1] = dp[i+1][1]
dp[i][2] = dp[i+1][2]
dp[i][3] = (dp[i+1][3]*2) % MOD
dp[i][4] = dp[i+1][4]
dp[i][5] = (dp[i+1][1] + dp[i+1][3] + dp[i+1][5]*3) % MOD
elif (y>>i) & 1 == 1 and (x>>i) & 1 == 1:
dp[i][0] = 0
dp[i][1] = 0
dp[i][2] = (dp[i+1][0] + dp[i+1][2]) % MOD
dp[i][3] = (dp[i+1][1] + dp[i+1][3]) % MOD
dp[i][4] = (dp[i+1][4]*2) % MOD
dp[i][5] = (dp[i+1][4] + dp[i+1][5]*3) % MOD
elif (y>>i) & 1 == 1 and (x>>i) & 1 == 0:
dp[i][0] = 0
dp[i][1] = (dp[i+1][0] + dp[i+1][1]) % MOD
dp[i][2] = dp[i+1][2]
dp[i][3] = (dp[i+1][2] + dp[i+1][3]*2) % MOD
dp[i][4] = (dp[i+1][0] + dp[i+1][2] + dp[i+1][4]*2) % MOD
dp[i][5] = (dp[i+1][1] + dp[i+1][3] + dp[i+1][4] + dp[i+1][5]*3) % MOD
elif (y>>i) & 1 == 0 and (x>>i) & 1 == 1:
dp[i][0] = 0
dp[i][1] = 0
dp[i][2] = 0
dp[i][3] = (dp[i+1][1] + dp[i+1][3]) % MOD
dp[i][4] = dp[i+1][4]
dp[i][5] = (dp[i+1][5]*3) % MOD
return (sum(dp[0]))%MOD
print(func(l, r)) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
ll L,R;
ll f[64][2][2][2][2];
int l[100],r[100];
ll dfs(int x,int fl,int fr,int s,int ht)
{
if(x<0)return 1;
// x：第x位 （二进制转十进制下权值是1的为第0位）
// fl： y之前是否有填过小于 R的某一位的数字 例如：R:11000 y:10??? fl即为1 表示后面数字可以往大了填
// fr： x之前是否有填过大于 L的某一位的数字 类似fl
// s： 之前x是否有某一位 < y的某一位 如果有，后面的位数就没有相对大小限制了 (确保 x<=y)
//ht：确保 x，y 最高位一样都为1 （满足推导的要求）
ll &ans=f[x][fl][fr][s][ht];
if(ans)return ans;
for(int i=0;i<2;i++)//y
{
for(int j=0;j<2;j++)//x
{
int tl=fl,tr=fr,ts=s,tt=ht;
if(!tt&&i)if(i!=j)continue;
if((j&i)!=j)continue;
if(!tl&&j<l[x])continue;
if(!tr&&i>r[x])continue;
if(!ts&&j>i)continue;
if(j>l[x])tl=1;if(i<r[x])tr=1;if(j<i)ts=1;if(j)tt=1;
ans+=dfs(x-1,tl,tr,ts,tt);
}
}
ans%=mod;
return ans;
}
int main()
{
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
cin>>L>>R;
for(int i=63;i>=0;i--)
{
l[i]=(bool)((1ll<<i)&L);r[i]=(bool((1ll<<i)&R));
}
cout<<dfs(63,0,0,0,0);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
#define int long long
using namespace std;
const int MOD=1e9+7;
int L,R,Mx[101],Mi[101],dp[101][2][2][2];
int DFS(int pos,bool limit,bool lead,bool is) {
if(pos==0)return !lead;
if(dp[pos][limit][lead][is]!=-1)return dp[pos][limit][lead][is];
int up=limit?Mx[pos]:1;
int tmp=0;
if(is||Mi[pos]==0)tmp+=DFS(pos-1,limit&&Mx[pos]==0,lead,is);
if(up){
tmp+=DFS(pos-1,limit,false,is||(Mi[pos]==0));
if(!lead&&(is||Mi[pos]==0))tmp+=DFS(pos-1,limit&&Mx[pos]==1,false,is);
}
tmp%=MOD;
dp[pos][limit][lead][is]=tmp;
return tmp;
}
int solve() {
while(R)Mx[++Mx[0]]=R&1,R=R>>1;
while(L)Mi[++Mi[0]]=L&1,L=L>>1;
// for(int i=1; i<=Mx[0]; i++)cout<<Mx[i]<<" ";
// cout<<endl;
// for(int i=1; i<=Mi[0]; i++)cout<<Mi[i]<<" ";
// cout<<endl;
return DFS(Mx[0],true,true,false);
}
signed main() {
memset(dp,-1,sizeof(dp));
scanf("%lld %lld",&L,&R);
cout<<solve();
// for(int i=1; i<=Mx[0]; i++)cout<<dp[i][0]<<" "<<dp[i][1]<<endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import sys
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 7)
from functools import lru_cache
MOD = 10 ** 9 + 7
"""
桁数が異なると矛盾
したがって商が1であることが必要
すると、xがyのsubsetであることが必要
一番上の桁を含むsubsetがx
"""
@lru_cache()
def F_naive(L,R):
#
ret = 0
ret += F((L+1)//2,(R+1)//2) # 2x,2y
answer = 0
for x in range(L,R+1):
for y in range(x,R+1):
if x.bit_length() == y.bit_length() and (x^y)+x == y:
answer += 1
return answer
@lru_cache()
def F(L,R):
# x subset y かつ 2x > y
if L < 0:
L = 0
if R < L:
return 0
if R == 0:
return 0
ret = 0
# 下一桁が0,0ととる場合
ret += F((L+1)//2, R//2)
# 下一桁が0,1ととる場合
# 2(2x) > (2y+1) iff 2x > y
ret += F((L+1)//2,(R-1)//2)
# 下一桁が1,0ととる場合
# これはsubsetにならない
# 下一桁が1,1ととる場合
# 2(2x+1) > 2y+1 iff 2x >= y
ret += G(L//2,(R-1)//2)
return ret
@lru_cache()
def G(L,R):
# x subset y かつ 2x >= y
if L < 0:
L = 0
if R < L:
return 0
if R == 0:
return 1
ret = 0
# 下一桁が0,0ととる場合
ret += G((L+1)//2, R//2)
# 下一桁が0,1ととる場合
# 2(2x) >= (2y+1) iff 2x > y
ret += F((L+1)//2,(R-1)//2)
# 下一桁が1,0ととる場合
# これはsubsetにならない
# 下一桁が1,1ととる場合
# 2(2x+1) >= 2y+1 iff 2x >= y
ret += G(L//2,(R-1)//2)
return ret
L,R = map(int,input().split())
answer = F(L,R) % MOD
print(answer) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int OO = 1e9;
const int MOD = 1e9 + 7;
LL l, r;
LL dp[64][2][2][2];
LL dfs(int t, int x, int y, int z) {
if (t == -1) return 1;
LL& ans = dp[t][x][y][z];
if (ans != -1) return ans;
ans = 0;
for (int a = 0; a < 2; ++a) {
for (int b = 0; b <= a; ++b) {
int t1 = x, t2 = y, t3 = z;
if (x == 1 && (1LL << t) * a > ((1LL << t) & r))
continue;
if (y == 1 && (1LL << t) * b < ((1LL << t) & l))
continue;
if (z == 0 && a != b)
continue;
if ((1LL << t) * a < ((1LL << t) & r))
x = 0;
if ((1LL << t) * b > ((1LL << t) & l))
y = 0;
if (z == 0 && a == b && a == 1)
z = 1;
ans += dfs(t - 1, x, y, z);
ans %= MOD;
x = t1, y = t2, z = t3;
}
}
return ans;
}
void solve() {
scanf("%lld%lld", &l, &r);
memset(dp, -1, sizeof(dp));
printf("%lld\n", dfs(63, 1, 1, 0));
}
int32_t main() {
// #ifndef ONLINE_JUDGE
// freopen("input.txt", "r", stdin);
// clock_t begin = clock();
// #endif
solve();
// #ifndef ONLINE_JUDGE
// clock_t end = clock();
// printf("ELAPSED TIME: %f\n", (double) (end - begin) / CLOCKS_PER_SEC);
// #endif
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
#define pb push_back
#define ll long long
#define ii pair<int,int>
#define st first
#define nd second
#define N 200005
#define inf 1000000000
#define MOD 1000000007
#define sz(x) ((int)x.size())
#define umax(x,y) x=max(x,y)
#define umin(x,y) x=min(x,y)
using namespace std;
int dp[62][2][2][2];
int mul(int x,int y) {
return (ll)x*y%MOD;
}
int add(int x,int y) {
x+=y;
if(x>=MOD) x-=MOD;
if(x<0) x+=MOD;
return x;
}
int solve(ll l,ll r) {
dp[61][0][0][0]=1;
for(int i=61;i>0;i--) {
for(int a=0;a<2;a++) {
for(int b=0;b<2;b++) {
for(int c=0;c<2;c++) {
// put 1,1
if(a || (r>>(i-1))&1)
dp[i-1][a][b || !((l>>(i-1))&1)][1]=add(dp[i-1][a][b || !((l>>(i-1))&1)][1],dp[i][a][b][c]);
// put 1,0
if(c && (b || !((l>>(i-1))&1)) && (a || (r>>(i-1))&1))
dp[i-1][a][b][c]=add(dp[i-1][a][b][c],dp[i][a][b][c]);
// put 0,0
if((b || !((l>>(i-1))&1)))
dp[i-1][(a || (r>>(i-1))&1)][b][c]=add(dp[i-1][(a || (r>>(i-1))&1)][b][c],dp[i][a][b][c]);
}
}
}
}
int res=0;
for(int a=0;a<2;a++) {
for(int b=0;b<2;b++) {
res=add(res,dp[0][a][b][1]);
}
}
return res;
}
int main(int arg,char* argv[]) {
ll l,r;
cin>>l>>r;
cout<<solve(l,r);
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define finish(x) return cout << x << endl, 0
#define ll long long
ll l, r;
int dp[61][2][2][2], mod = 1e9 + 7;
int solve(int bit, bool a, bool b, bool s){
if(bit == -1) return 1;
int &ret = dp[bit][a][b][s];
if(ret != -1) return ret;
ret = 0;
// 0 0
if(a || ((l >> bit) & 1) == 0) ret += solve(bit - 1, a, b || ((r >> bit) & 1), s);
if(ret >= mod) ret -= mod;
// 0 1
if(s && (a || ((l >> bit) & 1) == 0) && (b || ((r >> bit) & 1) == 1)) ret += solve(bit - 1, a, b, s);
if(ret >= mod) ret -= mod;
// 1 1
if(b || ((r >> bit) & 1) == 1) ret += solve(bit - 1, a || ((l >> bit) & 1) == 0, b, 1);
if(ret >= mod) ret -= mod;
return ret;
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
memset(dp, -1, sizeof dp);
cin >> l >> r;
cout << solve(60, 0, 0, 0) << endl;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#include <iostream>
using namespace std;
typedef long long ll;
ll L, R;
ll mod = 1000000007;
ll dp[63][2][2][2]; //digit, deviated from l, deviated from r, initial bit has been set
ll solve(int dig, int x, int y, int firstBit){
if (dig == -1) return 1LL;
if (dp[dig][x][y][firstBit] != -1) return dp[dig][x][y][firstBit];
dp[dig][x][y][firstBit] = 0;
if (x == 1 || !(1LL<<dig & L)){ // x 0, y 0
dp[dig][x][y][firstBit] += (1LL<<dig & R) ? solve(dig-1, x, 1, firstBit) : solve(dig-1, x, y, firstBit);
}
if ((x == 1 || !(1LL<<dig & L)) && (y == 1 || (1LL<<dig & R)) && firstBit){ // x 0, y 1
dp[dig][x][y][firstBit] += solve(dig-1, x, y, firstBit);
}
if (y == 1 || (1LL<<dig & R)){ // x 1, y 1
dp[dig][x][y][firstBit] += !(1LL<<dig & L) ? solve(dig-1, 1, y, 1) : solve(dig-1, x, y, 1);
}
dp[dig][x][y][firstBit] %= mod;
return dp[dig][x][y][firstBit];
}
int main() {
cin >> L >> R;
memset(dp, -1, sizeof(dp));
cout << solve(62, 0, 0, 0) << endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int mod = 1000*1000*1000 + 7;
int memo[60][2][2][2];
ll L, R;
void add_self(int& a, int b) { a += b; if(a >= mod) a -= mod; }
int dp(int pos, int flagX, int flagY, int flagZ) {
if(pos == -1) {
return 1;
}
int& ans = memo[pos][flagX][flagY][flagZ];
if(ans != -1) {
return ans;
}
ans = 0;
// x 0, y 0
if(flagX || (L>>pos & 1) == 0) {
add_self(ans, dp(pos-1, flagX, (R>>pos & 1) ? 1 : flagY, flagZ));
}
// x 0, y 1
if((flagX || (L>>pos & 1) == 0) && (flagY || (R>>pos & 1)) && flagZ) {
add_self(ans, dp(pos-1, flagX, flagY, flagZ));
}
// x 1, y 1
if(flagY || (R>>pos & 1)) {
add_self(ans, dp(pos-1, (L>>pos & 1) == 0 ? 1 : flagX, flagY, 1));
}
return ans;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
memset(memo, -1, sizeof memo);
cin >> L >> R;
cout << dp(59, 0, 0, 0) << endl;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define f first
#define s second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
const int N = (int)2e5 + 123, inf = 1e9, mod = 1e9 + 7;
const ll INF = 1e18;
ll l, r, p3[100], ans;
ll solvex(ll bit, ll x, ll l){
if(x >= l)
return p3[bit + 1];
if(x + (1ll << (bit + 1)) - 1 < l)
return 0;
return (solvex(bit - 1, x, l) * 2 + solvex(bit - 1, x | (1ll << bit), l)) % mod;
}
ll solvey(ll bit, ll y, ll r){
if(y + (1ll << (bit + 1)) - 1 <= r)
return p3[bit + 1];
if(y > r)
return 0;
return (solvey(bit - 1, y, r) + solvey(bit - 1, y | (1ll << bit), r) * 2) % mod;
}
ll solve(ll bit, ll x, ll y, ll l, ll r){
if(x + (1ll << (bit + 1)) - 1 < l)
return 0;
if(y > r)
return 0;
if(y + (1ll << (bit + 1)) - 1 <= r)
return solvex(bit, x, l);
if(x >= l)
return solvey(bit, y, r);
return (solve(bit - 1, x, y, l, r) + solve(bit - 1, x , y | (1ll << bit), l, r) + solve(bit - 1, x | (1ll << bit), y | (1ll << bit), l, r)) % mod;
}
int main()
{
p3[0] = 1;
for(int i = 1; i < 100; i++)
p3[i] = p3[i - 1] * 3 % mod;
cin >> l >> r;
for(int i = 0; i <= 60; i++)
ans = (ans + solve(i - 1, (1ll << i), (1ll << i), l, r)) % mod;
cout << ans;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 |
L,R = map(int,input().split())
dp = []
for i in range(62):
nowA = []
for i in range(2):
nowB = []
for i in range(2):
nowC = []
for i in range(2):
nowC.append(0)
nowB.append(nowC)
nowA.append(nowB)
dp.append(nowA)
dp[0][0][0][0] = 1
#print (dp[3][0][0][0])
for i in range(61):
i += 1
keta = 62-i
lb = (L >> (keta-1)) % 2
rb = (R >> (keta-1)) % 2
for j in range(2):
for k in range(2):
for m in range(2):
pre = dp[i-1][j][k][m]
for x in range(2):
for y in range(2):
nj = j
nk = k
nm = m
if x == 1 and y == 0:
continue;
if x != y and m == 0:
continue;
if j == 0 and lb == 1 and x == 0:
continue;
if k == 0 and rb == 0 and y == 1:
continue;
if j == 0 and lb == 0 and x == 1:
nj = 1
if k == 0 and rb == 1 and y == 0:
nk = 1
if m == 0 and x == 1 and y == 1:
nm = 1
dp[i][nj][nk][nm] += dp[i-1][j][k][m]
dp[i][nj][nk][nm] %= 10 ** 9 +7
ans = 0
for j in range(2):
for k in range(2):
ans += dp[-1][j][k][1]
print (ans % (10 ** 9 + 7)) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define INF 2e9
#define ll long long
#define MINUS(a) memset(a,0xff,sizeof(a))
ll L, R;
ll mo = 1000000007;
ll dp[2][2][2][61];
/* dp(d, msb, l, r)内のret: x, yのd桁目のより下の値の組み合わせ数
msb: d桁目より上ですでにMSBとなる桁があったかどうか
l : d桁目より上を決めた段階で、L<=xか
r : d桁目より上を決めた段階で、y<=Rか
上を前提にcom()でx, yのd桁目として0, 1を取りうるか見ていく。
*/
ll com(int d, int S, int LT, int RT){
if(d == -1){
return S == 1;
}
/* メモ */
if(dp[S][LT][RT][d]>=0){
return dp[S][LT][RT][d];
}
ll ret = 0;
/* aがxのd桁目、bがyのd桁目 */
int a, b;
for(int a=0; a<2; a++){
for(int b=0; b<2; b++){
/* x>=Lになっているか */
if(a==0 && LT && (L&(1LL<<d))>0){
continue;
}
/* y<=Rになっているか */
if(b==1 && RT && (R&(1LL<<d))==0){
continue;
}
/* y-xで繰り下がりは起こさせない */
if(a>b){
continue;
}
/* MSBがなくてa==1かつb==1はだめ */
if(S==0 && a != b){
continue;
}
int NLT = LT && (((L>>d)&1)==a);
int NRT = RT && (((R>>d)&1)==b);
int NS = S || (a==1 && b==1);
ret += com(d-1, NS, NLT, NRT);
}
}
return dp[S][LT][RT][d] = ret % mo;
}
int main(){
cin>>L>>R;
MINUS(dp);
cout<<com(61, 0, 1, 1)<<endl;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <numeric>
#include <functional>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <memory>
#include <thread>
#include <tuple>
#include <limits>
using namespace std;
constexpr int mod = 1000000007;
map<tuple<int, long long, long long>, int> result;
long long solve(int sz, long long s, long long e) {
if (sz == 1) return 1;
if (sz == 2) return s == e ? 1 : 3;
auto ccheck = result.emplace(make_tuple(sz, s, e), 0);
if (!ccheck.second) { return ccheck.first->second; }
long long len = (1ull << (sz-1));
long long sublen = len / 2;
long long ans = 0;
if (e - s >= sublen) ans += solve(sz - 1, s, e - sublen);
if (s < sublen) ans += solve(sz - 1, s, min(sublen-1, e));
if (e >= sublen) ans += solve(sz - 1, max(0ll, s - sublen), e - sublen);
ccheck.first->second = ans % mod;
return ans % mod;
}
int main() {
long long L, R;
scanf("%lld%lld", &L, &R);
long long ans = 0;
for (int sz = 1; sz <= 63; sz++) {
long long beg = (1ull << (sz - 1));
long long end = (1ull << sz) - 1;
if (beg > R) continue;
if (end < L) continue;
ans += solve(sz, max(0ll, L-beg), min(end - beg, R - beg));
ans %= mod;
}
printf("%lld\n", ans);
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import sys
sys.setrecursionlimit(10**9)
from functools import lru_cache
mod=1000000007
@lru_cache()
def calc(l,r):
if l>r:
return 0
if r==0:
return 1
b_l,b_r=l.bit_length(),r.bit_length()
if b_l==b_r:
return calc(l-(1<<(b_l-1)),r-(1<<(b_r-1)))
if r==(1<<(b_r-1))-1 and l==0:
return pow(3,b_r,mod)
return (calc(l,r-(1<<(b_r-1)))+calc(l,(1<<(b_r-1))-1)+calc(0,r-(1<<b_r-1)))%mod
@lru_cache()
def solve(l,r):
if l>r:
return 0
b_l,b_r=l.bit_length(),r.bit_length()
if b_r>b_l:
return (solve(l,(1<<b_r-1)-1)+calc(0,r-(1<<(b_r-1))))%mod
a=1<<(b_l-1) if l else 0
return calc(l-a,r-a)
l,r=map(int,input().split())
print(solve(l,r)) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L,R = map(int,input().split())
mod = 10**9+7
m = 64 +1
fac = [1]*m
ninv = [1]*m
finv = [1]*m
for i in range(2,m):
fac[i] = fac[i-1]*i%mod
ninv[i] = (-(mod//i)*ninv[mod%i])%mod
finv[i] = finv[i-1]*ninv[i]%mod
def comb(n,k):
return (fac[n]*finv[k]%mod)*finv[n-k]%mod
def f(L,R):
if L>R : return 0
R = bin(R)[2:]
N = len(R)
ret = f(L,int("0"+"1"*(N-1),2))
L = bin(L)[2:]
if len(L) != N : L = "1"+"0"*(N-1)
for i in range(N):
if R[i] == "0" : continue
R2 = R[:i] + "0" + "?"*(N-i-1)
if i==0: R2 = R
for j in range(N):
if L[j] == "1" and j!=0 : continue
L2 = L[:j] + "1" + "?"*(N-j-1)
if j==0 : L2 = L
if L2[0] == "0" : break
tmp = 1
for r,l in zip(R2[1:],L2[1:]):
if r=="0" and l=="1" : tmp *= 0
if r=="?" and l=="?" : tmp *= 3
if r=="?" and l=="0" : tmp *= 2
if r=="1" and l=="?" : tmp *= 2
tmp %= mod
ret += tmp
ret %= mod
return ret%mod
print(f(L,R)) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
using namespace std;
typedef long long ll;
const int LGN = 63;
const int Z = 1e9+7;
int main() {
ios::sync_with_stdio(false);
ll l, r;
cin >> l >> r;
int p3[LGN];
for (int i = 0; i < LGN; i++) {
p3[i] = i == 0 ? 1 : 3ll * p3[i-1] % Z;
}
int dp[LGN+1][5];
dp[0][0] = dp[0][1] = dp[0][2] = dp[0][3] = dp[0][4] = 1;
for (int i = 1; i <= LGN; i++) {
bool bl = l >> (i - 1) & 1, br = r >> (i - 1) & 1;
if (!bl && !br) {
dp[i][0] = dp[i-1][0];
dp[i][1] = dp[i-1][1];
} else if (!bl && br) {
dp[i][0] = (dp[i-1][2] + dp[i-1][4]) % Z;
dp[i][1] = (0ll + dp[i-1][3] + dp[i-1][4] + dp[i-1][1]) % Z;
} else if (bl && !br) {
dp[i][0] = dp[i][1] = 0;
} else {
dp[i][0] = dp[i][1] = dp[i-1][1];
}
if (!bl) {
dp[i][2] = (dp[i-1][2] + p3[i-1]) % Z;
dp[i][3] = (2ll * dp[i-1][3] + p3[i-1]) % Z;
} else {
dp[i][2] = dp[i][3] = dp[i-1][3];
}
if (!br) {
dp[i][4] = dp[i-1][4];
} else {
dp[i][4] = (2ll * dp[i-1][4] + p3[i-1]) % Z;
}
}
cout << dp[LGN][0] << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
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);
FCoincidence solver = new FCoincidence();
solver.solve(1, in, out);
out.close();
}
static class FCoincidence {
long l;
long r;
long[][][][] dp = new long[2][2][2][60];
public void solve(int testNumber, InputReader in, OutputWriter out) {
l = in.readLong();
r = in.readLong();
ArrayUtils.fill(dp, -1);
out.printLine(rec(0, 0, 0, 59));
}
long rec(int x, int y, int z, int i) {
if (i == -1) return 1;
if (dp[x][y][z][i] >= 0) return dp[x][y][z][i];
long ret = 0;
boolean flg1 = (l >> i & 1) == 0;
boolean flg2 = (r >> i & 1) == 1;
if (x == 1 || flg1) ret += rec(x, flg2 ? 1 : y, z, i - 1);
if ((x == 1 || flg1) && (y == 1 || flg2) && z == 1) ret += rec(x, y, z, i - 1);
if (y == 1 || flg2) ret += rec(flg1 ? 1 : x, y, 1, i - 1);
ret %= MiscUtils.MOD7;
return dp[x][y][z][i] = ret;
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public long readLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static class ArrayUtils {
public static void fill(long[][] array, long value) {
for (long[] row : array) {
Arrays.fill(row, value);
}
}
public static void fill(long[][][] array, long value) {
for (long[][] row : array) {
fill(row, value);
}
}
public static void fill(long[][][][] array, long value) {
for (long[][][] row : array) {
fill(row, value);
}
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void close() {
writer.close();
}
public void printLine(long i) {
writer.println(i);
}
}
static class MiscUtils {
public static final int MOD7 = (int) (1e9 + 7);
}
}
| JAVA |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
typedef long long ll;
ll dp [61][2][2][2];
int main() {
cin.tie(nullptr);
ios::sync_with_stdio(false);
cout << setprecision(20);
ll l, r;
cin >> l >> r;
constexpr ll mod = 1e9 + 7;
dp[60][0][0][0] = 1;
for (int i = 59; i >= 0; --i) {
int lb = (l >> i) & 1;
int rb = (r >> i) & 1;
rep(j, 2) {
rep(k, 2) {
rep(s, 2) {
rep(x, 2) {
ll pre = dp[i+1][j][k][s];
rep(y, 2) {
if (x && !y) continue;
int nj = j, nk = k, ns = s;
if (!s && x != y) continue;
if (x && y) ns = 1;
if (!j && !x && lb) continue;
if (x && !lb) nj = 1;
if (!k && y && !rb) continue;
if (!y && rb) nk = 1;
dp[i][nj][nk][ns] += pre;
dp[i][nj][nk][ns] %= mod;
}
}
}
}
}
}
ll ans = 0;
rep(j, 2) {
rep(k, 2) {
rep(s, 2) {
ans += dp[0][j][k][s];
ans %= mod;
}
}
}
cout << ans << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L, R = map(int, input().split())
MOD = 10 ** 9 + 7
l = '{:060b}'.format(L)[::-1]
r = '{:060b}'.format(R)[::-1]
memo = [[[[-1 for l in range(2)] for k in range(2)] for j in range(2)] for i in range(60)]
def f(pos, flagX, flagY, flagZ):
if pos == -1:
return 1
if memo[pos][flagX][flagY][flagZ] != -1:
return memo[pos][flagX][flagY][flagZ]
ret = 0
# x 0, y 0
if flagX or l[pos] == '0':
ret += f(pos - 1, flagX, 1 if r[pos] == '1' else flagY, flagZ)
# x 0, y 1
if (flagX or l[pos] == '0') and (flagY or r[pos] == '1') and flagZ:
ret += f(pos - 1, flagX, flagY, flagZ)
# x 1, y 1
if flagY or r[pos] == '1':
ret += f(pos - 1, 1 if l[pos] == '0' else flagX, flagY, 1)
ret %= MOD
memo[pos][flagX][flagY][flagZ] = ret
return ret
ans = f(59, 0, 0, 0)
print(ans)
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include "bits/stdc++.h"
using namespace std;
typedef long long ll;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);
#define l_ength size
void mul_mod(ll& a, ll b){
a *= b;
a %= MOD;
}
void add_mod(ll& a, ll b){
a = (a<MOD)?a:(a-MOD);
b = (b<MOD)?b:(b-MOD);
a += b;
a = (a<MOD)?a:(a-MOD);
}
ll dp[63][2][2][2][2][2];
int a[63],b[63],f[10];
int main(void){
int i,j,k;
ll l,r,ans=0ll;
cin >> l >> r;
for(i=0; i<60; ++i){
b[i] = l&1; l >>= 1;
a[i] = r&1; r >>= 1;
}
reverse(a,a+60);
reverse(b,b+60);
dp[0][1][1][1][0][1] = 1ll;
for(i=0; i<60; ++i){
for(j=0; j<128; ++j){
for(k=0; k<7; ++k){
f[k] = ((j&(1<<k))?1:0);
}
if(f[5] == 0 && f[6] == 1){
continue;
}
if(f[0] && f[5]<f[6]){
continue;
}
if(f[1] && f[5]>a[i]){
continue;
}
if(f[2] && f[6]<b[i]){
continue;
}
if(f[4] && f[3]>f[6]){
continue;
}
add_mod(dp[i+1][((f[0]&&(f[5]==f[6]))?1:0)][((f[1]&&(f[5]==a[i]))?1:0)][((f[2]&&(f[6]==b[i]))?1:0)][f[5]][((f[4] && (f[3]==f[6]))?1:0)],dp[i][f[0]][f[1]][f[2]][f[3]][f[4]]);
}
}
for(j=0; j<32; ++j){
for(k=0; k<5; ++k){
f[k] = (j&(1<<k))?1:0;
}
if(f[4]){
continue;
}
add_mod(ans,dp[60][f[0]][f[1]][f[2]][f[3]][f[4]]);
}
cout << ans << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import java.io.IOException;
import java.io.InputStream;
import java.util.NoSuchElementException;
public class Main {
static long[][][][] dp = new long[64][2][2][2];
static long l, r;
public static void main(String[] args) {
FastScanner fsc = new FastScanner();
l = fsc.nextLong();
r = fsc.nextLong();
for (int i = 0; i < 64; i++) {
for (int gt = 0; gt < 2; gt++) {
for (int lt = 0; lt < 2; lt++) {
for (int msb = 0; msb < 2; msb++) {
dp[i][gt][lt][msb] = -1;
}
}
}
}
System.out.println(dp(0, 0, 0, 0));
}
public static long dp(int i, int gt, int lt, int msb) {
if (i >= 64) {
return 1;
}
if (dp[i][gt][lt][msb] >= 0) {
return dp[i][gt][lt][msb];
}
long ret = 0;
boolean x0 = gt == 1 || ((1l << (63 - i)) & l) == 0;
boolean y1 = lt == 1 || ((1l << (63 - i)) & r) != 0;
// x: 0, y: 0
if (x0) {
ret += dp(i + 1, gt, ((1l << (63 - i)) & r) != 0 ? 1 : lt, msb);
}
// x: 0, y: 1
if (msb == 1 && x0 && y1) {
ret += dp(i + 1, gt, lt, msb);
}
// x: 1, y: 1
if (y1) {
ret += dp(i + 1, ((1l << (63 - i)) & l) == 0 ? 1 : gt, lt, 1);
}
return dp[i][gt][lt][msb] = ret % Const.MOD;
}
}
class FastScanner {
private final InputStream in = System.in;
private final byte[] buffer = new byte[1024];
private int ptr = 0;
private int buflen = 0;
private boolean hasNextByte() {
if (ptr < buflen) {
return true;
} else {
ptr = 0;
try {
buflen = in.read(buffer);
} catch (IOException e) {
e.printStackTrace();
}
if (buflen <= 0) {
return false;
}
}
return true;
}
private int readByte() {
if (hasNextByte()) {
return buffer[ptr++];
} else {
return -1;
}
}
private static boolean isPrintableChar(int c) {
return 33 <= c && c <= 126;
}
public boolean hasNext() {
while (hasNextByte() && !isPrintableChar(buffer[ptr])) {
ptr++;
}
return hasNextByte();
}
public String next() {
if (!hasNext()) {
throw new NoSuchElementException();
}
StringBuilder sb = new StringBuilder();
int b = readByte();
while (isPrintableChar(b)) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public long nextLong() {
if (!hasNext()) {
throw new NoSuchElementException();
}
long n = 0;
boolean minus = false;
int b = readByte();
if (b == '-') {
minus = true;
b = readByte();
}
if (b < '0' || '9' < b) {
throw new NumberFormatException();
}
while (true) {
if ('0' <= b && b <= '9') {
n *= 10;
n += b - '0';
} else if (b == -1 || !isPrintableChar(b)) {
return minus ? -n : n;
} else {
throw new NumberFormatException();
}
b = readByte();
}
}
public int nextInt() {
long nl = nextLong();
if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) {
throw new NumberFormatException();
}
return (int) nl;
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
class Const {
public static final long MOD7 = 1_000_000_007;
public static final long MOD9 = 1_000_000_009;
public static final long MOD99 = 998_244_353;
public static final long LINF = 1_000_000_000_000_000_000l;
public static final int IINF = 1_000_000_000;
public static final double DINF = 1e150;
public static final double DELTA = 0.000_000_000_1;
public static final double LDELTA = 0.000_001;
public static final String YES = "YES";
public static final String NO = "NO";
public static final String Yes = "Yes";
public static final String No = "No";
public static final String POSSIBLE = "POSSIBLE";
public static final String IMPOSSIBLE = "IMPOSSIBLE";
public static final String Possible = "Possible";
public static final String Impossible = "Impossible";
public static final int[] dx8 = {1, 0, -1, 0, 1, -1, -1, 1};
public static final int[] dy8 = {0, 1, 0, -1, 1, 1, -1, -1};
public static final int[] dx = {1, 0, -1, 0};
public static final int[] dy = {0, 1, 0, -1};
public static long MOD = MOD7;
public static void setMod(long mod) {
MOD = mod;
}
private Const(){}
}
| JAVA |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | mod = 10 ** 9 + 7
memo = {}
def solve(L, R):
if (L, R) in memo:
return memo[(L, R)]
if L > R:
res = 0
elif L == 1:
res = 1 + solve(2, R)
else:
res = (
solve(L // 2, (R - 1) // 2) +
solve((L + 1) // 2, R // 2) +
solve((L + 1) // 2, (R - 1) // 2)
)
res %= mod
memo[(L, R)] = res
return res
L, R = map(int, input().split())
print(solve(L, R) % mod) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L, R = map(int, raw_input().split())
"""
R ????????1??????
y 0001????0??????
^d ^rd
x 0001???????1??
^ld
L ???????????0???
"""
ans = 0
for d in range(70):
if d<len(bin(L))-3:
LD = []
elif d==len(bin(L))-3:
LD = [i for i in range(-1,d) if i==-1 or (L>>i&1)==0]
else:
LD = [d]
if d<len(bin(R))-3:
RD = [d]
elif d==len(bin(R))-3:
RD = [i for i in range(-1,d) if i==-1 or (R>>i&1)==1]
else:
RD = []
for ld in LD:
for rd in RD:
a = 1
for i in range(d):
# i<ld: x==0 or 1
# i==ld: x==1
# i>ld: x==L
# i<rd: y==0 or 1
# i==rd: y==0
# i>rd: y==R
# y>=x
if i<ld and i<rd:
a *= 3
if i<ld and i==rd:
a *= 1
if i<ld and i>rd:
a *= (R>>i&1)+1
if i==ld and i<rd:
a *= 1
if i==ld and i==rd:
a *= 0
if i==ld and i>rd:
a *= (R>>i&1)
if i>ld and i<rd:
a *= 2-(L>>i&1)
if i>ld and i==rd:
if (L>>i&1)==0:
a *= 1
else:
a *= 0
if i>ld and i>rd:
if (L>>i&1)==(R>>i&1):
a *= 1
if (L>>i&1)==0 and (R>>i&1)==1:
a *= 1
if (L>>i&1)==1 and (R>>i&1)==0:
a *= 0
ans += a
print ans%(10**9+7)
| PYTHON |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000
pair<long long,long long> get(long long Lx,long long Rx,int x,long long d){
long long lx,rx;
if(x==0){
if((Lx>>d)&1LL)return {-1,-1};
else lx = Lx;
if((Rx>>d)&1LL)rx = (1LL<<d)-1LL;
else rx=Rx;
}
else{
if((Lx>>d)&1LL)lx=Lx^(1LL<<d);
else lx=0LL;
if((Rx>>d)&1LL)rx=Rx^(1LL<<d);
else return {-1,-1};
}
return {lx,rx};
}
int Get(long long Lx,long long Rx,long long Ly,long long Ry,long long d,int f){
if(d<0){
if(Lx==0&&Rx==0&&Ly==0&&Ry==0)return 1;
else return 0;
}
static map<vector<long long>,int> mp;
vector<long long> args = {Lx,Rx,Ly,Ry,d,f};
if(mp.count(args))return mp[args];
int ret = 0;
for(int y=0;y<2;y++){
for(int x=0;x<2;x++){
if(y==0&&x==1)continue;
if((f==0)&&x!=y)continue;
auto [lx,rx] = get(Lx,Rx,x,d);
auto [ly,ry] = get(Ly,Ry,y,d);
if(lx==-1||rx==-1)continue;
if(x==1&&y==1)ret = mod(ret +Get(lx,rx,ly,ry,d-1,1));
else ret = mod(ret + Get(lx,rx,ly,ry,d-1,f));
}
}
mp[args] = ret;
return ret;
}
int main(){
long long L,R;
cin>>L>>R;
int ans = Get(L,R,L,R,60,0);
cout<<ans<<endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define LL long long
#define mo 1000000007
LL L,R;int f[61][2][2][2];
int cal(int x,int a,int b,int c){
/* a±íʾǰÃæÊÇ·ñÓÐijλx>L
b±íʾÊÇ·ñÓÐijλy<R
c±íʾǰÃæÊÇ·ñÓÐ1
*/
if(x<0)return 1;
if(f[x][a][b][c])return f[x][a][b][c];
int l=(L>>x)&1,r=(R>>x)&1,&ans=f[x][a][b][c];
if(a)l=0;if(b)r=1;
for(int i=l;i<2;i++)for(int j=i;j<=r;j++){
if(c)(ans+=cal(x-1,a|(i>l),b|(j<r),c))%=mo;
else if((i^j)==0)(ans+=cal(x-1,a|(i>l),b|(j<r),(i==1)))%=mo;
}
return ans;
}
int main(){
scanf("%lld%lld",&L,&R);
printf("%d",cal(60,0,0,0));
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
const long mod = 1000000007;
int main()
{
long l, r;
cin >> l >> r;
long dp[61][2][2][2] = { }; // 上からi桁確定、L<=x確定?、y<=R確定?、桁数同じ?
dp[0][0][0][0] = 1;
for (int i = 1; i <= 60; i++) { // 上からi桁確定
for (int j = 0; j < 2; j++) { // L <= x ?
for (int k = 0; k < 2; k++) { // y <= R ?
for (int s = 0; s < 2; s++) { // 桁数同じ?
for (int x = 0; x < 2; x++) { // xの上からi桁目
for (int y = 0; y < 2; y++) { // yの上からi桁目
if (x == 1 && y == 0) continue;
if (s == 0 && x != y) continue; // 桁数が合わない
if (j == 0 && (l >> (60 - i)) % 2 == 1 && x == 0) continue; // L > x
if (k == 0 && (r >> (60 - i)) % 2 == 0 && y == 1) continue; // y > R
int jnew = j, knew = k, snew = s;
if ((l >> (60 - i)) % 2 == 0 && x == 1) jnew = 1;
if ((r >> (60 - i)) % 2 == 1 && y == 0) knew = 1;
if (x == 1 && y == 1) snew = 1;
dp[i][jnew][knew][snew] += dp[i - 1][j][k][s];
dp[i][jnew][knew][snew] %= mod;
}
}
}
}
}
}
long ans = 0;
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
for (int s = 0; s < 2; s++) {
ans = (ans + dp[60][j][k][s]) % mod;
}
}
}
cout << ans << endl;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | # coding: utf-8
# Your code here!
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline #文字列入力のときは注意
#n = int(input())
l,r = [int(i) for i in readline().split()]
def make_dp(init, size):
res = "[{}]*{}".format(init,size[-1])
for i in reversed(size[:-1]):
res = "[{} for _ in [0]*{}]".format(res,i)
return eval(res)
MOD = 10**9 + 7
R = bin(r)[2:]
L = bin(l)[2:]
L = "0"*(len(R)-len(L)) + L
dp = make_dp(0,(len(R)+1,2,2,2))
dp[0][0][0][0] = 1
for i, (r,l) in enumerate(zip(R,L)): #i桁目からi+1桁目に遷移
ri = int(r)
li = int(l)
for is_less in range(2):
for is_more in range(2):
for is_nonzero in range(2):
for dl in range(0 if is_more else li,2):
for dr in range(dl, 2 if is_less else ri+1): # d: i+1桁目の数字
dp[i+1][is_nonzero or dr != 0][is_more or li < dl][is_less or dr < ri] += dp[i][is_nonzero][is_more][is_less]*(dr==dl or is_nonzero)
dp[i+1][is_nonzero or dr != 0][is_more or li < dl][is_less or dr < ri] %= MOD
#print(dp[-1])
ans = 0
for i in range(2):
for j in range(2):
for k in range(2):
ans += dp[-1][i][j][k]
print(ans%MOD)
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import sys
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7
input=lambda:sys.stdin.readline().rstrip()
from itertools import product
def resolve():
L,R=map(int,input().split())
L=map(int,bin(L)[2:].zfill(60))
R=map(int,bin(R)[2:].zfill(60))
dp=[[[0]*2 for _ in range(2)] for _ in range(2)]
dp[0][0][0]=1 # dp[msb][mt][lt]
for l,r in zip(L,R):
ndp=[[[0]*2 for _ in range(2)] for _ in range(2)]
for msb,mt,lt,x,y in product(range(2),repeat=5):
if((not msb) and (x!=y)): continue
if((not mt) and l>x): continue
if((not lt) and y>r): continue
if(x>y): continue
ndp[max(msb,x*y==1)][max(mt,l<x)][max(lt,y<r)]+=dp[msb][mt][lt]
ndp[max(msb,x*y==1)][max(mt,l<x)][max(lt,y<r)]%=MOD
dp=ndp
print(sum(dp[msb][mt][lt] for msb,mt,lt in product(range(2),repeat=3))%MOD)
resolve() | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;
ll L, R, res, pow3[60];
int main() {
cin >> L >> R;
pow3[0] = 1;
for (int i = 1; i < 60; ++i) {
pow3[i] = pow3[i - 1] * 3 % mod;
}
for (ll i = 0; i < 60; ++i) {
if (L <= (1ll << i + 1) - 1 && (1ll << i) <= R) {
ll l = max(L, (1ll << i));
ll r = min(R, (1ll << i + 1) - 1);
ll dp[3];
dp[0] = dp[2] = 0;
dp[1] = 1;
for (ll j = i - 1; j >= 0; --j) {
if (r&(1ll << j)) {
if (l&(1ll << j)) {
res += dp[0] * pow3[j];
dp[0] *= 2;
dp[0] %= mod;
}
else {
res += dp[0] * pow3[j];
res += dp[2] * pow3[j];
dp[0] *= 2;
dp[2] *= 2;
dp[0] += dp[1];
dp[2] += dp[1];
dp[0] %= mod;
dp[2] %= mod;
}
}
else {
if (l&(1ll << j)) {
dp[1] = 0;
}
else {
res += dp[2] * pow3[j];
dp[2] *= 2;
dp[2] %= mod;
}
}
res %= mod;
}
res += dp[0] + dp[1] + dp[2];
}
}
cout << res % mod << endl;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define int ll
#define inf 0x3f3f3f3f
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define ll long long
#define vi vector <int>
#define pii pair <int, int>
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#define rep(i,n) for (int i = 0; i < (int) (n); ++ i)
#define foreach(it,c) for (__typeof(c.begin()) it = c.begin(); it != c.end(); ++ it)
inline int read() {
int x = 0, f = 1, c = getchar();
for (;!isdigit(c);c = getchar()) if (c == '-') f ^= 1;
for (; isdigit(c);c = getchar()) x = x * 10 + c - '0';
return f ? x : -x;
}
string s, t;
string to2(int x) {string str; rep(i, 60) str = char('0' + (x >> i & 1)) + str; return str;}
int dp[60][2][2][2];
int DP(int id, int sml, int lgr, int fst) {
if (id == 60) return 1;
int & rem = dp[id][sml][lgr][fst];
if (~rem) return rem;
int res = 0;
for (int a = lgr ? 0 : s[id] - '0'; a <= (sml ? 1 : t[id] - '0'); ++ a) {
for (int b = lgr ? 0 : s[id] - '0'; b <= (sml ? 1 : t[id] - '0'); ++ b) {
if (!fst && b && !a) continue;
if (a > b) continue;
res += DP(id + 1, sml | b < t[id] - '0', lgr | a > s[id] - '0', fst | b);
if (res >= mod) res -= mod;
}
}
// printf("%d %d %d %d = %d\n", id, sml, lgr, fst, res);
return rem = res;
}
signed main() {
int l, r;
scanf("%lld %lld", &l, &r);
s = to2(l), t = to2(r);
memset(dp, -1, sizeof dp);
printf("%lld\n", DP(0, 0, 0, 0));
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define PII pair<int, int>
#define LL long long
using namespace std;
const int LOG = 65;
const LL MOD = 1000000007;
const LL INF = (LL) 1e9 + 5;
LL L, R, dp[LOG][2][2][2];
bool vis[LOG][2][2][2];
vector<int> d1, d2;
void trans(LL x, vector<int> &d) {
d.clear();
for (int i = 0; i < LOG; i++) {
d.push_back(x % 2);
x /= 2;
}
// cout << "Get:"; for (int x : d) cout << x; cout << '\n';
}
LL dfs(int pos, bool lim1, bool lim2, bool first) {
if (pos < 0) return 1;
if (vis[pos][lim1][lim2][first]) return dp[pos][lim1][lim2][first];
vis[pos][lim1][lim2][first] = true;
int up1 = lim1 ? d1[pos] : 1;
LL res = 0;
for (int i = 0; i <= up1; i++) {
int up2 = lim2 ? min(d2[pos], i) : i;
for (int j = 0; j <= up2; j++) {
if (first && i == 1 && j == 0) continue;
res = (res + dfs(pos - 1, lim1 && i == d1[pos], lim2 && j == d2[pos], first && i == 0)) % MOD;
}
}
return dp[pos][lim1][lim2][first] = res;
}
LL solve(LL r1, LL r2) {
trans(r1, d1);
trans(r2, d2);
memset(vis, 0, sizeof(vis));
return dfs((int) d1.size() - 1, 1, 1, 1);
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> L >> R;
cout << (solve(R, R) - solve(R, L - 1) + MOD) % MOD << '\n';
// LL bf = 0;
// for (LL i = L; i <= R; i++) {
// for (LL j = L; j <= R; j++) {
// int m1 = 0, m2 = 0, t1 = i, t2 = j;
// while (t1) t1 >>= 1, m1++;
// while (t2) t2 >>= 1, m2++;
// if (m1 != m2) continue;
// if ((i | j) == i) bf++;
// }
// }
// cout << "BF say " << bf << '\n';
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
#define ll long long
#define fornum(A,B,C) for(A=B;A<C;A++)
#define mp make_pair
#define pii pair<int,int>
#define pll pair<ll,ll>
using namespace std;
/////////////////////////////////////////////////////
#define MOD (ll)(1e9+7)
ll L, R;
ll bL[90], bR[90];
ll mk[90][2][2][2];
ll i, j, k,l, ans;
ll solve(ll p,ll a,ll b,ll c){
if(p==-1){
return 1;
}
if(mk[p][a][b][c]!=-1)
return mk[p][a][b][c];
ll ret = 0;
// 0 0
if(a==1||bL[p]==0){
ret += solve(p - 1, a, bR[p] == 1 ? 1 : b, c);
// 0 1
if(c==1&&(b==1||bR[p]==1)){
ret += solve(p - 1, a, b, c);
}
}
// 1 1
if(b==1||bR[p]==1){
ret += solve(p - 1, bL[p] == 0 ? 1 : a, b, 1);
}
mk[p][a][b][c] = ret % MOD;
return ret % MOD;
}
int main(){
scanf("%lld%lld", &L,&R);
fornum(i,0,70){
fornum(j,0,2){
fornum(k,0,2){
fornum(l,0,2){
mk[i][j][k][l] = -1;
}
}
}
}
fornum(i,0,62){
if(R&(1ll<<i)){
bR[i] = 1;
}
if(L&(1ll<<i)){
bL[i] = 1;
}
}
printf("%lld", solve(60,0,0,0));
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L, R = map(int, input().split())
MOD = 10 ** 9 + 7
l = '{:060b}'.format(L)[::-1]
r = '{:060b}'.format(R)[::-1]
memo = [[[[-1 for l in range(2)] for k in range(2)] for j in range(2)] for i in range(60)]
def f(pos, flagX, flagY, flagZ):
if pos == -1:
return 1
if memo[pos][flagX][flagY][flagZ] != -1:
return memo[pos][flagX][flagY][flagZ]
ret = 0
# x 0 y 0
if l[pos] == '0' or flagX:
ret += f(pos - 1, flagX, 1 if flagY or r[pos] == '1' else 0, flagZ)
# x 0 y 1
if (l[pos] == '0' or flagX) and (r[pos] == '1' or flagY) and flagZ:
ret += f(pos - 1, flagX, flagY, flagZ)
# x 1 y 1
if r[pos] == '1' or flagY:
ret += f(pos - 1, 1 if l[pos] == '0' or flagX else 0, flagY, 1)
ret %= MOD
memo[pos][flagX][flagY][flagZ] = ret
return ret
ans = f(59, 0, 0, 0)
print(ans) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;
long long dp[61][2][2][2];
int main(){
long long l, r;
cin >> l >> r;
//dp[桁(i)][L(j)][R(k)][最上位(s)]
dp[60][0][0][0] = 1;
for (int i = 59; i >= 0; i--){
int lb = (l >> i) & 1;
int rb = (r >> i) & 1;
for (int j = 0; j < 2; j++){
for (int k = 0; k < 2; k++){
for (int s = 0; s < 2; s++){
for (int x = 0; x < 2; x++){
for (int y = 0; y < 2; y++){
if (x && !y) continue;
int nj = j; int nk = k; int ns = s;
if (!s && x != y) continue;
if (x && y) ns = 1;
if (!j && lb && !x) continue;
if (!lb && x) nj = 1;
if (!k && !rb && y) continue;
if (rb && !y) nk = 1;
dp[i][nj][nk][ns] += dp[i + 1][j][k][s];
dp[i][nj][nk][ns] %= mod;
}
}
}
}
}
}
long long ans = 0;
for (int j = 0; j < 2; j++){
for (int k = 0; k < 2; k++){
for (int s = 0; s < 2; s++) {
ans += dp[0][j][k][s];
ans %= mod;
}
}
}
cout << ans << endl;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#define ll long long
using namespace std;
ll mod=1000000007, dp[62][2][4][4];
int main() {
ll L, R;
cin >> L >> R;
dp[61][0][0][0]=1;
for(int i=60; i>=0; --i){
for(int j=0; j<32; ++j){
int a=(j/4)%4, b=j%4, t=j/16;
if(((a&1) || !(L&((ll) 1<<i))) && ((b&1) || !(L&((ll) 1<<i)))){
int c=a, d=b;
if((R&((ll) 1<<i))){
c |= 2;
d |= 2;
}
dp[i][t][c][d]=(dp[i][t][c][d]+dp[i+1][t][a][b])%mod;
}
if(((a&2) || (R&((ll) 1<<i))) && ((b&2) || (R&((ll) 1<<i)))){
int e=a, f=b;
if(!(L&((ll) 1<<i))){
e |= 1;
f |= 1;
}
dp[i][1][e][f]=(dp[i][1][e][f]+dp[i+1][t][a][b])%mod;
}
if((((a&1) || !(L&((ll) 1<<i))) && ((b&2) || (R&((ll) 1<<i)))) && t){
int g=a, h=b;
if((R&((ll) 1<<i))) g |= 2;
if(!(L&((ll) 1<<i))) h |= 1;
dp[i][t][g][h]=(dp[i][t][g][h]+dp[i+1][t][a][b])%mod;
}
}
}
ll ans=0;
for(int i=0; i<16; ++i){
ans=(ans+dp[0][1][i/4][i%4])%mod;
}
cout << ans << endl;
return 0;
}
| CPP |
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