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name
stringlengths 2
112
| description
stringlengths 29
13k
| source
int64 1
7
| difficulty
int64 0
25
| solution
stringlengths 7
983k
| language
stringclasses 4
values |
|---|---|---|---|---|---|
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | n,q = map(int,raw_input().split())
s = raw_input()
x = [int(s[i]) for i in range(n)]
p = [1]
c = 1
MOD = 10**9+7
for i in range(1,10**5+1):
c = (c*2)%MOD
p.append(c)
cum_sum = [0]*(n+1)
c = 0
for i in range(n):
c += x[i]
cum_sum[i+1] = c
for i in range(q):
l,r = map(int,raw_input().split())
n1 = cum_sum[r] - cum_sum[l-1]
n2 = r+1-l-n1
# print n1,n2
ans = (p[n1]-1)*(p[n2])%MOD
print ans | PYTHON |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline long long int __gcd(long long int a, long long int b);
void pdash(int n = 1);
int bitcount(long long int u);
long long int power(long long int x, long long int y);
long long int power(long long int x, long long int y, long long int z);
long long int modInverse(long long int n, long long int p);
long long int nCrF(long long int n, long long int r, long long int p);
void cordinate_compression(vector<int>& v);
void make_unique(vector<int>& vec);
const long long int mod = 1e9 + 7;
void solve() {
int n, q;
cin >> n >> q;
string str;
cin >> str;
vector<int> vec(n);
for (int i = 0; i < n; i++) {
vec[i] = (str[i] - '0');
}
for (int i = 1; i < n; i++) {
vec[i] += vec[i - 1];
}
while (q--) {
int zero, one;
int a, b;
cin >> a >> b;
a--;
b--;
one = vec[b] - (((a - 1) >= 0) ? vec[a - 1] : 0);
zero = (b - a) + 1 - one;
cout << ((power(2, zero, mod) * ((power(2, one, mod) - 1 + mod) % mod)) %
mod)
<< "\n";
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
while (t--) {
solve();
}
}
inline long long int __gcd(long long int a, long long int b) {
if (a == 0 || b == 0) {
return max(a, b);
}
long long int tempa, tempb;
while (1) {
if (a % b == 0)
return b;
else {
tempa = a;
tempb = b;
a = tempb;
b = tempa % tempb;
}
}
}
void pdash(int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < 30; j++) {
cout << "-";
}
cout << "\n";
}
}
long long int power(long long int x, long long int y) {
long long int result = 1;
while (y > 0) {
if (y & 1) {
result = (result * x);
}
y = y >> 1;
x = (x * x);
}
return result;
}
long long int power(long long int x, long long int y, long long int z) {
long long int result = 1;
x = x % z;
while (y > 0) {
if (y & 1) {
result = (result * x) % z;
}
y = y >> 1;
x = (x * x) % z;
}
return result;
}
long long int modInverse(long long int n, long long int p) {
return power(n, p - 2, p);
}
long long int nCrF(long long int n, long long int r, long long int p) {
if (r == 0) return 1;
long long int f[n + 1];
f[0] = 1;
for (long long int i = 1; i <= n; i++) f[i] = f[i - 1] * i % p;
return (f[n] * modInverse(f[r], p) % p * modInverse(f[n - r], p) % p) % p;
}
void cordinate_compression(vector<int>& v) {
vector<int> p = v;
make_unique(p);
for (int i = 0; i < (int)((v).size()); i++)
v[i] = (int)(lower_bound(p.begin(), p.end(), v[i]) - p.begin());
}
void make_unique(vector<int>& vec) {
sort(vec.begin(), vec.end());
vec.erase(unique(vec.begin(), vec.end()), vec.end());
}
int bitcount(long long int u) {
int cnt = 0;
while (u) {
u = u & (u - 1);
cnt++;
}
return cnt;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long MOD = (long long)1e9 + 7;
const long double PI = 3.141592653589793238462643383279502884197;
long long fac[1] = {1}, inv[1] = {1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long mp(long long a, long long b) {
long long ret = 1;
while (b) {
if (b & 1) ret = ret * a % MOD;
a = a * a % MOD;
b >>= 1;
}
return ret;
}
long long cmb(long long r, long long c) {
return fac[r] * inv[c] % MOD * inv[r - c] % MOD;
}
priority_queue<int, vector<int>, greater<int>> pq;
vector<int> v;
char s[100002];
int psum[100002];
int main() {
int n, m;
scanf("%d %d", &n, &m);
scanf("%s", s + 1);
for (int i = 1; i <= n; i++) {
psum[i] = psum[i - 1] + s[i] - '0';
}
while (m--) {
int a, b;
scanf("%d %d", &a, &b);
long long aa = psum[b] - psum[a - 1];
long long bb = b - a + 1 - aa;
if (aa == 0) {
printf("0\n");
continue;
}
long long ans = 0;
ans = mp(2, aa + bb) - 1 + MOD;
ans = (ans - mp(2, bb) + 1) + MOD + MOD;
printf("%lld\n", ans % MOD);
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | from sys import stdin, stdout
import cProfile, math
from collections import Counter
from bisect import bisect_left,bisect,bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
printHeap = str()
memory_constrained = False
P = 10**9+7
import sys
sys.setrecursionlimit(10000000)
class Operation:
def __init__(self, name, function, function_on_equal, neutral_value=0):
self.name = name
self.f = function
self.f_on_equal = function_on_equal
def add_multiple(x, count):
return x * count
def min_multiple(x, count):
return x
def max_multiple(x, count):
return x
sum_operation = Operation("sum", sum, add_multiple, 0)
min_operation = Operation("min", min, min_multiple, 1e9)
max_operation = Operation("max", max, max_multiple, -1e9)
class SegmentTree:
def __init__(self,
array,
operations=[sum_operation, min_operation, max_operation]):
self.array = array
if type(operations) != list:
raise TypeError("operations must be a list")
self.operations = {}
for op in operations:
self.operations[op.name] = op
self.root = SegmentTreeNode(0, len(array) - 1, self)
def query(self, start, end, operation_name):
if self.operations.get(operation_name) == None:
raise Exception("This operation is not available")
return self.root._query(start, end, self.operations[operation_name])
def summary(self):
return self.root.values
def update(self, position, value):
self.root._update(position, value)
def update_range(self, start, end, value):
self.root._update_range(start, end, value)
def __repr__(self):
return self.root.__repr__()
class SegmentTreeNode:
def __init__(self, start, end, segment_tree):
self.range = (start, end)
self.parent_tree = segment_tree
self.range_value = None
self.values = {}
self.left = None
self.right = None
if start == end:
self._sync()
return
self.left = SegmentTreeNode(start, start + (end - start) // 2,
segment_tree)
self.right = SegmentTreeNode(start + (end - start) // 2 + 1, end,
segment_tree)
self._sync()
def _query(self, start, end, operation):
if end < self.range[0] or start > self.range[1]:
return None
if start <= self.range[0] and self.range[1] <= end:
return self.values[operation.name]
self._push()
left_res = self.left._query(start, end,
operation) if self.left else None
right_res = self.right._query(start, end,
operation) if self.right else None
if left_res is None:
return right_res
if right_res is None:
return left_res
return operation.f([left_res, right_res])
def _update(self, position, value):
if position < self.range[0] or position > self.range[1]:
return
if position == self.range[0] and self.range[1] == position:
self.parent_tree.array[position] = value
self._sync()
return
self._push()
self.left._update(position, value)
self.right._update(position, value)
self._sync()
def _update_range(self, start, end, value):
if end < self.range[0] or start > self.range[1]:
return
if start <= self.range[0] and self.range[1] <= end:
self.range_value = value
self._sync()
return
self._push()
self.left._update_range(start, end, value)
self.right._update_range(start, end, value)
self._sync()
def _sync(self):
if self.range[0] == self.range[1]:
for op in self.parent_tree.operations.values():
current_value = self.parent_tree.array[self.range[0]]
if self.range_value is not None:
current_value = self.range_value
self.values[op.name] = op.f([current_value])
else:
for op in self.parent_tree.operations.values():
result = op.f(
[self.left.values[op.name], self.right.values[op.name]])
if self.range_value is not None:
bound_length = self.range[1] - self.range[0] + 1
result = op.f_on_equal(self.range_value, bound_length)
self.values[op.name] = result
def _push(self):
if self.range_value is None:
return
if self.left:
self.left.range_value = self.range_value
self.right.range_value = self.range_value
self.left._sync()
self.right._sync()
self.range_value = None
def __repr__(self):
ans = "({}, {}): {}\n".format(self.range[0], self.range[1],
self.values)
if self.left:
ans += self.left.__repr__()
if self.right:
ans += self.right.__repr__()
return ans
def display(string_to_print):
stdout.write(str(string_to_print) + "\n")
def primeFactors(n): #n**0.5 complex
factors = dict()
for i in range(2,math.ceil(math.sqrt(n))+1):
while n % i== 0:
if i in factors:
factors[i]+=1
else: factors[i]=1
n = n // i
if n>2:
factors[n]=1
return (factors)
def isprime(n):
"""Returns True if n is prime."""
if n < 4:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False
i = 5
w = 2
while i * i <= n:
if n % i == 0:
return False
i += w
w = 6 - w
return True
def test_print(*args):
if test:
print(args)
def display_list(list1, sep=" "):
stdout.write(sep.join(map(str, list1)) + "\n")
def get_int():
return int(stdin.readline().strip())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
memory = dict()
def clear_cache():
global memory
memory = dict()
def cached_fn(fn, *args):
global memory
if args in memory:
return memory[args]
else:
result = fn(*args)
memory[args] = result
return result
# ----------------------------------------------------------------------------------- MAIN PROGRAM
TestCases = False
test = False
def main():
n, q = get_tuple()
li = list(map(int,list(input())))
ones = [0]
k = 0
for i in li:
k+=i
ones.append(k)
for i in range(q):
l,r = get_tuple()
sm = ones[r]-ones[l-1]
print(((pow(2,sm,P)-1)*pow(2,r-l+1-sm,P))%P)
# --------------------------------------------------------------------------------------------- END
if TestCases:
for _ in range(get_int()):
cProfile.run('main()') if test else main()
else: cProfile.run('main()') if test else main() | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long poww(int n) {
long long ans = 1;
long long a = 2;
while (n) {
if (n & 1) ans *= a, ans %= 1000000000 + 7;
a *= a, a %= 1000000000 + 7;
n >>= 1;
}
return ans;
}
int main(void) {
int n, m;
string s;
int p[100005];
cin >> n >> m;
cin >> s;
int x = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '1') x++;
p[i + 1] = x;
}
p[0] = 0;
for (int i = 1; i <= m; i++) {
int l, r;
cin >> l >> r;
int num1 = p[r] - p[l - 1];
int num0 = r - l + 1 - num1;
int x = poww(num1) - 1;
long long y = (long long)x * (poww(num0) - 1);
y %= 1000000000 + 7;
long long sum = x + y;
sum %= 1000000000 + 7;
cout << sum << endl;
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long zeros[100000 + 342], ones[100000 + 342];
long long add(long long a, long long b) { return (a + b) % mod; }
long long sub(long long a, long long b) {
return ((a % mod) - (b % mod) + mod) % mod;
}
long long mul(long long a, long long b) {
return ((a % mod) * (b % mod)) % mod;
}
long long poww(long long a, long long b) {
if (b == 0LL) {
return 1;
}
long long aa = poww(a, b / 2LL);
aa = mul(aa, aa);
if (b % 2LL == 1LL) {
aa = mul(aa, a);
}
return aa;
}
int main() {
ios::sync_with_stdio(false);
int n, q;
string str;
cin >> n >> q;
cin >> str;
for (int i = 0; i < str.length(); i++) {
if (str[i] == '0') {
zeros[i + 1]++;
} else {
ones[i + 1]++;
}
zeros[i + 1] += zeros[i];
ones[i + 1] += ones[i];
}
while (q--) {
int l, r;
cin >> l >> r;
long long no_of_ones = ones[r] - ones[l - 1];
long long no_of_zeros = zeros[r] - zeros[l - 1];
long long byone = poww(2LL, no_of_ones), byzero, mlplr, ans;
byone = sub(byone, 1);
byzero = byone;
mlplr = sub(poww(2LL, no_of_zeros), 1);
byzero = mul(byzero, mlplr);
cout << add(byone, byzero) << "\n";
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import sys
r = sys.stdin.readlines()
M = 10 ** 9 + 7
n, q = map(int, r[0].strip().split())
s = list(map(int, r[1].strip()))
p = [0]
for i in range(n):
p.append(p[i] + int(s[i]))
for k in range(q):
a, b = map(int, r[k + 2].strip().split())
l = b - a + 1
one = p[b] - p[a - 1]
zero = l - one
sys.stdout.write(str(((pow(2, one, M) - 1) * pow(2, zero, M)) % M) + "\n") | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | from sys import stdin, stdout
n,q = tuple(map(int,stdin.readline().split()))
arr = list(str(stdin.readline()))[:-1]
arr[0] = int(arr[0])
for i in range(1,n):
arr[i] = int(arr[i]) + arr[i-1]
arr.insert(0,0)
mod = int(1e9+7)
inputs = [ tuple(map(int,line.split())) for line in stdin ]
ans = []
for l,r in inputs:
o = (arr[r] - arr[l-1])
z = (r - l + 1 - o)
e_o = (pow(2,o,mod) - 1)%mod
e_z = e_o * (pow(2,z,mod)-1)%mod
ans.append(str(int((e_o + e_z)%mod)))
stdout.write("\n".join(ans)) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import os
import sys
def init_input():
it = iter(os.read(sys.stdin.fileno(), 10 ** 7).split())
return lambda: next(it).decode(), lambda: int(next(it)), lambda: float(next(it))
ns, ni, nf = init_input()
MOD = 10 ** 9 + 7
n, q = ni(), ni()
s = 'x' + ns()
c = [0] * (n + 1)
for i in range(1, n + 1):
c[i] = c[i - 1] + (s[i] == '1')
p2 = [1] * (2 * n + 1)
for i in range(1, 2 * n + 1):
p2[i] = p2[i - 1] * 2 % MOD
out = []
for qq in range(q):
l, r = ni(), ni()
o = c[r] - c[l - 1]
z = (r - l + 1) - o
ans = (p2[o + z] - 1 - p2[z] + 1) % MOD
out.append(str(ans))
sys.stdout.write("\n".join(out))
| PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #!/usr/bin/env python
"""
This file is part of https://github.com/Cheran-Senthil/PyRival.
Copyright 2018 Cheran Senthilkumar all rights reserved,
Cheran Senthilkumar <[email protected]>
Permission to use, modify, and distribute this software is given under the
terms of the MIT License.
"""
from __future__ import division, print_function
import cmath
import itertools
import math
import operator as op
# import random
import sys
from atexit import register
from bisect import bisect_left, bisect_right
# from collections import Counter, MutableSequence, defaultdict, deque
# from copy import deepcopy
# from decimal import Decimal
# from difflib import SequenceMatcher
# from fractions import Fraction
# from heapq import heappop, heappush
if sys.version_info[0] < 3:
# from cPickle import dumps
from io import BytesIO as stream
# from Queue import PriorityQueue, Queue
else:
# from functools import reduce
from io import StringIO as stream
from math import gcd
# from pickle import dumps
# from queue import PriorityQueue, Queue
if sys.version_info[0] < 3:
class dict(dict):
"""dict() -> new empty dictionary"""
def items(self):
"""D.items() -> a set-like object providing a view on D's items"""
return dict.iteritems(self)
def keys(self):
"""D.keys() -> a set-like object providing a view on D's keys"""
return dict.iterkeys(self)
def values(self):
"""D.values() -> an object providing a view on D's values"""
return dict.itervalues(self)
def gcd(x, y):
"""gcd(x, y) -> int
greatest common divisor of x and y
"""
while y:
x, y = y, x % y
return x
input = raw_input
range = xrange
filter = itertools.ifilter
map = itertools.imap
zip = itertools.izip
def sync_with_stdio(sync=True):
"""Set whether the standard Python streams are allowed to buffer their I/O.
Args:
sync (bool, optional): The new synchronization setting.
"""
global input, flush
if sync:
flush = sys.stdout.flush
else:
sys.stdin = stream(sys.stdin.read())
input = lambda: sys.stdin.readline().rstrip('\r\n')
sys.stdout = stream()
register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
def main():
n, q = map(int, input().split())
d = input()
one_cnt = [0] * (n + 1)
for i in range(n):
one_cnt[i] = one_cnt[i-1]
if d[i] == '1':
one_cnt[i] = one_cnt[i-1] + 1
for _ in range(q):
l, r = map(int, input().split())
ones = one_cnt[r - 1] - one_cnt[l - 2]
zeros = (r - (l - 1)) - ones
one_sol = pow(2, ones, 1000000007) - 1
zero_sol = one_sol * (pow(2, zeros, 1000000007) - 1)
print((one_sol + zero_sol) % 1000000007)
if __name__ == '__main__':
sync_with_stdio(False)
if 'PyPy' in sys.version:
from _continuation import continulet
def bootstrap(c):
callable, arg = c.switch()
while True:
to = continulet(lambda _, f, x: f(x), callable, arg)
callable, arg = c.switch(to=to)
c = continulet(bootstrap)
c.switch()
main()
else:
import threading
sys.setrecursionlimit(2097152)
threading.stack_size(134217728)
main_thread = threading.Thread(target=main)
main_thread.start()
main_thread.join()
| PYTHON |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;
public class Main {
private static class Solution {
int[] arr;
int[] pre;
int n;
int[] p2;
int mod = 1_000_000_000 + 7;
public Solution(int[] arr) {
this.arr = arr;
n = arr.length - 1;
pre = new int[n + 1];
for (int i = 1; i <= n; ++i) {
pre[i] = pre[i - 1] + arr[i];
}
p2 = new int[n + 1];
p2[0] = 1;
for (int i = 1; i <= n; ++i) {
p2[i] = (p2[i - 1] * 2) % mod;
}
}
public int query(int low, int high) {
int cnt1 = pre[high] - pre[low - 1];
if (cnt1 == 0)
return 0;
int cnt0 = high - low + 1 - cnt1;
long ret = (long) ((p2[cnt1] - 1)) * p2[cnt0];
ret %= mod;
if (ret < 0)
ret += mod;
return (int) ret;
}
}
public static void main(String[] args) throws Exception {
MyScanner in = new MyScanner();
int n = in.nextInt();
int q = in.nextInt();
String str = in.next();
int[] arr = new int[n + 1];
for (int i = 1; i <= n; ++i) {
arr[i] = str.charAt(i - 1) - '0';
}
Solution sol = new Solution(arr);
StringBuilder sb = new StringBuilder();
while (q-- > 0) {
int low = in.nextInt();
int high = in.nextInt();
sb.append(sol.query(low, high) + "\n");
}
System.out.print(sb.toString());
}
// -----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
// --------------------------------------------------------
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
static Scanner in;
// static StreamTokenizer in;
// static int nextInt() throws IOException {
// in.nextToken();
// return (int)in.nval;
// }
static Random rand = new Random();
static long gcd(long a, long c) {
BigInteger q = BigInteger.valueOf(a);
BigInteger w = BigInteger.valueOf(c);
q = q.gcd(w);
return q.longValue();
}
static int[] w, d;
public static void main(String args[]) throws IOException {
// in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = in.nextInt();
int q = in.nextInt();
String s = in.next();
int[] a = new int[n];
long[] pow = new long[n];
pow[0] = 1;
long two = 1;
int mod = (int) (1e9+7);
for(int i = 0;i<n;i++) {
a[i] = s.charAt(i) -'0';
}
for(int i=1;i<n;i++) {
a[i] = a[i]+a[i-1];
two*=2;
two%=mod;
pow[i] = (pow[i-1] + two)%mod;
}
for(int h=0;h<q;h++) {
int l = in.nextInt()-1;
int r = in.nextInt()-1;
int k =0;
if(l==0) k = a[r];
else k = a[r] - a[l-1];
long ans = 0;
if(r - k - l == -1) ans = pow[r-l];
else ans = (pow[r-l] - pow[r-k-l] +mod) % mod;
out.println(ans);
}
out.close();
}
}
class Pair implements Comparable{
int a;
int b;
public Pair() {
this.a = 0;
this.b = 0;
}
public Pair(int a, int b) {
this.a = a;
this.b = b;
}
public int compareTo(Object arg0) {
if(Integer.compare(this.a, ((Pair) arg0).a) == 0) return Integer.compare(this.b, ((Pair) arg0).b);
return Integer.compare(this.a, ((Pair) arg0).a);
}
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long cnt[100002];
string s;
long long pre[100002];
long long fi(long long a, long long n) {
long long b = n - a;
long long i, cn = 0, ans = 0;
if (a == 0) return 0;
ans = pre[b];
ans *= (pre[a] - 1);
ans %= 1000000007;
if (ans < 0) ans += 1000000007;
return ans;
}
int main() {
long long n, q, i, j;
cin >> n >> q;
pre[0] = 1;
for (i = 1; i < 100001; i++) {
pre[i] = (pre[i - 1]) * 2;
pre[i] %= 1000000007;
if (pre[i] < 1000000007) pre[i] += 1000000007;
}
cin >> s;
for (i = 0; i < n; i++) {
if (s[i] == '0')
cnt[i + 1] = cnt[i];
else
cnt[i + 1] = cnt[i] + 1;
}
while (q--) {
long long l, r;
cin >> l >> r;
cout << fi(-cnt[l - 1] + cnt[r], r - l + 1) << endl;
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | /* Rajkin Hossain */
import java.io.*;
import java.util.*;
public class C {
static FastInput k = new FastInput(System.in);
static FastOutput z = new FastOutput();
static int n, q;
static char [] y;
static Pair [] data;
static long mod = (long) 1e9 + 7l;
static long pow(long a, long b){
if(b == 0) return 1l % mod;
long x = pow(a, b / 2);
x = (x * x) % mod;
if(b % 2 == 1) x = (x * a) % mod;
return x;
}
static void startAlgorithm(){
while(q-->0){
int left = k.nextInt();
int right = k.nextInt();
Pair p2 = data[right];
Pair p1 = data[left-1];
long one = p2.one - p1.one;
long zero = p2.zero - p1.zero;
long oneSum = 0l, zeroSum = 0l;
if(one % 2l == 0){
oneSum = pow(4l, one/2) - 1l;
}
else{
oneSum = (1l % mod) + ((2l % mod) * (pow(4l, (one-1)/2) - 1l)) % mod;
oneSum %= mod;
}
if(zero % 2 == 0){
zeroSum = pow(4l, zero/2) - 1l;
}
else{
zeroSum = (1l % mod) + ((2l % mod) * (pow(4l, (zero-1)/2) - 1l)) % mod;
zeroSum %= mod;
}
zeroSum = ((oneSum % mod) * (zeroSum % mod)) % mod;
long ans = ((oneSum % mod) + (zeroSum % mod)) % mod;
z.println(ans);
}
}
public static void main(String[] args) throws Exception {
while(k.hasNext()){
n = k.nextInt();
q = k.nextInt();
y = k.next().toCharArray();
data = new Pair[n+1];
data[0] = new Pair(0,0);
for(int i = 1; i<=n; i++){
if(y[i-1] == '0'){
data[i] = new Pair(data[i-1].zero + 1, data[i-1].one);
}
else{
data[i] = new Pair(data[i-1].zero, data[i-1].one + 1);
}
}
startAlgorithm();
}
z.flush();
System.exit(0);
}
static class Pair {
int zero, one;
public Pair(int zero, int one) {
this.zero = zero;
this.one = one;
}
}
public static class FastInput {
BufferedReader reader;
StringTokenizer tokenizer;
public FastInput(InputStream stream){
reader = new BufferedReader(new InputStreamReader(stream));
}
public FastInput(String path){
try {
reader = new BufferedReader(new FileReader(path));
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
tokenizer = null;
}
public String next() {
return nextToken();
}
public String nextLine() {
try {
return reader.readLine();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return null;
}
public boolean hasNext(){
try {
return reader.ready();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return false;
}
public String nextToken() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
String line = null;
try {
line = reader.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
if (line == null) {
return null;
}
tokenizer = new StringTokenizer(line);
}
return tokenizer.nextToken();
}
public Integer nextInt() {
return Integer.parseInt(nextToken());
}
public Long nextLong() {
return Long.valueOf(nextToken());
}
}
public static class FastOutput extends PrintWriter {
public FastOutput() {
super(new BufferedOutputStream(System.out));
}
}
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.util.*;
public class ProblemC {
public static final long M = 1_000_000_000 + 7;
/* Iterative Function to calculate (x^y)%p in O(log y) */
static long power(long x, long y, long p) {
long res = 1;
x = x % p;
while (y > 0) {
if ((y & 1l) > 0) res = (res * x) % p;
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
/* INPUT */
int n = scan.nextInt(), q = scan.nextInt();
String s = scan.next();
int[] accu = new int[n];
int curr = 0;
for (int i = 0; i < n; i++) {
int ai = s.charAt(i) - '0';
curr += ai;
accu[i] = curr;
}
// System.out.println(Arrays.toString(accu));
long[] pow2 = new long[n + 1];
pow2[0] = 1;
long p2 = 1;
for (int i = 1; i <= n; i++) {
p2 *= 2;
p2 %= M;
pow2[i] = p2;
}
// System.out.println(Arrays.toString(pow2));
StringBuilder res = new StringBuilder();
while (q-- > 0) {
int l = scan.nextInt() - 1, r = scan.nextInt() - 1;
int len = r - l + 1;
int ones = accu[r];
if (l > 0) ones -= accu[l - 1];
int zero = len - ones;
// System.out.println(ones + " 1:0 " + zero);
long onesMax = power(2l, ones - 1, M);
// System.out.println(onesMax);
// a * (1 - r^n)/ ( 1 - r)
// r = 2
long onesSum = pow2[ones]; //power(2l, ones, M);
onesSum--;
if (onesSum <= 0) onesSum += M;
// onesSum * (1 - r ^2) / -1
long zeroSum = pow2[zero]; //power(2l, zero, M);
zeroSum--;
if (zeroSum <= 0) onesSum += M;
zeroSum *= onesSum;
zeroSum %= M;
res.append(((onesSum + zeroSum) % M)).append("\n");
}
System.out.println(res);
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
const long long mo = 1e9 + 7;
int n, m, k;
long long a[maxn], sum[maxn];
long long c[maxn];
long long ans, ct, cnt, tmp, flag;
char s[maxn];
long long power(long long a, long long n) {
long long ans = 1;
a = a % mo;
while (n) {
if (n & 1) ans = (ans * a) % mo;
n >>= 1;
a = (a * a) % mo;
}
return ans;
}
int main() {
int T, cas = 1;
while (scanf("%d%d", &n, &m) != EOF) {
ans = 0;
flag = 1;
memset(c, 0, sizeof(c));
scanf("%s", s + 1);
for (int i = 1; i <= n; i++) {
if (s[i] == '0')
c[i] = c[i - 1];
else
c[i] = c[i - 1] + 1;
}
while (m--) {
long long x, y;
scanf("%lld%lld", &x, &y);
if (c[y] - c[x - 1] == 0) {
puts("0");
continue;
}
long long k = c[y] - c[x - 1];
long long tp = (power(2, k) - 1 + mo) % mo;
long long kk = (y - x + 1 - k);
long long tmp = (tp * (power(2, kk) - 1 + mo) % mo) % mo;
ans = (tmp + tp) % mo;
printf("%lld\n", ans);
}
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import sys
r = sys.stdin.readlines()
M = 10 ** 9 + 7
n, q = r[0].strip().split()
n = int(n)
q = int(q)
s = r[1]
p = [0]
k = 0
for v in range(n):
d = int(s[v])
k += d
p.append(k)
ans = []
for k in range(q):
a, b = r[k + 2].strip().split()
a = int(a)
b = int(b)
l = b - a + 1
one = p[b] - p[a - 1]
ans.append((pow(2, l, M) - pow(2, l - one, M) + M) % M)
sys.stdout.write("\n".join(map(str, ans))) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int s[100001];
long long p[100001];
long long mod = 1e9 + 7;
int cnt(int left, int right) { return s[right] - s[left - 1]; }
int main() {
int n, q;
cin >> n >> q;
string a;
cin >> a;
s[1] = (int)(a[0] - '0');
for (int i = 1; i < a.length(); i++) {
s[i + 1] = s[i] + (a[i] == '1');
}
p[0] = 1;
for (int i = 1; i <= 100000; i++) {
p[i] = p[i - 1] * 2 % mod;
}
while (q--) {
int left, right;
cin >> left >> right;
int one = cnt(left, right);
int zero = right - left + 1 - one;
cout << ((p[one] - 1) * p[zero]) % mod << "\n";
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | from sys import stdin, stdout
from itertools import repeat
def main():
n, q = map(int, stdin.readline().split())
s = stdin.readline().strip()
dat = map(int, stdin.read().split(), repeat(10, 2 * q))
a = [0] * (n + 1)
for i, x in enumerate(s):
a[i+1] = a[i] + (x == '1')
ans = [0] * q
mod = 1000000007
for i in xrange(q):
l, r = dat[i*2] - 1, dat[i*2+1]
o = a[r] - a[l]
z = r - l - o
ans[i] = ((pow(2, o, mod) - 1) * pow(2, z, mod)) % mod
stdout.write('\n'.join(map(str, ans)))
main()
| PYTHON |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
using namespace std;
string s;
long long tree_bit[200100];
int lim;
void update_bit(int idx, int val) {
while (idx <= lim) {
tree_bit[idx] += val;
if (tree_bit[idx] >= 1000000007) tree_bit[idx] %= 1000000007;
idx += (idx & -idx);
}
}
long long query_bit(int idx) {
if (!idx) return 0;
long long sum = 0;
while (idx != 0) {
sum += tree_bit[idx];
idx -= (idx & -idx);
if (sum >= 1000000007) sum %= 1000000007;
}
return sum;
}
long long pw2[100100];
long long get_progression(long long a, long long pod) {
if (a >= 1000000007) a %= 1000000007;
long long ret = a * (pw2[pod] - 1);
if (ret >= 1000000007) ret %= 1000000007;
return ret;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n, q;
pw2[0] = 1;
for (int i = 1; i <= 100010; i++) {
pw2[i] = (pw2[i - 1] << 1);
if (pw2[i] >= 1000000007) pw2[i] %= 1000000007;
}
cin >> n >> q;
cin >> s;
lim = n;
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
update_bit(i + 1, 1);
}
}
while (q--) {
int l, r;
cin >> l >> r;
l--;
int range = r - l;
int ones = query_bit(r) - query_bit(l);
long long ans = 0;
ans = get_progression(1, ones);
ans += get_progression(ans, range - ones);
if (ans >= 1000000007) ans %= 1000000007;
cout << ans << endl;
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int delta[4][2] = {{-1, -0}, {0, 1}, {1, 0}, {0, -1}};
long long calc_pow(long long a, long long p, long long m = 1000000007) {
long long res = 1;
while (p > 0) {
if (p & 1) res = (res * a) % m;
a = (a * a) % m;
p >>= 1;
}
return res;
}
long long calc_pow_without_mod(long long a, long long p) {
long long res = 1;
while (p > 0) {
if (p & 1) res = res * a;
a = a * a;
p >>= 1;
}
return res;
}
void func() {
long long n, q;
cin >> n >> q;
string s;
cin >> s;
char arr[n + 1];
for (int i = 0; i < n; i++) arr[i + 1] = s[i];
long long pre[n + 1];
for (int i = 0; i <= n; i++) pre[i] = 0;
for (int i = 1; i <= (n); i++) {
pre[i] = pre[i - 1];
if (arr[i] == '1') pre[i]++;
}
for (int i = 1; i <= (q); i++) {
int l, r;
cin >> l >> r;
long long ones = pre[r] - pre[l - 1];
long long zeroes = r - l + 1 - ones;
long long val = (calc_pow(2, ones) - 1 + 1000000007) % 1000000007;
long long new_val = (calc_pow(2, zeroes) - 1 + 1000000007) % 1000000007;
new_val = (new_val * val) % 1000000007;
cout << (val + new_val) % 1000000007 << endl;
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
func();
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, q;
cin >> n >> q;
vector<long long> powe(n + 1, 0);
powe[0] = 1;
for (long long i = 1; i <= n; i++) powe[i] = (powe[i - 1] * 2) % 1000000007;
vector<long long> A(n, 0), pre(n, 0);
string s;
cin >> s;
for (long long i = 0; i < n; i++)
if (s[i] == '1') A[i] = 1;
for (long long i = 0; i < n; i++)
pre[i] = A[i] + (i - 1 >= 0 ? pre[i - 1] : 0);
for (long long i = 0; i < q; i++) {
long long l, r;
cin >> l >> r;
l--;
r--;
long long num = r - l + 1, one = pre[r] - (l - 1 >= 0 ? pre[l - 1] : 0);
long long zer = num - one;
long long val = powe[one] - 1;
long long val2 = powe[zer] - 1;
val = ((val + ((val * val2) % 1000000007) % 1000000007) % 1000000007 +
2 * 1000000007) %
1000000007;
cout << val << endl;
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAX = 100005;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
string str;
long long n, q;
long long l, r;
long long cum[100005];
long long p[100005];
void make() {
for (int i = 1; i <= n; i++) {
if (str[i - 1] == '1')
cum[i] = cum[i - 1] + 1;
else
cum[i] = cum[i - 1];
}
p[0] = 0;
p[1] = 1;
long long temp = 2;
for (long long i = 2; i < 100005; i++) {
temp *= 2;
temp %= mod;
p[i] = temp - 1;
}
}
int main() {
cin >> n >> q;
cin >> str;
make();
long long n1, n0;
long long temp1, temp2, ans;
while (q--) {
cin >> l >> r;
n1 = cum[r] - cum[l - 1];
n0 = r - l + 1 - n1;
temp1 = p[n1];
temp2 = p[n0];
temp2 *= temp1;
temp2 %= mod;
ans = temp1 + temp2;
ans %= mod;
cout << ans << endl;
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
n,m=map(int,input().split())
t=list(map(int,list(input())))
p=[0]+t[:]
mod=10**9+7
o=[]
for i in range(n):
p[i+1]+=p[i]
for _ in range(m):
l,r=map(int,input().split())
a=p[r]-p[l-1]
b=(r-l+1)-a
o.append(str((pow(2,a+b,mod)-pow(2,b,mod))%mod))
print('\n'.join(map(str,o))) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | n,q=map(int,input().split())
a=list(map(int,list(input())))
p=[0]+a[:]
M=10**9+7
for i in range(1,n+1):p[i]+=p[i-1]
o=[]
for _ in range(q):
l,r=map(int,input().split())
w=p[r]-p[l-1]
z=(r-l+1)-w
o.append(str((pow(2,w+z,M)-pow(2,z,M))%M))
print('\n'.join(map(str,o))) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long INF = (long long)0x3f3f3f3f3f3f3f, MAX = 9e18, MIN = -9e18;
const int N = 1e6 + 10, M = 2e6 + 10, mod = 1e9 + 7, inf = 0x3f3f3f3f;
long long a[N], sum[N];
long long q_pow(long long a, long long k) {
long long s = 1;
while (k) {
if (k & 1) s = s * a % mod;
a = a * a % mod;
k >>= 1;
}
return s;
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _;
_ = 1;
while (_--) {
long long n, m, inv2 = q_pow(2, mod - 2);
string s;
cin >> n >> m >> s;
s = " " + s;
for (int i = 1; (i) <= (n); i++) sum[i] = sum[i - 1] + s[i] - '0';
while (m--) {
long long l, r;
cin >> l >> r;
long long sum1 = sum[r] - sum[l - 1];
if (sum1 == 0) {
cout << 0 << '\n';
continue;
}
long long sum0 = r - l + 1 - sum1;
long long ans = q_pow(2, sum1) - 1;
long long now = ans;
ans = (ans + now * (q_pow(2, sum0) - 1) % mod) % mod;
cout << ans << '\n';
}
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
namespace fast_IO {
inline int read_int() {
register int ret = 0, f = 1;
register char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
ret = (ret << 1) + (ret << 3) + int(c - 48);
c = getchar();
}
return ret * f;
}
} // namespace fast_IO
using namespace fast_IO;
int N, Q;
namespace BIT {
int c[200005];
inline int lowbit(int x) { return x & (-x); }
inline void add(int pos, int w) {
while (pos <= N) c[pos] += w, pos += lowbit(pos);
}
inline int query(int pos) {
int res = 0;
while (pos) res += c[pos], pos -= lowbit(pos);
return res;
}
} // namespace BIT
long long w[200005];
char s[200005];
inline void init() {
N = read_int(), Q = read_int();
scanf("%s", s + 1);
for (register int i = 1; i <= N; i++) BIT::add(i, s[i] - '0');
w[1] = 1;
for (register int i = 2; i <= N + 1; i++)
w[i] = w[i - 1] << 1, w[i] %= 1000000007;
for (register int i = 1; i <= N + 1; i++)
w[i] += w[i - 1], w[i] %= 1000000007;
for (register int i = 1; i <= N + 1; i++)
w[i] += w[i - 1], w[i] %= 1000000007;
}
inline void calc() {
register int l, r;
register int zero, one;
while (Q--) {
l = read_int(), r = read_int();
one = BIT::query(r) - BIT::query(l - 1);
zero = r - l + 1 - one;
int len = r - l + 1;
long long ans = w[len - 1] - w[len - one - 1] + one;
ans = (ans % 1000000007 + 1000000007) % 1000000007;
printf("%lld\n", ans);
}
}
int main() {
init();
calc();
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.*;
import java.util.*;
public class Main {
public static long MOD = 1000_000_007;
public static void solve(FastIO io) {
int n = io.nextInt();
int q = io.nextInt();
String parts = io.nextLine();
int[] ones = new int[n+1];
for(int i = 1; i <= n; ++i){
if(parts.charAt(i-1) == '1')
ones[i] = 1;
}
for(int i = 1; i <= n; ++i)
ones[i] += ones[i-1];
for(int i = 0, l, r; i < q; ++i){
l = io.nextInt();
r = io.nextInt();
int oCnt = ones[r] - ones[l-1], zCnt = r-l+1 - oCnt;
if(oCnt > 0) {
long r1 = (binPow(2, oCnt, MOD) + MOD - 1)%MOD;
long r2 = r1 * ((binPow(2, zCnt, MOD) + MOD - 1)%MOD) %MOD;
io.println((r1 + (zCnt > 0 ? r2 : 0))%MOD);
continue;
}
io.println(0);
}
}
public static long binPow(long a, long b, long m){
a %= m;
long res = 1;
while (b > 0) {
if ((b & 1) != 0)
res = res * a % m;
a = a * a % m;
b >>= 1;
}
return res;
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
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 int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int 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 int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
public static void main(String[] args) throws FileNotFoundException {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | from sys import stdin
out = []
n,q = map(int, raw_input().split())
s = raw_input()
pref = [0]
for i in s:
if i == '0':
pref.append(pref[-1])
else:
pref.append(pref[-1]+1)
MOD = pow(10,9)+7
for line in stdin.readlines():
l,r = map(int, line.split())
num1 = pref[r] - pref[l-1]
num0 = r - l + 1 - num1
k = num0
n = num1
out.append(str((pow(2,n,MOD)-1 + (pow(2,n, MOD)-1)*(pow(2,k,MOD)-1))%MOD))
print('\n'.join(out))
| PYTHON |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import math
if __name__ == '__main__':
n,q = [int(x) for x in input().split()]
qq = str(input())
s = [ int(x) for x in qq]
prefix = [0]*n
prefix[0]= s[0]
temp = [0]*(n+1)
temp[0]=1
mod = (pow(10,9)//1)+7
for i in range(1,n):
prefix[i] += prefix[i-1] + s[i]
temp[i] =( 2*(temp[i-1]%mod) )%mod
temp[n] = (2*(temp[n-1]%mod))%mod
ansarray=[]
while q> 0:
q-=1
l,r = [int(x)-1 for x in input().split()]
a = prefix[r]-prefix[l]+s[l]
d = r-l+1
val1 = temp[d]
val2 = temp[d-a]
# val2 = val2%mod
ansarray.append((val1-val2)%mod)
print(*ansarray) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions,statistics
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
from sys import stdout
from heapq import heappush, heappop, heapify ,_heapify_max,_heappop_max,nsmallest,nlargest
def main():
mod=1000000007
f=[1]
t=[1]
for i in range(1,100005):
t.append((t[-1]*2)%mod)
f.append((f[-1]+t[i])%mod)
# InverseofNumber(mod)
# InverseofFactorial(mod)
# factorial(mod)
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
tc=1
for _ in range(tc):
n,q=ria()
s=rs()
cs1=[0]
cs0=[0]
ns1=0
ns0=0
for i in range(n):
if s[i]=='1':
ns1+=1
else:
ns0+=1
cs1.append(ns1)
cs0.append(ns0)
for i in range(q):
l,r=ria()
n1q=cs1[r]-cs1[l-1]
n0q=cs0[r]-cs0[l-1]
ans=0
if n1q>0:
ans+=f[n1q-1]
if n0q>0:
ans+=(t[n1q]-1)*f[n0q-1]
wi(ans%mod)
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
| PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e5 + 5;
const int MOD = 1e9 + 7;
string a;
long long sum[MAX_N];
long long fpow(long long a, long long b) {
long long aa = 1;
while (b > 0) {
if (b & 1) aa = aa * a % MOD;
a = a * a % MOD;
b /= 2;
}
return aa;
}
int main() {
long long n, m;
cin >> n >> m;
cin >> a;
sum[0] = 0;
for (int i = 1; i <= a.length(); i++) {
sum[i] = sum[i - 1];
if (a[i - 1] == '1') sum[i]++;
}
long long p, q;
while (m--) {
cin >> p >> q;
long long cn1 = sum[q] - sum[p - 1];
long long cn0 = q - p + 1 - cn1;
long long ans = 0;
ans = fpow(2, cn1) - 1;
if (ans < 0) ans += MOD;
ans = ans * fpow(2, cn0) % MOD;
cout << ans << endl;
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.util.*;
import java.io.*;
import java.math.*;
// Solution to PCS Problem 89: Banh Mi
// https://pcs.cs.cloud.vt.edu/contests/89/problems/C
// Solution by: jLink23
public class BanhMi {
private static PrintWriter out;
public static final int MOD = 1000000007;
public static void main(String[] args) {
out = new PrintWriter(System.out);
FastScanner in = new FastScanner();
int n = in.nextInt();
int q = in.nextInt();
String banhmi = in.next();
// Create prefix sum array
int[] prefix = new int[n + 1];
for (int i = 1; i <= n; i++) {
if (banhmi.charAt(i-1) == '0') {
prefix[i] = prefix[i-1];
} else {
prefix[i] = prefix[i-1] + 1;
}
}
// Precompute powers of 2 for quicker access
Long[] pow2 = new Long[100001];
pow2[0] = 1L;
for (int i = 1; i <= 100000; i++) {
pow2[i] = pow2[i-1] * 2L % MOD;
}
for (int i = 0; i < q; i++) {
int l = in.nextInt();
int r = in.nextInt();
int k = r-l+1; // Length of segment
int x = prefix[r] - prefix[l-1]; // Number of 1's
int y = k - x; // Number of 0's
out.println(((pow2[x] - 1) * pow2[y]) % MOD);
}
out.flush();
}
public static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(Reader in) {
br = new BufferedReader(in);
}
public FastScanner() {
this(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String readNextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readIntArray(int n) {
int[] a = new int[n];
for (int idx = 0; idx < n; idx++) {
a[idx] = nextInt();
}
return a;
}
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = (int)1e5 + 15;
const int MOD = int(1e9) + 7;
int n, q;
vector<long long> ans;
int one_cnt[MAXN];
long long bin_pow[MAXN];
long long bin_pref[MAXN];
void in() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> n >> q;
ans.resize(q + 5);
string s;
cin >> s;
one_cnt[1] = (s[0] == '1');
for (int i = 1; i < n; ++i) {
one_cnt[i + 1] = one_cnt[i] + (s[i] == '1');
}
bin_pow[0] = 1;
bin_pref[0] = 1;
for (int i = 1; i <= n; ++i) {
bin_pow[i] = (bin_pow[i - 1] * 2) % MOD;
bin_pref[i] = (bin_pref[i - 1] + bin_pow[i]) % MOD;
}
for (int i = 0; i < q; ++i) {
int L, R;
cin >> L >> R;
int one = one_cnt[R] - one_cnt[L - 1];
int nul_cnt = (R - L + 1) - one;
long long answer = bin_pref[one - 1];
long long tmp = (bin_pref[one - 1] * bin_pref[nul_cnt - 1]) % MOD;
answer = (answer + tmp) % MOD;
ans[i] = answer;
}
}
void out() {
for (int i = 0; i < q; ++i) {
cout << ans[i] << "\n";
}
}
int main() {
in();
out();
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | ###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
# from itertools import permutations
# from functools import cmp_to_key # for adding custom comparator
# from fractions import Fraction
from collections import *
from sys import stdin
from bisect import *
from heapq import *
from math import log2
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
rr = lambda x : reversed(range(x))
mod = int(1e9)+7
inf = float("inf")
n, q = gil()
a = gbs()
sp = [[0 for _ in range(n)] for _ in range(int(log2(n))+1)]
sp[0] = a
m = len(sp)
for i in range(1, m):
d = 1<<(i-1)
for j in range(n):
sp[i][j] = sp[i-1][j] + (sp[i-1][j+d] if j+d < n else 0)
def getSum(l, r):
ans = 0
while l <= r:
bt = int(log2(r-l+1))
ans += sp[bt][l]
delta = (1<<bt)
l += delta
return ans
for _ in range(q):
l, r = gil()
one = getSum(l-1, r-1)
zero = r-l+1 - one
ans = ((pow(2, one, mod) - 1 + mod)%mod)*pow(2, zero, mod)
if ans >= mod:
ans %= mod
print(ans) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.util.*;
import java.io.*;
public class Main {
static int n, q;
static boolean[] arr;
static query[] qs;
static long[] out, pows;
static long mod = (long)(1e9 + 7), modi = 500000004L, ans;
public static void main (String[] args) throws IOException {
//pre comp pows of two mod
pows = new long[100_000];
pows[0] = 1;
for(int i = 1; i < 100_000; ++i){
pows[i] = (pows[i - 1] * 2) % mod;
}
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
String[] nq = br.readLine().split(" ");
n = p(nq[0]);
q = p(nq[1]);
arr = new boolean[n];
qs = new query[q];
out = new long[q];
char[] in = br.readLine().toCharArray();
for(int i = 0; i < n; ++i) arr[i] = in[i] == '1';
for(int i = 0; i < q; ++i){
String[] curr = br.readLine().split(" ");
int ll = p(curr[0]) - 1, rr = p(curr[1]) - 1;
qs[i] = new query(ll, rr, i);
}
Arrays.sort(qs);
int cl = 0, cr = 0;
ans = (arr[0] ? 1 : 0);
for(int i = 0; i < q; ++i){
int l = qs[i].l, r = qs[i].r;
while(cl > l){
--cl;
if(!arr[cl]) mul2();
else add(cr - cl + 1);
}
while(cr < r){
++cr;
if(!arr[cr]) mul2();
else add(cr - cl + 1);
}
while(cl < l){
if(!arr[cl]) div2();
else sub(cr - cl + 1);
++cl;
}
while(cr > r){
if(!arr[cr]) div2();
else sub(cr - cl + 1);
--cr;
}
out[qs[i].id] = ans;
}
for(int i = 0; i < q; ++i) pw.println(out[i]);
pw.close();
}
static void add(int size){
ans = (ans + pows[size - 1]) % mod;
}
static void sub(int size){
ans -= pows[size - 1];
if(ans < 0) ans += mod;
}
static void mul2(){
ans = (ans * 2) % mod;
}
static void div2(){
ans = (ans * modi) % mod;
}
static int p(String in){
return Integer.parseInt(in);
}
static class query implements Comparable<query> {
int l, r, id;
query(int ll, int rr, int ii){
l = ll;
r = rr;
id = ii;
}
public int compareTo(query in){
int lbuck1 = (int)(l / Math.sqrt(n));
int lbuck2 = (int)(in.l / Math.sqrt(n));
if(lbuck1 != lbuck2) return lbuck1 - lbuck2;
if(lbuck1 % 2 == 0) return r - in.r;
return in.r - r;
}
}
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.ByteArrayInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.nio.charset.StandardCharsets;
import java.util.Arrays;
import java.util.InputMismatchException;
public class C1062 {
static class Solver {
int N, Q, cs[];
long del[], sum[], MOD = 1_000_000_007;
long add(long a, long b) { return (a + b) % MOD; }
long mul(long a, long b) { return (a * b) % MOD; }
long pow(long b, long e) {
if (e == 0) return 1;
if (e == 1) return b;
long sq = pow(b, e >> 1);
sq = mul(sq, sq);
return (e & 1) == 0 ? sq : mul(sq, b);
}
void solve(int testNumber, FastScanner s, PrintWriter out) {
N = s.nextInt();
Q = s.nextInt();
cs = new int[N + 1];
char[] ch = s.next().toCharArray();
for (int i = 1; i <= N; i++)
cs[i] += cs[i - 1] + ch[i - 1] - '0';
while (Q-- > 0) {
int l = s.nextInt(), r = s.nextInt(), len = r - l + 1;
long z = cs[r] - cs[l - 1], o = len - z;
long oz = add(MOD - 1, pow(2, z)); // ones
long tot = add(oz, mul(oz, add(MOD - 1, pow(2, o))));
out.println(tot);
}
}
}
final static boolean cases = false;
public static void main(String[] args) {
FastScanner s = new FastScanner(System.in);
PrintWriter out = new PrintWriter(System.out);
Solver solver = new Solver();
for (int t = 1, T = cases ? s.nextInt() : 1; t <= T; t++)
solver.solve(t, s, out);
out.close();
}
static int min(int a, int b) {
return a < b ? a : b;
}
static int max(int a, int b) {
return a > b ? a : b;
}
static long min(long a, long b) {
return a < b ? a : b;
}
static long max(long a, long b) {
return a > b ? a : b;
}
static int swap(int a, int b) {
return a;
}
static Object swap(Object a, Object b) {
return a;
}
static String ts(Object... o) {
return Arrays.deepToString(o);
}
static class FastScanner {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastScanner(InputStream stream) {
this.stream = stream;
}
public FastScanner(File f) throws FileNotFoundException {
this(new FileInputStream(f));
}
public FastScanner(String s) {
this.stream = new ByteArrayInputStream(s.getBytes(StandardCharsets.UTF_8));
}
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++];
}
boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
boolean isEndline(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public String next() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = read();
while (isEndline(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndline(c));
return res.toString();
}
// Jacob Garbage
public int[] nextIntArray(int N) {
int[] ret = new int[N];
for (int i = 0; i < N; i++)
ret[i] = this.nextInt();
return ret;
}
public int[][] next2DIntArray(int N, int M) {
int[][] ret = new int[N][];
for (int i = 0; i < N; i++)
ret[i] = this.nextIntArray(M);
return ret;
}
public long[] nextLongArray(int N) {
long[] ret = new long[N];
for (int i = 0; i < N; i++)
ret[i] = this.nextLong();
return ret;
}
public long[][] next2DLongArray(int N, int M) {
long[][] ret = new long[N][];
for (int i = 0; i < N; i++)
ret[i] = nextLongArray(M);
return ret;
}
public double[] nextDoubleArray(int N) {
double[] ret = new double[N];
for (int i = 0; i < N; i++)
ret[i] = this.nextDouble();
return ret;
}
public double[][] next2DDoubleArray(int N, int M) {
double[][] ret = new double[N][];
for (int i = 0; i < N; i++)
ret[i] = this.nextDoubleArray(M);
return ret;
}
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.util.*;
import java.io.*;
public class Main
{
public static int gcd(int a , int b)
{
int divs=0;
int divd=0;
if(a>=b)
{
divd=a;
divs=b;
}
else
{
divd=b;
divs=a;
}
int rem=divd%divs;
while(rem!=0)
{
int temp=divs;
divs=rem;
divd=temp;
rem=divd%divs;
}
return divs;
}
static class spComp implements Comparator<long[]>
{
public int compare(long[] a, long[] b)
{
if(a[0]>b[0])
return 1;
else if(a[0]<b[0])
return -1;
else if(a[1]>b[1])
return 1;
else if(a[1]>b[1])
return -1;
else return 1;
}
}
class unionFind
{
int parent[];
int height[];
int size=0;
unionFind(int n)
{
size=n;
parent = new int[n];
height = new int[n];
// Arrays.fill(parent, -1);
for(int i=0;i<n;i++)
parent[i]=i;
}
int find(int a)
{
if(parent[a]==a)
return a;
parent[a]=find(a);
return parent[a];
}
void add(int a, int b)
{
a = find(a);
b = find(b);
if(height[a] > height[b])
{
parent[b] = a;
}
else if(height[ b ] > height[ a ])
parent[ a ] = b;
else if(height[ a ] == height[ b ])
{
height[ a ] = height[ a ]+1;
parent[ b ] = a;
}
}
}
static ArrayList<Integer> largestpow(int val, ArrayList<Integer> primes)
{
int n=primes.size();
ArrayList<Integer> powers= new ArrayList<Integer>();
for(int i=0;i<n;i++)
{
int p=primes.get(i);
while(val%p==0)
{
p=p*p;
}
powers.add(p);
}
return powers;
}
public static int twoPow(int a)
{
int val=1;
int ans=0;
if(a==1)
return 1;
while(val<a)
{
val*=2;
ans++;
}
if(val==a)
return ans;
else return ans;
// return ans;
}
static long pow2(int a)
{
long modd=1000000007;
long ans=1;
for(int i=0;i<a;i++)
{
ans*=2;
ans%=modd;
}
return ans;
}
public static void main(String args[])
{
Scanner sc= new Scanner(System.in);
int n= sc.nextInt();
int q= sc.nextInt();
int arr[]= new int[n];
String str= sc.next();
for(int i=0;i<n;i++)
{
arr[i]= str.charAt(i)-'0';
}
int zeros[]= new int[n];
int ones[]= new int[n];
long modd= 1000000007;
long powers[]= new long[n+1];
for(int i=0;i<=n;i++)
{
if(i==0)
powers[i]=1;
else
powers[i]= (powers[i-1]*2)%modd;
}
for(int i=0;i<n;i++)
{
if(i==0)
{
if(arr[i]==0)
zeros[i]=1;
if(arr[i]==1)
ones[i]=1;
}
else
{
if(arr[i]==0)
{
zeros[i]=zeros[i-1]+1;
ones[i]=ones[i-1];
}
if(arr[i]==1)
{
ones[i]= ones[i-1]+1;
zeros[i]=zeros[i-1];
}
}
}
for(int i=0;i<q;i++)
{
int l= sc.nextInt()-1;
int r= sc.nextInt()-1;
if(l!=0)
{
int zer=zeros[r]-zeros[l-1];
int one= ones[r]-ones[l-1];
long ans= (powers[one+zer]-powers[zer] + modd)%modd;
System.out.println(ans);
}
else
{
int zer=zeros[r];
int one= ones[r];
long ans= (powers[one+zer]-powers[zer] + modd)%modd;
System.out.println(ans);
}
}
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
long double PI = 4 * atan(1);
long long n, q, dp[100001], po[100001], x, cur, ans, l, r;
string second;
void init() {
po[0] = 1;
for (int i = 1; i < 100001; i++) {
po[i] = (po[i - 1] * 2LL) % MOD;
}
}
int main() {
ios::sync_with_stdio(0), cin.tie(0);
init();
cin >> n >> q >> second;
for (int i = 1; i < n + 1; i++) {
if (second[i - 1] == '1') {
dp[i] = 1;
}
dp[i] += dp[i - 1];
}
while (q--) {
cin >> l >> r;
x = dp[r] - dp[l - 1];
ans = po[x] - 1;
if (ans < 0) ans += MOD;
cur = po[r - l + 1 - x] - 1;
if (cur < 0) cur += MOD;
cur = (ans * cur) % MOD;
ans = (ans + cur) % MOD;
cout << ans << "\n";
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;
import java.io.*;
public class codeforces
{
static class Student{
int x,y;
Student(int x,int y){
this.x=x;
this.y=y;
//this.z=z;
}
}
static int prime[];
static void sieveOfEratosthenes(int n)
{
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
int pos=0;
prime= new int[n+1];
for(int p = 2; p*p <=n; p++)
{
// If prime[p] is not changed, then it is a prime
if(prime[p] == 0)
{
// Update all multiples of p
prime[p]=p;
for(int i = p*p; i <= n; i += p)
prime[i] = p;
}
}
}
static class Sortbyroll implements Comparator<Student>
{
// Used for sorting in ascending order of
// roll number
public int compare(Student c, Student b)
{
return c.x-b.x;
}
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
static class Edge{
int a,b;
Edge(int a,int b){
this.a=a;
this.b=b;
}
}
static class Trie{
HashMap<Character,Trie>map;
int c;
Trie(){
map=new HashMap<>();
//z=null;
//o=null;
c=0;
}
}
//static long ans;
static int parent[];
static int rank[];
static int b[][];
static int bo[];
static int ho[];
static int seg[];
//static int pos;
// static long mod=1000000007;
//static int dp[][];
static HashMap<Integer,Integer>map;
static PriorityQueue<Student>q=new PriorityQueue<>();
//static Stack<Integer>st;
// static ArrayList<Character>ans;
static ArrayList<ArrayList<Integer>>adj;
//static long ans;
static Trie root;
static long fac[];
static int gw,gb;
static long mod=(long)(1000000007);
//static int ans;
static void solve()throws IOException{
FastReader sc=new FastReader();
int n,q,l,r,o,z,i;
long ans,t;
n=sc.nextInt();
q=sc.nextInt();
String s=sc.nextLine();
int a[]=new int[n+1];
for(i=1;i<=n;i++){
a[i]=s.charAt(i-1)-'0';
a[i]+=a[i-1];
}
StringBuilder sb=new StringBuilder();
for(i=0;i<q;i++){
l=sc.nextInt();
r=sc.nextInt();
o=a[r]-a[l-1];
z=r-l+1-o;
t=0;
ans=0;
t=(power(2,o,mod)-(long)1+mod)%mod;
ans+=t;
t=(t*(power(2,z,mod)-(long)1+mod)%mod)%mod;
ans=(ans+t)%mod;
sb.append(ans+"\n");
}
System.out.println(sb.toString());
}
static long nCr(long n, long r,
long p)
{
return (fac[(int)n]* modInverse(fac[(int)r], p)
% p * modInverse(fac[(int)(n-r)], p)
% p) % p;
}
//static int prime[];
//static int dp[];
public static void main(String[] args){
//long sum=0;
try {
codeforces.solve();
} catch (Exception e) {
e.printStackTrace();
}
}
static long modInverse(long n, long p)
{
return power(n, p-(long)2,p);
}
static long power(long x, long y, long p)
{
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y %(long)2!=0)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res%p;
}
/*public static long power(long x,long a) {
if(a==0) return 1;
if(a%2==1)return (x*1L*power(x,a-1))%mod;
return (power((int)((x*1L*x)%mod),a/2))%mod;
}*/
static int find(int x)
{
// Finds the representative of the set
// that x is an element of
while(parent[x]!=x)
{
// if x is not the parent of itself
// Then x is not the representative of
// his set,
x=parent[x];
// so we recursively call Find on its parent
// and move i's node directly under the
// representative of this set
}
return x;
}
static void union(int x, int y)
{
// Find representatives of two sets
int xRoot = find(x), yRoot = find(y);
// Elements are in the same set, no need
// to unite anything.
if (xRoot == yRoot)
return;
// If x's rank is less than y's rank
if (rank[xRoot] < rank[yRoot])
// Then move x under y so that depth
// of tree remains less
parent[xRoot] = yRoot;
// Else if y's rank is less than x's rank
else if (rank[yRoot] < rank[xRoot])
// Then move y under x so that depth of
// tree remains less
parent[yRoot] = xRoot;
else // if ranks are the same
{
// Then move y under x (doesn't matter
// which one goes where)
parent[yRoot] = xRoot;
// And increment the the result tree's
// rank by 1
rank[xRoot] = rank[xRoot] + 1;
}
}
static long gcd(long a, long b)
{
if (a == 0){
//ans+=b;
return b;
}
return gcd(b % a, a);
}
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 |
import com.sun.scenario.effect.impl.sw.sse.SSEBlend_SRC_OUTPeer;
import org.omg.CORBA.INTERNAL;
import javax.swing.plaf.basic.BasicTreeUI;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;
public class Main {
private static Scanner sc;
private static Printer pr;
static boolean []visited;
static int []color;
static int []pow=new int[(int)3e5+1];
static int []count;
static boolean check=true;
static boolean checkBip=true;
static TreeSet<Integer>[]list;
private static long aLong=(long)(Math.pow(10,9)+7);
static final long div=998244353;
static int answer=0;
static int[]countSubTree;
static int[]colorNode;
static boolean con=true;
static ArrayList<Integer>dfsPath=new ArrayList<>();
static ArrayList<Integer>ans=new ArrayList<>();
static boolean[]rec;
static int min=Integer.MAX_VALUE;
static ArrayList<Task>tasks=new ArrayList<>();
static boolean []checkpath;
private static void solve() throws IOException {
int n=sc.nextInt(),q=sc.nextInt();
int []sum=new int[n+1];
StringBuilder builder=new StringBuilder(sc.next());
if (builder.charAt(0)=='1')sum[1]=1;
for (int i=1;i<builder.length();++i){
if (builder.charAt(i)=='1') sum[i+1]=sum[i]+1;
else sum[i+1]=sum[i];
}
//System.out.println("sum:"+Arrays.toString(sum));
while (q-->0){
int l=sc.nextInt(),r=sc.nextInt();
int one=sum[r]-sum[l-1];
int zero=(r-l+1)-one;
pr.println(((power(2,one,aLong)-1)*power(2,zero,aLong))%aLong);
}
}
public static class dsu{
int []rank;
int[]parent;
long []sum;
int n;
public dsu(int n) {
this.n = n;
parent=new int[n];
rank=new int[n];
sum=new long[n];
for (int i=0;i<n;++i){
rank[i]=1;
parent[i]=i;
sum[i]=0;
}
}
public int dad(int child){
if (parent[child]==child)return child;
else return dad(parent[child]);
}
public void merge(int u,int v){
int dadU=dad(u);
int dadV=dad(v);
if (dadU==dadV)return ;
if (u!=v){
if (rank[u]<rank[v]){
parent[u]=v;
sum[v]+=sum[u];
}
else if (rank[u]==rank[v]){
parent[u]=v;
sum[v]+=sum[u];
}
else{
parent[v]=u;
sum[u]+=sum[v];
}
}
}
}
public static long power(long x,long y,long mods){
if (y==0) return 1%mods;
long u=power(x,y/2,mods);
u=((u)*(u))%mods;
if (y%2==1) u=(u*(x%mods))%mods;
return u;
}
public static class Task implements Comparable<Task>{
long l,r;
public Task(long w,long v) {
this.l=w;
this.r=v;
}
@Override
public int compareTo(Task o) {
if (o.l<this.l)return 1;
else if (o.l>this.l)return -1;
else {
if (o.r>this.r)return 1;
else if (o.r<this.r)return -1;
}
return 0;
}
}
public static void printSolve(StringBuilder str, int[] colors, int n,int color){
for(int i = 1;i<=n;++i)
if(colors[i] == color)
str.append(i + " ");
str.append('\n');
}
public static class pairTask{
int val;
int size;
public pairTask(int x, int y) {
this.val = x;
this.size = y;
}
}
public static void subTree(int src,int parent,int x){
countSubTree[src]=1;
if (src==x) {
checkpath[src]=true;
//System.out.println("src:"+src);
}
else checkpath[src]=false;
for (int v:list[src]){
if (v==parent)
continue;
subTree(v,src,x);
countSubTree[src]+=countSubTree[v];
checkpath[src]|=checkpath[v];
}
}
public static boolean prime(long src){
for (int i=2;i*i<=src;i++){
if (src%i==0)
return false;
}
return true;
}
public static void bfsColor(int src){
Queue<Integer>queue = new ArrayDeque<>();
queue.add(src);
while (!queue.isEmpty()){
boolean b=false;
int vertex=-1,p=-1;
int poll=queue.remove();
for (int v:list[poll]){
if (color [v]==0){
vertex=v;
break;
}
}
for (int v:list[poll]){
if (color [v]!=0&&v!=vertex){
p=v;
break;
}
}
for (int v:list[p]){
if (color [v]!=0){
color[vertex]=color[v];
b=true;
break;
}
}
if (!b){
color[vertex]=color[poll]+1;
}
}
}
static int add(int a,int b ){
if (a+b>=div)
return (int)(a+b-div);
return (int)a+b;
}
public static int isBipartite(ArrayList<Integer>[]list,int src){
color[src]=0;
Queue<Integer>queue=new LinkedList<>();
int []ans={0,0};
queue.add(src);
while (!queue.isEmpty()){
ans[color[src=queue.poll()]]++;
for (int v:list[src]){
if (color[v]==-1){
queue.add(v);
color[v]=color[src]^1;
}else if (color[v]==color[src])
check=false;
}
}
return add(pow[ans[0]],pow[ans[1]]);
}
public static int powerMod(long b, long e){
long ans=1;
while (e-->0){
ans=ans*b%div;
}
return (int)ans;
}
public static int dfs(int s){
int ans=1;
visited[s]=true;
for (int k:list[s]){
if (!visited[k]){
ans+=dfs(k);
}
}
return ans;
}
public static int[] radixSort(int[] f) {
int[] to = new int[f.length];
{
int[] b = new int[65537];
for (int i = 0; i < f.length; i++) b[1 + (f[i] & 0xffff)]++;
for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
for (int i = 0; i < f.length; i++) to[b[f[i] & 0xffff]++] = f[i];
int[] d = f;
f = to;
to = d;
}
{
int[] b = new int[65537];
for (int i = 0; i < f.length; i++) b[1 + (f[i] >>> 16)]++;
for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
for (int i = 0; i < f.length; i++) to[b[f[i] >>> 16]++] = f[i];
int[] d = f;
f = to;
to = d;
}
return f;
}
public static long []primeFactor(int n){
long []prime=new long[n+1];
prime[1]=1;
for (int i=2;i<=n;i++)
prime[i]=((i&1)==0)?2:i;
for (int i=3;i*i<=n;i++){
if (prime[i]==i){
for (int j=i*i;j<=n;j+=i){
if (prime[j]==j)
prime[j]=i;
}
}
}
return prime;
}
public static StringBuilder binaryradix(long number){
StringBuilder builder=new StringBuilder();
long remainder;
while (number!=0) {
remainder = number % 2;
number >>= 1;
builder.append(remainder);
}
builder.reverse();
return builder;
}
public static int binarySearch(long[] a, int index,long target) {
int l = index;
int h = a.length - 1;
while (l<=h) {
int med = l + (h-l)/2;
if(a[med] - target <= target) {
l = med + 1;
}
else h = med - 1;
}
return h;
}
public static int val(char c){
return c-'0';
}
public static long gcd(long a,long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
private static class Pair implements Comparable<Pair> {
long x;
long y;
Pair() {
this.x = 0;
this.y = 0;
}
Pair(long x, long y) {
this.x = x;
this.y = y;
}
@Override
public boolean equals(Object obj) {
if (this == obj) {
return true;
}
if (obj == null) return false;
Pair other = (Pair) obj;
if (this.x == other.x && this.y == other.y) {
return true;
}
return false;
}
@Override
public int compareTo(Pair other) {
if (this.x != other.x) return Long.compare(this.x, other.x);
return Long.compare(this.y*other.x, this.x*other.y);
}
}
public static void main(String[] args) throws IOException {
sc = new Scanner(System.in);
pr = new Printer(System.out);
solve();
pr.close();
// sc.close();
}
private static class Scanner {
BufferedReader br;
Scanner (InputStream in) {
br = new BufferedReader(new InputStreamReader(in));
}
private boolean isPrintable(int ch) {
return ch >= '!' && ch <= '~';
}
private boolean isCRLF(int ch) {
return ch == '\n' || ch == '\r' || ch == -1;
}
private int nextPrintable() {
try {
int ch;
while (!isPrintable(ch = br.read())) {
if (ch == -1) {
throw new NoSuchElementException();
}
}
return ch;
} catch (IOException e) {
throw new NoSuchElementException();
}
}
String next() {
try {
int ch = nextPrintable();
StringBuilder sb = new StringBuilder();
do {
sb.appendCodePoint(ch);
} while (isPrintable(ch = br.read()));
return sb.toString();
} catch (IOException e) {
throw new NoSuchElementException();
}
}
int nextInt() {
try {
// parseInt from Integer.parseInt()
boolean negative = false;
int res = 0;
int limit = -Integer.MAX_VALUE;
int radix = 10;
int fc = nextPrintable();
if (fc < '0') {
if (fc == '-') {
negative = true;
limit = Integer.MIN_VALUE;
} else if (fc != '+') {
throw new NumberFormatException();
}
fc = br.read();
}
int multmin = limit / radix;
int ch = fc;
do {
int digit = ch - '0';
if (digit < 0 || digit >= radix) {
throw new NumberFormatException();
}
if (res < multmin) {
throw new NumberFormatException();
}
res *= radix;
if (res < limit + digit) {
throw new NumberFormatException();
}
res -= digit;
} while (isPrintable(ch = br.read()));
return negative ? res : -res;
} catch (IOException e) {
throw new NoSuchElementException();
}
}
long nextLong() {
try {
// parseLong from Long.parseLong()
boolean negative = false;
long res = 0;
long limit = -Long.MAX_VALUE;
int radix = 10;
int fc = nextPrintable();
if (fc < '0') {
if (fc == '-') {
negative = true;
limit = Long.MIN_VALUE;
} else if (fc != '+') {
throw new NumberFormatException();
}
fc = br.read();
}
long multmin = limit / radix;
int ch = fc;
do {
int digit = ch - '0';
if (digit < 0 || digit >= radix) {
throw new NumberFormatException();
}
if (res < multmin) {
throw new NumberFormatException();
}
res *= radix;
if (res < limit + digit) {
throw new NumberFormatException();
}
res -= digit;
} while (isPrintable(ch = br.read()));
return negative ? res : -res;
} catch (IOException e) {
throw new NoSuchElementException();
}
}
float nextFloat() {
return Float.parseFloat(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
try {
int ch;
while (isCRLF(ch = br.read())) {
if (ch == -1) {
throw new NoSuchElementException();
}
}
StringBuilder sb = new StringBuilder();
do {
sb.appendCodePoint(ch);
} while (!isCRLF(ch = br.read()));
return sb.toString();
} catch (IOException e) {
throw new NoSuchElementException();
}
}
void close() {
try {
br.close();
} catch (IOException e) {
// throw new NoSuchElementException();
}
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class List {
String Word;
int length;
List(String Word, int length) {
this.Word = Word;
this.length = length;
}
}
private static class Printer extends PrintWriter {
Printer(PrintStream out) {
super(out);
}
}
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import sys
i = int
r = sys.stdin.readlines()
M = 10 ** 9 + 7
n, q = r[0].strip().split()
n = i(n)
q = i(q)
s = r[1]
p = [0]
k = 0
for v in range(n):
d = i(s[v])
k += (1 - d)
p.append(k)
ans = []
for k in range(q):
a, b = r[k + 2].strip().split()
a = i(a)
b = i(b)
l = b - a + 1
zero = p[b] - p[a - 1]
ans.append(str((pow(2, l, M) - pow(2, zero, M) + M) % M))
sys.stdout.write("\n".join(ans)) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.net.Inet4Address;
import java.util.*;
import java.io.*;
public class Did_I_Deserve_It{
public static long power(long x, long y, long p)
{
// Initialize result
long res = 1;
// Update x if it is more
// than or equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply x
// with result
if((y & 1)==1)
res = (res * x) % p;
// y must be even now
// y = y / 2
y = y >> 1;
x = (x * x) % p;
}
return res;
}
public static void main(String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
StringTokenizer st1 = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st1.nextToken());
int q = Integer.parseInt(st1.nextToken());
String ss = br.readLine();
point arr[] = new point[n];
long zeros = 0l, ones = 0l;
for(int i = 0 ; i < ss.length() ; i++)
{
if(ss.charAt(i) == '1')
ones+=1l;
else
zeros+=1l;
arr[i] = new point(zeros , ones);
}
while(q-- > 0)
{
st1 = new StringTokenizer(br.readLine());
int l = Integer.parseInt(st1.nextToken())-1;
int r = Integer.parseInt(st1.nextToken())-1;
ones = arr[r].ones - arr[l].ones;
zeros = arr[r].zeros - arr[l].zeros;
if(ss.charAt(l) == '0')
zeros+=1l;
else
ones+=1l;
long sum = (long)(power(2,ones,1000000007)-1l)*(power(2,zeros,1000000007));
out.println(sum % (1000000007));
}
out.flush();
out.close();
}
static class point{
long zeros , ones;
public point(long z , long o)
{
zeros = z;
ones = o;
}
}
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
if ("" == "input") {
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
} else if ("" != "") {
freopen(
""
".in",
"r", stdin);
freopen(
""
".out",
"w", stdout);
}
int t = 1;
if (0) {
cin >> t;
}
while (t--) solve();
return 0;
}
const int mod = (int)1e9 + 7;
void solve() {
int n, q;
cin >> n >> q;
string s;
cin >> s;
vector<long long> p(n + 1);
for (int i = 0; i < n; ++i) {
p[i + 1] = p[i] + (s[i] - '0');
}
vector<long long> pw(n + 1);
pw[0] = 1;
for (int i = 1; i <= n; ++i) {
pw[i] = 2ll * pw[i - 1] % mod;
}
vector<long long> G(n + 1);
G[1] = 1;
for (int i = 2; i <= n; ++i) {
G[i] = (pw[i] - 1 + mod) % mod;
}
for (int i = 0; i <= n; ++i) {
42;
}
while (q--) {
int l, r;
cin >> l >> r;
--l, --r;
int N = r - l + 1;
int cnt = p[r + 1] - p[l];
42;
cout << (G[N] - 1ll * G[N - cnt] % mod + mod) % mod << "\n";
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long power(long long n) {
if (n == 0) return 1;
long long a = power(n / 2);
a *= a;
a %= 1000000007;
if (n % 2) a *= 2;
a %= 1000000007;
return a;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, q;
cin >> n >> q;
string s;
cin >> s;
int pre[n + 1][2];
pre[0][0] = 0;
pre[0][1] = 0;
for (int i = int(1); i <= int(n); i++) {
if (s[i - 1] == '1') {
pre[i][0] = pre[i - 1][0];
pre[i][1] = pre[i - 1][1] + 1;
} else {
pre[i][0] = pre[i - 1][0] + 1;
pre[i][1] = pre[i - 1][1];
}
}
while (q--) {
int l, r;
cin >> l >> r;
long long ct1 = pre[r][1] - pre[l - 1][1];
long long ct2 = pre[r][0] - pre[l - 1][0];
long long val1 = power(ct1) - 1;
val1 += 1000000007;
val1 %= 1000000007;
long long val2 = power(ct2) - 1;
val2 += 1000000007;
val2 %= 1000000007;
long long ans = 0;
ans += val1;
val2 *= val1;
val2 %= 1000000007;
ans += val2;
ans %= 1000000007;
cout << ans << "\n";
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
if (v.size() == 0) {
os << "empty vector\n";
return os;
}
for (auto element : v) os << element << " ";
return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, pair<T, second>& p) {
os << "(" << p.first << ", " << p.second << ")";
return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T>& v) {
if (v.size() == 0) {
os << "empty set\n";
return os;
}
auto endit = v.end();
endit--;
os << "[";
for (auto it = v.begin(); it != v.end(); it++) {
os << *it;
if (it != endit) os << ", ";
}
os << "]";
return os;
}
template <typename T>
ostream& operator<<(ostream& os, multiset<T>& v) {
if (v.size() == 0) {
os << "empty multiset\n";
return os;
}
auto endit = v.end();
endit--;
os << "[";
for (auto it = v.begin(); it != v.end(); it++) {
os << *it;
if (it != endit) os << ", ";
}
os << "]";
return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, map<T, second>& v) {
if (v.size() == 0) {
os << "empty map\n";
return os;
}
auto endit = v.end();
endit--;
os << "{";
for (auto it = v.begin(); it != v.end(); it++) {
os << "(" << (*it).first << " : " << (*it).second << ")";
if (it != endit) os << ", ";
}
os << "}";
return os;
}
template <typename T>
ostream& operator<<(ostream& os, vector<vector<T>>& v) {
for (auto& subv : v) {
for (auto& e : subv) os << e << " ";
os << "\n";
}
return os;
}
bool do_debug = false;
vector<long long> fact(200001), ifact(200001);
long long power(long long x, long long n) {
if (!n) return 1;
if (n % 2)
return (x * (power((x * x) % 1000000007, n / 2) % 1000000007)) % 1000000007;
return power((x * x) % 1000000007, n / 2) % 1000000007;
}
long long inverse(long long n) { return power(n, 1000000007 - 2) % 1000000007; }
void pre_factorials() {
fact[0] = 1;
for (long long i = 1; i < 200001; i++)
fact[i] = (fact[i - 1] * i) % 1000000007;
for (long long i = 0; i < 200001; i++) ifact[i] = inverse(i);
}
long long ncr(long long n, long long r) {
if (n < r)
return 0;
else if (n == r)
return 1;
else if (r == 0)
return 0;
long long ans = fact[n];
ans = (ans * ifact[r]) % 1000000007;
ans = (ans * ifact[n - r]) % 1000000007;
return ans;
}
void Runtime_Terror() {
long long n, q;
cin >> n >> q;
string s;
cin >> s;
vector<long long> v(n);
for (long long i = 0; i < n; i++) v[i] = s[i] - '0';
vector<pair<long long, long long>> cnt(n);
if (v[0] == 0)
cnt[0] = {1, 0};
else
cnt[0] = {0, 1};
for (long long i = 1; i < n; i++) {
if (v[i] == 0)
cnt[i] = {cnt[i - 1].first + 1, cnt[i - 1].second};
else
cnt[i] = {cnt[i - 1].first, cnt[i - 1].second + 1};
}
for (long long i = 0; i < q; i++) {
long long l, r;
cin >> l >> r;
if (l == 1) {
long long one = cnt[r - 1].second;
long long zero = cnt[r - 1].first;
long long ans = 0;
long long tem =
((power(2, one) - 1) % 1000000007 + 1000000007) % 1000000007;
ans += tem;
long long dtem = tem;
tem *= ((power(2, zero) - 1) % 1000000007 + 1000000007) % 1000000007;
tem %= 1000000007;
ans += tem;
ans %= 1000000007;
cout << ans << endl;
} else {
long long one = cnt[r - 1].second - cnt[l - 2].second;
long long zero = cnt[r - 1].first - cnt[l - 2].first;
long long ans = 0;
long long tem =
((power(2, one) - 1) % 1000000007 + 1000000007) % 1000000007;
ans += tem;
long long dtem = tem;
tem *= ((power(2, zero) - 1) % 1000000007 + 1000000007) % 1000000007;
tem %= 1000000007;
ans += tem;
ans %= 1000000007;
cout << ans << endl;
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
for (long long i = 0; i < t; i++) Runtime_Terror();
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long n, i, j, ans = 0;
long long M = (long long)1E9 + 7;
long long a[1000000], zero[1000000];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long sum = 0;
a[1] = 1, a[2] = 2;
sum = 3;
for (i = 3; i <= 100000 + 5; i++) {
a[i] = (1 + sum + M) % M;
sum += a[i];
sum = (sum + M) % M;
}
for (i = 1; i <= 100000 + 5; i++) a[i] = (a[i] + a[i - 1] + M) % M;
long long q, l, r;
cin >> n >> q;
string s;
cin >> s;
s = "k" + s;
for (i = 1; i <= n; i++) zero[i] = zero[i - 1] + (s[i] == '0');
while (q--) {
cin >> l >> r;
long long x = zero[r] - zero[l - 1];
long long ans = a[r - l + 1] % M - (a[x]) % M;
cout << (ans + M) % M << endl;
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 5;
int pre[N], pw[N];
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
long long n, q, l, r, a, b;
char ch;
cin >> n >> q;
int sum = 0, temp, ans;
pw[0] = 1, pre[0] = 0;
for (int i = 1; i <= n; i++) {
cin >> ch;
pre[i] += (pre[i - 1] + (ch == '0'));
pw[i] = ((pw[i - 1] << 1) % 1000000007);
}
for (int i = 0; i < q; i++) {
cin >> l >> r;
b = pre[r] - pre[l - 1];
a = r - l + 1 - b;
ans = (1ll * pw[a] - 1ll) * 1ll * pw[b] % 1000000007;
cout << ans << endl;
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
static void precompute_pow(long arr[],int n){
arr[0]=1l;
for(int i=1;i<=n;i++) arr[i]=(arr[i-1]*2l)%mod;
}
public static void process()throws IOException
{
int n=ni(),q=ni();
long pow[]=new long[n+1];
char arr[]=(" "+nln()).toCharArray();
precompute_pow(pow,n);
int pref_1[]=new int[n+1],pref_0[]=new int[n+1];
for(int i=1;i<=n;i++){
pref_1[i]=pref_1[i-1]+(arr[i]-'0');
pref_0[i]=pref_0[i-1]+(1-(arr[i]-'0'));
}
while(q-- > 0){
int l=ni(),r=ni();
long res=pow[pref_1[r]-pref_1[l-1]]-1l;
res%=mod;
res=res+res*(pow[pref_0[r]-pref_0[l-1]]-1l)%mod;
res%=mod;
pn(res);
}
}
static FastReader sc;
static PrintWriter out;
public static void main(String[]args)throws IOException
{
out = new PrintWriter(System.out);
sc=new FastReader();
long s = System.currentTimeMillis();
int t=1;
//t=ni();
while(t-->0)
process();
out.flush();
System.err.println(System.currentTimeMillis()-s+"ms");
}
static void pn(Object o){out.println(o);}
static void p(Object o){out.print(o);}
static int ni()throws IOException{return Integer.parseInt(sc.next());}
static long nl()throws IOException{return Long.parseLong(sc.next());}
static double nd()throws IOException{return Double.parseDouble(sc.next());}
static String nln()throws IOException{return sc.nextLine();}
static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}
static long mod=(long)1e9+7l;
static<T> void r_sort(T arr[],int n){
Random r = new Random();
for (int i = n-1; i > 0; i--){
int j = r.nextInt(i+1);
T temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
Arrays.sort(arr);
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////
static class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while (st == null || !st.hasMoreElements()){
try{ st = new StringTokenizer(br.readLine()); } catch (IOException e){ e.printStackTrace(); }
}
return st.nextToken();
}
String nextLine(){
String str = "";
try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); }
return str;
}
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////////
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 1;
long long l, r, n, q, p[MAXN], a[MAXN], k, m, mod = 1e9 + 7;
string s;
int main() {
cin >> n >> q >> s;
for (int i = 0; i < n; i++) {
a[i + 1] += a[i];
if (s[i] == '1') a[i + 1]++;
}
p[0] = 1;
for (int i = 1; i <= 1e5; i++) p[i] = (p[i - 1] * 2) % mod;
for (int i = 0; i < q; i++) {
cin >> l >> r;
cout << (p[r - l + 1] - p[r - l + 1 - a[r] + a[l - 1]] + mod) % mod << "\n";
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int maxN = 1e6 + 5, MOD = 1e9 + 7, eps = 1e-7, INF = 1e9;
const double PI = acos(-1);
int n, q, a, b;
string str;
int acum[maxN];
long long mulmod(long long a, long long b) {
long long ret = 0;
while (b) {
if (b & 1) ret = (ret + a) % MOD;
b >>= 1, a = (a << 1) % MOD;
}
return ret;
}
long long fastPow(long long x, long long n) {
long long ret = 1;
while (n) {
if (n & 1) ret = ret * x % MOD;
n >>= 1, x = x * x % MOD;
}
return ret;
}
int main() {
cin >> n >> q >> str;
acum[0] = 0;
for (int i = int(0); i < int(n); i++) {
char c = str[i];
if (c == '0') {
acum[i + 1] = acum[i];
} else {
acum[i + 1] = acum[i] + 1;
}
}
for (int i = int(0); i < int(q); i++) {
cin >> a >> b;
long long tot = b - a + 1, ceros = tot - (acum[b] - acum[a - 1]);
long long ans = (fastPow(2, tot) + MOD - 1) % MOD;
ans -= (fastPow(2, ceros) + MOD - 1) % MOD;
ans += MOD;
ans %= MOD;
cout << ans << '\n';
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.util.Scanner;
public class Main {
private static long MOD = 1000000007;
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt(), q = s.nextInt(), curSum = 0;
s.nextLine();
String str = s.nextLine();
long[] pows = new long[100001];
pows[0] = 1;
for (int i = 1; i <= 100000; i++) {
pows[i] = (pows[i - 1] << 1) % MOD;
}
int[] prefixSums = new int[n];
for (int i = 0; i < n; i++) {
if (str.charAt(i) == '1') {
curSum++;
}
prefixSums[i] = curSum;
}
for (int i = 0; i < q; i++) {
int l = s.nextInt() - 1, r = s.nextInt() - 1;
int numOnes = prefixSums[r];
if (l > 0) {
numOnes -= prefixSums[l - 1];
}
int numZeros = (r - l + 1) - numOnes;
long num = (pows[numOnes] - 1) * pows[numZeros];
num %= MOD;
System.out.println(num);
}
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
public class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve() {
int mod= 1000000007;
int n = ni(), Q = ni();
char[] s = ns(n);
int[] ones = new int[n+1];
int[] zeros = new int[n+1];
for(int i = 0; i < n; i++) {
ones[i+1] = ones[i] + (s[i] == '0' ? 0 : 1);
zeros[i+1] = zeros[i] + (s[i] == '0' ? 1 : 0);
}
for(int q = 0; q < Q; q++) {
int l = ni(), r = ni();
int o = ones[r] - ones[l-1];
int z = zeros[r] - zeros[l-1];
long ans = pow(2, o, mod) - 1;
ans = ans * pow(2, z, mod);
out.println(ans % mod);
}
}
public static long pow(long a, long n, long mod) {
// a %= mod;
if(n < 0) return 0;
long ret = 1;
int x = 63 - Long.numberOfLeadingZeros(n);
for (; x >= 0; x--) {
ret = ret * ret % mod;
if (n << 63 - x < 0)
ret = ret * a % mod;
}
return ret;
}
void run() throws Exception {
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if (!INPUT.isEmpty())
tr(System.currentTimeMillis() - s + "ms");
}
public static void main(String[] args) throws Exception {
new Main().run();
}
private byte[] inbuf = new byte[1024];
public int lenbuf = 0, ptrbuf = 0;
private int readByte() {
if (lenbuf == -1)
throw new InputMismatchException();
if (ptrbuf >= lenbuf) {
ptrbuf = 0;
try {
lenbuf = is.read(inbuf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (lenbuf <= 0)
return -1;
}
return inbuf[ptrbuf++];
}
private boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
private int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b))
;
return b;
}
private double nd() {
return Double.parseDouble(ns());
}
private char nc() {
return (char) skip();
}
private String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while (p < n && !(isSpaceChar(b))) {
buf[p++] = (char) b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private char[][] nm(int n, int m) {
char[][] map = new char[n][];
for (int i = 0; i < n; i++)
map[i] = ns(m);
return map;
}
private int[] na(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = ni();
return a;
}
private int ni() {
int num = 0, b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
private long nl() {
long num = 0;
int b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o) {
System.out.println(Arrays.deepToString(o));
}
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 |
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.InputMismatchException;
public class Q3 {
public static int mod = (int) 1e9 + 7;
public static long[] pow = new long[100003];
public static void main(String[] args) {
InputReader s = new InputReader(System.in);
PrintWriter out = new PrintWriter(System.out);
int t = 1;
// t = s.nextInt();
pow[0] = 1;
for (int i = 1; i < 100003; i++) {
pow[i] = pow[i - 1] * 2;
if (pow[i] >= mod) {
pow[i] -= mod;
}
}
nexttest:
while (t-- > 0) {
int n = s.nextInt();
int m = s.nextInt();
char[] xx = s.nextLine().toCharArray();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
if (xx[i] == '1') {
a[i] = 1;
}
}
Query[] q = new Query[m];
for (int i = 0; i < m; i++) {
q[i] = new Query(s.nextInt() - 1, s.nextInt() - 1);
}
long[] res = processQueries(a, q);
printArray(res, out);
}
out.close();
}
public static class Query {
int index;
int a;
int b;
public Query(int a, int b) {
this.a = a;
this.b = b;
}
}
static int count = 0;
static long invmod = pow(2, mod - 2, mod);
static long add(int[] a, int i, long cur) {
count++;
if (a[i] == 1) {
return (cur + pow[count - 1]) >= mod ? cur + pow[count - 1] - mod : cur + pow[count - 1];
}
return (cur * 2) >= mod ? cur * 2 - mod : cur * 2;
}
static long remove(int[] a, int i, long cur) {
count--;
if (a[i] == 1) {
return (cur - pow[count]) < 0 ? (cur - pow[count] + mod) : (cur - pow[count]);
}
return cur * invmod % mod;
}
public static long[] processQueries(int[] a, Query[] queries) {
for (int i = 0; i < queries.length; i++) {
queries[i].index = i;
}
int sqrtn = (int) Math.sqrt(a.length);
Arrays.sort(queries, Comparator.comparingInt((Query q) -> q.a / sqrtn).thenComparingInt(q -> q.b));
long[] res = new long[queries.length];
int L = 0;
int R = 0;
long cur = a[0];
count = 1;
for (Query query : queries) {
while (L > query.a) {
cur = add(a, --L, cur);
// System.out.println(2 +" " + cur);
}
while (R < query.b) {
cur = add(a, ++R, cur);
// System.out.println(3 + " " + cur);
}
while (L < query.a) {
cur = remove(a, L++, cur);
// System.out.println(1 + " " + cur);
}
while (R > query.b) {
cur = remove(a, R--, cur);
// System.out.println(4 + " " + cur);
}
res[query.index] = cur;
}
return res;
}
public static void printArray(long[] a, PrintWriter out) {
for (int i = 0; i < a.length; i++) {
out.println(a[i]);
}
}
public static int pow(long x, long n, int mod) {
long res = 1;
for (long p = x; n > 0; n >>= 1, p = (p * p) % mod) {
if ((n & 1) != 0) {
res = (res * p % mod);
}
}
return (int) res;
}
static long gcd(long n1, long n2) {
long r;
while (n2 != 0) {
r = n1 % n2;
n1 = n2;
n2 = r;
}
return n1;
}
static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int snext() {
if (snumChars == -1) {
throw new InputMismatchException();
}
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int nextInt() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String readString() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isEndOfLine(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MAXN = 1e5, MOD = 1e9 + 7;
int n, q;
string s;
vector<int> pow2(MAXN + 1);
int main() {
ios::sync_with_stdio(0);
cin.tie(nullptr);
cout.tie(nullptr);
pow2[0] = 1;
for (int i = 1; i <= MAXN; ++i) {
pow2[i] = (pow2[i - 1] * 2) % MOD;
}
cin >> n >> q >> s;
vector<pii> pref(n + 1);
for (int i = 1; i <= n; ++i) {
pref[i] = pref[i - 1];
if (s[i - 1] == '0')
pref[i].second++;
else
pref[i].first++;
}
int l, r;
for (int i = 0; i < q; ++i) {
cin >> l >> r;
ll res = ((ll)(pow2[pref[r].first - pref[l - 1].first] - 1) *
pow2[pref[r].second - pref[l - 1].second]) %
MOD;
cout << res << endl;
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 100;
const int mod = 1e9 + 7;
inline pair<long long, long long> mp(long long a, long long b) {
pair<long long, long long> temp;
temp.first = a;
temp.second = b;
return temp;
}
long long read_int() {
char r;
bool start = false, neg = false;
long long ret = 0;
while (true) {
r = getchar();
if ((r - '0' < 0 || r - '0' > 9) && r != '-' && !start) {
continue;
}
if ((r - '0' < 0 || r - '0' > 9) && r != '-' && start) {
break;
}
if (start) ret *= 10;
start = true;
if (r == '-')
neg = true;
else
ret += r - '0';
}
if (!neg)
return ret;
else
return -ret;
}
long long p[N][2], comp[N];
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long n, q;
cin >> n >> q;
string st;
cin >> st;
p[0][0] = p[0][1] = 0;
for (long long i = 0; i < st.size(); i++) {
if (i != 0) {
p[i][0] = p[i - 1][0];
p[i][1] = p[i - 1][1];
}
long long cur = st[i] - '0';
p[i][cur]++;
}
comp[0] = 1;
for (long long i = 1; i < N; i++) {
comp[i] = comp[i - 1] * 2;
comp[i] %= mod;
}
while (q--) {
long long l, r;
cin >> l >> r;
l--;
r--;
long long c0;
if (l == 0)
c0 = p[r][0];
else
c0 = p[r][0] - p[l - 1][0];
long long ans = comp[r - l + 1] - comp[c0];
ans += mod;
ans %= mod;
cout << ans << endl;
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import sys
input_file = sys.stdin
C = (10**9+7)
[n, q] = list(int(i) for i in input_file.readline().split())
temp = input_file.readline()
lst = []
for char in temp[:-1]:
lst.append(int(char))
new_lst = [(0, 0)]
for i in lst:
if i == 0:
new_lst.append((new_lst[-1][0]+1, new_lst[-1][1]))
else:
new_lst.append((new_lst[-1][0], new_lst[-1][1]+1))
ls = [1]
for i in range(n):
ls.append(ls[-1]*2 % C)
for line in input_file:
[l, r] = list(int(i) for i in line[:-1].split())
q = (new_lst[r][0] - new_lst[l-1][0], new_lst[r][1] - new_lst[l-1][1])
print((ls[sum(q)] - ls[q[0]]) % C)
| PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.Arrays;
import java.util.ArrayList;
import java.math.*;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class TestClass {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskA solver = new TaskA();
int testCount = 1;
// testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class TaskA {
private final long MOD = 1000000007;
public void solve(int testNumber, InputReader in, PrintWriter out) {
long[] pows = new long[100001];
pows[0] = 1;
for (int i = 1; i <= 100000; i++) {
pows[i] = (pows[i - 1] << 1) % MOD;
}
int n = in.nextInt();
int q = in.nextInt();
String s = in.next();
int[] pre = new int[n+1];
pre[0] = 0;
int cnt = 0;
for(int i=0;i<n;i++)
{
if(s.charAt(i)=='1') cnt++;
pre[i] = cnt;
}
for(int i=0;i<q;i++)
{
int l = in.nextInt()-1;
int r = in.nextInt()-1;
int ones = pre[r];
if (l > 0) {
ones -= pre[l - 1];
}
int zeroes = (r - l + 1) - ones;
long ans = (pows[ones]-1)*(pows[zeroes]) ;
ans%=MOD;
out.println(ans);
}
}
/* private long pow(long x, int n)
{
if (n == 0)
{
return 1;
}
long x2 = x * x % MOD;
return pow(x2, n/2) * (n % 2 == 0 ? 1 : x) % MOD;
}
*/
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
mod=1000000007
n,m=map(int,input().split())
s=list(input())
l=[0]*(n+1)
for i in range(1,n+1):
l[i]=l[i-1]+int(s[i-1])
#print(l)
for i in range(m):
l1,r=map(int,input().split())
s=r-l1
no=l[r]-l[l1-1]
ans=pow(2,s+1,mod)-pow(2,s-no+1,mod)
print(ans%mod) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
char s[100005];
int cnt[100005];
const long long mod = 1000000007;
long long powm(long long a, long long b) {
long long ret = 1;
while (b) {
if (b & 1) ret = ret * a % mod;
a = a * a % mod;
b >>= 1;
}
return ret;
}
int main() {
int n, q, l, r, m, len;
scanf("%d%d", &n, &q);
scanf("%s", s + 1);
for (int i = 1; i <= n; i++)
if (s[i] == '1')
cnt[i] = cnt[i - 1] + 1;
else
cnt[i] = cnt[i - 1];
for (int i = 0; i < q; i++) {
scanf("%d%d", &l, &r);
m = cnt[r] - cnt[l - 1];
len = r - l + 1;
printf("%I64d\n", (powm(2, m) + mod - 1) % mod * powm(2, len - m) % mod);
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
const int inf = 1e8;
const int mod = 1e9 + 7;
int n, t;
int pre[maxn];
char s[maxn];
long long ksm(long long a, long long b) {
long long res = 1;
while (b) {
if (b & 1) {
res = (res * a) % mod;
}
a = (a * a) % mod;
b >>= 1;
}
return res % mod;
}
int main() {
cin >> n >> t;
cin >> (s + 1);
int len = strlen(s + 1);
for (int i = 1; i <= len; ++i) {
pre[i] = pre[i - 1] + (s[i] == '1');
}
while (t--) {
int l, r;
cin >> l >> r;
long long first = pre[r] - pre[l - 1], second = r - l + 1 - first;
long long res = ksm(2, first) - 1;
if (second >= 0) res += (ksm(2, second) - 1) % mod * res % mod;
cout << res % mod << '\n';
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int64_t const base = 1e9 + 7;
int n, q, l, r, cnt, res;
string s;
int64_t f[100005], i_th[100005];
int main() {
ios::sync_with_stdio(0);
cin >> n >> q >> s;
i_th[1] = 1;
for (int i = 2; i <= 100003; ++i) i_th[i] = (i_th[i - 1] << 1) % base;
for (int i = 0; i < s.size(); ++i) f[i + 1] = f[i] + s[i] - '0';
while (q--) {
cin >> l >> r;
cnt = f[r] - f[l - 1];
res = (base + i_th[cnt + 1] - 1) % base;
res = ((i_th[r - l + 1 - cnt + 1]) * res) % base;
cout << res << '\n';
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 300000 + 10;
const int mod = 1e9 + 7;
int n, q;
char s[N];
long long sum[N];
long long pw[N], fibo[N], pw2[N], pw3[N];
int main() {
scanf("%d%d", &n, &q);
scanf("%s", s + 1);
for (int i = 1; i <= n; ++i) {
sum[i] = sum[i - 1] + (s[i] == '1');
}
for (int i = 1; i <= 2e5; ++i) {
if (i == 1)
fibo[i] = 1;
else
fibo[i] = fibo[i - 1] * 2;
fibo[i] %= mod;
pw[i] = pw[i - 1] + fibo[i];
pw[i] %= mod;
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
long long num1 = sum[r] - sum[l - 1];
long long num0 = (r - l + 1) - num1;
long long ans = pw[num1];
ans %= mod;
if (num0) ans += pw[num1] * pw[num0] % mod;
ans %= mod;
printf("%lld\n", ans);
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.*;
import java.util.*;
public class Main {
private static final int MOD = 1000000007;
private static int add(long a, long b) {
return (int) ((a%MOD+b%MOD)%MOD);
}
private static int subtract(long a, long b) {
return add(a, -b);
}
private static int multiply(long a, long b) {
return (int) ((a%MOD*b%MOD)%MOD);
}
private static int power(long a, long b) {
long c = 1;
while(b > 0) {
if((b&1) > 0) {
c = multiply(c, a);
}
a = multiply(a, a);
b >>= 1;
}
return (int) c;
}
public static void main(String[] args) throws IOException{
//Scanner f = new Scanner(new File("uva.in"));
//Scanner f = new Scanner(System.in);
//BufferedReader f = new BufferedReader(new FileReader("uva.in"));
BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
StringTokenizer st = new StringTokenizer(f.readLine());
int n = Integer.parseInt(st.nextToken());
int q = Integer.parseInt(st.nextToken());
int[] prefixSum = new int[n+1];
char[] s = f.readLine().toCharArray();
for(int i = 1; i <= n; i++) {
prefixSum[i] = prefixSum[i-1]+(s[i-1] == '1' ? 1 : 0);
}
for(int i = 0; i < q; i++) {
st = new StringTokenizer(f.readLine());
int l = Integer.parseInt(st.nextToken());
int r = Integer.parseInt(st.nextToken());
out.println(multiply(subtract(power(2, prefixSum[r]-prefixSum[l-1]), 1), power(2, r-l-prefixSum[r]+prefixSum[l-1]+1)));
}
f.close();
out.close();
}
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <typename T>
using V = vector<T>;
template <typename T, typename U>
using P = pair<T, U>;
template <typename T>
void cout_join(vector<T> &v, string d = " ") {
for (int i = (0); i < (v.size()); ++i) {
if (i > 0) cout << d;
cout << v[i];
}
cout << endl;
}
template <typename T>
long long int power(T n, T p, T m) {
if (p == 0) return 1LL;
if (p == 1) return n;
long long int k = power(n, p / 2, m);
return ((k * k) % m * (p % 2 ? n : 1)) % m;
}
const long long int mod = 1e9 + 7;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
long long int n, q;
cin >> n >> q;
string s;
cin >> s;
vector<long long int> sum(n + 1, 0);
for (int i = (1); i <= (n); ++i) {
sum[i] = sum[i - 1] + (s[i - 1] - '0');
}
for (int i = (0); i < (q); ++i) {
long long int l, r;
cin >> l >> r;
long long int cnt1 = sum[r] - sum[l - 1], cnt0 = r - (l - 1) - cnt1;
long long int enj = (power(2LL, cnt1, mod) - 1 + mod) % mod;
enj += (enj * ((power(2LL, cnt0, mod) - 1 + mod) % mod)) % mod;
enj %= mod;
cout << enj << endl;
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | MOD = 10 ** 9 + 7
MAXN = 10 ** 5 + 1
N, Q = map(int, raw_input().strip().split())
arr = map(int, [ i for i in raw_input().strip()])
# calculate prefix sum
pre = [0 for _ in range(N+1)]
for ind, val in enumerate(arr):
pre[ind + 1] = pre[ind] + val
# calculate powers of two
pows = [1 for _ in range(MAXN)]
for i in range(1, MAXN):
pows[i] = (2 * pows[i - 1]) % MOD
for _ in range(Q):
L, R = map(int, raw_input().strip().split())
x = pre[R] - pre[L - 1]
y = R - L + 1 - x
result = (pows[y] * (pows[x] - 1)) % MOD
print result
| PYTHON |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import sys
r = sys.stdin.readlines()
M = 10 ** 9 + 7
n, q = r[0].split(' ')
n = int(n)
q = int(q)
s = r[1]
p = [0]
for v in range(n):
p.append(p[v] + int(s[v]))
ans = []
for k in range(q):
a, b = r[k + 2].split(' ')
a = int(a)
b = int(b)
l = b - a + 1
one = p[b] - p[a - 1]
ans.append(str((pow(2, l, M) - pow(2, l - one, M) + M) % M))
sys.stdout.write("\n".join(ans)) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.*;
import java.util.*;
public class Main {
static class Reader {
BufferedReader in;
Reader() throws IOException {
in = new BufferedReader(new InputStreamReader(System.in)); // initial size buffer C = 32768 ??? /// default C = 8192
}
Reader(String name) throws IOException {
in = new BufferedReader(new FileReader(name));
}
StringTokenizer tokin = new StringTokenizer("");
String next() throws IOException {
if (!tokin.hasMoreTokens()) {
tokin = new StringTokenizer(in.readLine());
}
return tokin.nextToken();
}
int nextInt() throws IOException {
return Integer.parseInt(next());
}
}
static class Writer {
PrintWriter cout;
Writer() throws IOException {
cout = new PrintWriter(System.out);
}
Writer(String name) throws IOException {
cout = new PrintWriter(new FileWriter(name));
}
StringBuilder out = new StringBuilder();
void print(Object a) {
out.append(a);
}
void close() {
cout.print(out.toString());
cout.close();
}
}
static long mod = Long.parseLong("1000000007");
static long pow(long a, long b) {
if (b == 0)
return 1;
if (b % 2 == 1) {
return (a * pow(a, b - 1)) % mod;
} else {
long c = pow(a, b / 2);
return (c * c) % mod;
}
}
public static void main(String args[]) throws IOException {
Reader in = new Reader();
Writer out = new Writer();
int n = in.nextInt();
int q = in.nextInt();
int col1[] = new int[n];
String s = in.next();
if (s.charAt(0) == '1')
col1[0] = 1;
for (int i = 1; i < n; i++) {
col1[i] = col1[i - 1];
if (s.charAt(i) == '1')
col1[i]++;
}
for (int i = 0; i < q; i++) {
int l = in.nextInt() - 1;
int r = in.nextInt() - 1;
long col = col1[r];
long size = r - l + 1;
if (l > 0)
col -= col1[l - 1];
long ans1 = pow(2, col) - 1;
if (ans1 < 0)
ans1 += mod;
long p1 = pow(2, size - col) - 1;
if (p1 < 0)
p1 += mod;
long ans = (ans1 + p1 * ans1) % mod;
out.print(ans + "\n");
}
out.close();
}
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import sys
from math import *
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
def add(a,b):
return (a+b)%1000000007
def mul(a,b):
return (a*b)%1000000007
def sub(a,b):
return (a-b+1000000007)%1000000007
def qpow(a, b):
r = 1
k = a
for i in range(17):
if b & (1<<i):
r = mul(r, k)
k = mul(k, k)
return r
n, q = mints()
a = list(minp())
c = [0]*(n+1)
for i in range(n):
c[i+1] = c[i] + int(a[i])
for i in range(q):
l, r = mints()
k = (r-l+1)
o = c[r]-c[l-1]
z = sub(qpow(2,o),1)
print(mul(z,qpow(2,k-o))) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 |
import java.util.*;
import java.io.*;
public class Banhmi {
// https://codeforces.com/contest/1062/problem/C
public static void main(String[] args) throws IOException, FileNotFoundException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
//BufferedReader in = new BufferedReader(new FileReader("Banhmi"));
StringTokenizer st = new StringTokenizer(in.readLine());
int n = Integer.parseInt(st.nextToken());
int q = Integer.parseInt(st.nextToken());
char[] arr = new char[n];
arr = in.readLine().toCharArray();
int[] one_count = new int[n+1];
for (int i=0; i<n; i++) {
one_count[i+1] = one_count[i];
if (arr[i] == '1') one_count[i+1]++;
}
long mod = (long)1e9 + 7;
long[] powers_of_two = new long[n+10];
powers_of_two[0] = 1;
for (int i=1; i<powers_of_two.length; i++) {
powers_of_two[i] = powers_of_two[i-1] * 2;
powers_of_two[i] %= mod;
}
for (int i=0; i<q; i++) {
st = new StringTokenizer(in.readLine());
int l = Integer.parseInt(st.nextToken());
int r = Integer.parseInt(st.nextToken());
int num_ones = one_count[r] - one_count[l-1];
int num_zeroes = r-l+1 - num_ones;
long curans=0;
curans += powers_of_two[num_ones]-1;
curans %= mod;
long others = curans;
others *= powers_of_two[num_zeroes]-1;
others %= mod;
curans += others;
curans = (curans%mod+mod)%mod;
System.out.println(curans);
}
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import sys
r = sys.stdin.readlines()
M = 10 ** 9 + 7
n, q = r[0].strip().split()
n = int(n)
q = int(q)
s = r[1]
p = [0]
k = 0
for i in range(n):
d = int(s[i])
k += d
p.append(k)
ans = []
for k in range(q):
a, b = r[k + 2].strip().split()
a = int(a)
b = int(b)
l = b - a + 1
one = p[b] - p[a - 1]
zero = l - one
ans.append(str((pow(2, l, M) - pow(2, zero, M) + M) % M))
sys.stdout.write("\n".join(ans)) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
if (a == 0) return b;
return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
T ans = 1;
while (b > 0) {
if (b % 2 == 1) ans = ((ans % m) * (a % m)) % m;
b /= 2;
a = ((a % m) * (a % m)) % m;
}
return ans % m;
}
const long double eps = 1e-10L;
long long one[300005], zero[300005];
vector<long long> v;
void pw() {
long long val = 1;
long long temp = 1;
v.push_back(val);
for (long long i = 1; i <= 300005; i++) {
temp *= 2;
temp %= (long long)(1000 * 1000 * 1000 + 7);
val += temp;
val %= (long long)(1000 * 1000 * 1000 + 7);
v.push_back(val);
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long n, q;
cin >> n >> q;
pw();
string s;
cin >> s;
for (long long i = 1; i <= n; i++) {
if (s[i - 1] == '1')
one[i]++;
else
zero[i]++;
one[i] += one[i - 1];
zero[i] += zero[i - 1];
}
while (q--) {
long long a, b;
cin >> a >> b;
long long o = one[b] - one[a - 1];
long long z = zero[b] - zero[a - 1];
long long ans = v[o + z - 1];
if (z > 0) ans -= v[z - 1];
ans += (long long)(1000 * 1000 * 1000 + 7);
ans %= (long long)(1000 * 1000 * 1000 + 7);
cout << ans << "\n";
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.*;
import java.util.*;
import java.math.*;
public class A implements Runnable {
public void run() {
long startTime = System.nanoTime();
int M = (int) (1e9 + 7);
int n = nextInt();
int m = nextInt();
char[] c = nextToken().toCharArray();
int[] s = new int[n + 1];
for (int i = 0; i < n; i++) {
s[i + 1] = s[i] + (c[i] - '0');
}
for (int j = 0; j < m; j++) {
int a = nextInt() - 1;
int b = nextInt();
int x = s[b] - s[a];
int y = b - a - x;
println((get(2, x, M) + M - 1) % M * get(2, y, M) % M);
}
if (fileIOMode) {
System.out.println((System.nanoTime() - startTime) / 1e9);
}
out.close();
}
private long get(long x, int pow, int m) {
if (pow == 0) {
return 1;
}
if ((pow & 1) == 0) {
return get(x * x % m, pow >> 1, m);
} else {
return x * get(x, pow - 1, m) % m;
}
}
//-----------------------------------------------------------------------------------
private static boolean fileIOMode;
private static BufferedReader in;
private static PrintWriter out;
private static StringTokenizer tokenizer;
public static void main(String[] args) throws Exception {
fileIOMode = args.length > 0 && args[0].equals("!");
if (fileIOMode) {
in = new BufferedReader(new FileReader("a.in"));
out = new PrintWriter("a.out");
} else {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
}
tokenizer = new StringTokenizer("");
new Thread(new A()).start();
}
private static String nextLine() {
try {
return in.readLine();
} catch (IOException e) {
return null;
}
}
private static String nextToken() {
while (!tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(nextLine());
}
return tokenizer.nextToken();
}
private static int nextInt() {
return Integer.parseInt(nextToken());
}
private static long nextLong() {
return Long.parseLong(nextToken());
}
private static double nextDouble() {
return Double.parseDouble(nextToken());
}
private static BigInteger nextBigInteger() {
return new BigInteger(nextToken());
}
private static void print(Object o) {
if (fileIOMode) {
System.out.print(o);
}
out.print(o);
}
private static void println(Object o) {
if (fileIOMode) {
System.out.println(o);
}
out.println(o);
}
private static void printf(String s, Object... o) {
if (fileIOMode) {
System.out.printf(s, o);
}
out.printf(s, o);
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | def solve():
mod=10**9+7
def _mp():
return map(int,raw_input().split())
n,q=_mp()
#n=10**5
count=[0 for _ in range(n+2)]
p2=[1 for _ in range(n+2)]
r=1
for i in range(1,n+1):
# r*=2
p2[i]=(p2[i-1]*2)%mod
s=raw_input()
for i in range(1,n+1):
count[i]=count[i-1]+int(s[i-1])
for i in range(q):
l,r=_mp()
n1=r-l+1
n2=n1-(count[r]-count[l-1])
#if p2[n1]<p2[n2]:
# p2[n1]=p2[n1]+mod
print (p2[n1]-p2[n2])%mod
#print (2**n1-2**n2)%mod
if __name__=="__main__":
solve(); | PYTHON |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
char str[1008611];
long long s1[1008611], s2[1008611];
long long pow_(long long a, long long b) {
long long ans = 1;
while (b) {
if (b & 1) ans = (ans * a) % 1000000007;
b >>= 1;
a = (a * a) % 1000000007;
}
return ans;
}
long long inv(long long x) { return pow_(x, 1000000007 - 2); }
int main() {
int n, q;
cin >> n >> q;
cin >> str;
if (str[0] == '0') {
s1[0] = 1;
s2[0] = 0;
} else {
s1[0] = 0;
s2[0] = 1;
}
for (int i = 1; i < n; i++) {
s1[i] = s1[i - 1];
s2[i] = s2[i - 1];
if (str[i] == '0') {
s1[i] = (s1[i] + 1) % 1000000007;
} else {
s2[i] = (s2[i] + 1) % 1000000007;
}
}
long long n1, n2;
while (q--) {
int l, r;
cin >> l >> r;
n1 = (s1[r - 1] - s1[l - 1]) % 1000000007;
n2 = (s2[r - 1] - s2[l - 1]) % 1000000007;
if (str[l - 1] == '0') {
n1 = (n1 + 1) % 1000000007;
} else {
n2 = (n2 + 1) % 1000000007;
}
long long a, b;
a = pow_(2, n2) - 1;
a = a % 1000000007;
b = ((pow_(2, n1) - 1) * a);
b = (b + 1000000007) % 1000000007;
long long sum = (a + b) % 1000000007;
cout << sum << endl;
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | # Causes TLE
# C++17 implemention -> 1062c.cpp
MOD = 1000000007
def main():
buf = input()
buflist = buf.split()
n = int(buflist[0])
q = int(buflist[1])
buf = input()
x = buf
sum_list = [0] # sentinel / one indexing
for i, deliciousness in enumerate(x):
sum_list.append(int(deliciousness) + sum_list[i])
enjoyment_list = [0]
for i in range(n):
enjoyment_list.append((enjoyment_list[i] * 2 + 1) % MOD)
query_list = []
for i in range(q):
buf = input()
buflist = buf.split()
l = int(buflist[0]) # one indexing
r = int(buflist[1]) # one indexing
query_list.append((l, r))
for i, query in enumerate(query_list):
l = query[0]
r = query[1]
banhmi_count = r - l + 1
delicious_count = sum_list[r] - sum_list[l - 1]
non_delicious_count = banhmi_count - delicious_count
enjoyment = 0
# main part
if delicious_count == 0:
enjoyment = 0
else:
enjoyment += enjoyment_list[delicious_count]
enjoyment += (enjoyment_list[banhmi_count] - enjoyment_list[delicious_count] - enjoyment_list[non_delicious_count])
enjoyment = enjoyment % MOD
print(enjoyment)
if __name__ == '__main__':
main()
| PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAX_N = 100002;
const int MOD = 1000000007;
int n, nQueries, ps[MAX_N];
int64_t pw2[MAX_N];
void solve() {
cin >> n >> nQueries;
pw2[0] = 1;
for (int i = 1; i <= n; ++i) {
char x;
cin >> x;
ps[i] = ps[i - 1] + x - '0';
pw2[i] = pw2[i - 1] * 2 % MOD;
}
while (nQueries--) {
int l, r;
cin >> l >> r;
int64_t tmp1 = pw2[ps[r] - ps[l - 1]] - 1;
int64_t tmp2 = pw2[r - l + 1 - ps[r] + ps[l - 1]] - 1;
cout << (tmp1 + tmp1 * tmp2 % MOD) % MOD << '\n';
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
solve();
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int LIMIT = 1e5 + 7;
const int MOD = 1e9 + 7;
const int MAX = 1 << 30;
int n, q, l, r, dp[LIMIT], f[LIMIT];
char c;
int main() {
scanf("%d %d\n", &n, &q);
f[0] = 1;
for (int i = 0; i < n; i++) {
scanf("%c", &c);
dp[i + 1] = dp[i] + (c - '0');
f[i + 1] = f[i] * 2ll % MOD;
}
while (q--) {
scanf("%d %d", &l, &r);
unsigned long long ones = dp[r] - dp[l - 1];
unsigned long long zeros = r - l + 1 - ones;
unsigned long long ans = (f[ones] - 1ll) % MOD;
ans = (ans * (unsigned long long)f[zeros]) % MOD;
printf("%lld\n", ans);
}
return EXIT_SUCCESS;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | # TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
from sys import stdin, stdout
import collections
#N = int(input())
#s = input()
N,Q = [int(x) for x in stdin.readline().split()]
s = input()
ones = [0]*N
t = 0
for i in range(N):
if s[i]=='1':
t += 1
ones[i] = t
find = [1]*(N+1)
for i in range(1,N+1):
find[i] = (find[i-1]*2)%1000000007
ans = [0]*Q
for i in range(Q):
L,R = [int(x) for x in stdin.readline().split()]
L -= 1
R -= 1
if L==0:
one = ones[R]
else:
one = ones[R] - ones[L-1]
if one==0:
ans[i] = 0
else:
size = R-L+1
a1 = find[size-1]
res = find[size] - find[size-one]
ans[i] = res%1000000007
print(*ans,sep='\n')
| PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import sys
r = sys.stdin.readlines()
n, q = r[0].split(' ')
n = int(n)
q = int(q)
M = 10 ** 9 + 7
twopow = [1] * (2 * n + 1)
for j in range(1, (2 * n + 1)):
twopow[j] = twopow[j - 1] * 2 % M
s = r[1]
p = [0] * (n + 1)
for v in range(n):
p[v + 1] = p[v] + (s[v] == '1')
ans = [0] * q
for k in range(q):
a, b = r[k + 2].split(' ')
a = int(a)
b = int(b)
l = b - a + 1
one = p[b] - p[a - 1]
v = (twopow[l] - twopow[l - one] + M) % M
ans[k] = str(v)
sys.stdout.write("\n".join(ans))
| PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.DataInputStream;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author xwchen
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskC solver = new TaskC();
solver.solve(1, in, out);
out.close();
}
static class TaskC {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
int q = in.nextInt();
char[] s = in.next().toCharArray();
int[] sum = new int[s.length + 1];
long[] e2 = new long[s.length + 1];
long mod = 1000000007L;
e2[0] = 1;
for (int i = 1; i <= n; ++i) {
sum[i] = sum[i - 1] + ((s[i - 1] == '1') ? 1 : 0);
e2[i] = e2[i - 1] * 2;
e2[i] %= mod;
}
while (q-- > 0) {
int l = in.nextInt();
int r = in.nextInt();
int len = r - l + 1;
int one = sum[r] - sum[l - 1];
int zero = len - one;
long ans = e2[one] - 1;
long base = ans;
ans = ans + (e2[zero] - 1) * base % mod;
ans %= mod;
out.println(ans);
}
}
}
static class InputReader {
final private int BUFFER_SIZE = 1 << 10;
final private int LINE_SIZE = 1 << 20;
private DataInputStream in;
private byte[] buffer;
private int bufferPointer;
private int bytesRead;
public InputReader(InputStream inputStream) {
this.in = new DataInputStream(inputStream);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String next() {
byte[] buf = new byte[LINE_SIZE]; // line length
int cnt = 0, c;
c = read();
while (c == ' ' || c == '\n' || c == '\r')
c = read();
do {
if (c == ' ' || c == '\n' || c == '\r')
break;
buf[cnt++] = (byte) c;
} while ((c = read()) != -1);
return new String(buf, 0, cnt);
}
public int nextInt() {
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
private void fillBuffer() {
try {
bytesRead = in.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
} catch (IOException e) {
throw new InputMismatchException();
}
}
private byte read() {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
public class Main {
static class Task {
int NN = 1000006;
int MOD = 1000000007;
int [] zeros = new int[NN];
long [] exp2 = new long[NN];
public void solve(InputReader in, PrintWriter out) {
int n = in.nextInt(), q = in.nextInt();
String s = in.next();
for(int i=0;i<n;++i) {
zeros[i] = 0;
if(i > 0) {
zeros[i] = zeros[i - 1];
}
if(s.charAt(i) == '0') {
++zeros[i];
}
}
exp2[0] = 1L;
for(int i=1;i<NN;++i) {
exp2[i] = (exp2[i - 1] * 2L)%MOD;
}
while(q-- > 0) {
int l = in.nextInt(), r = in.nextInt();
--l;--r;
int zero = zeros[r];
if(l > 0) {
zero -= zeros[l - 1];
}
int len = r - l + 1;
long ans = exp2[len];
ans = (ans - exp2[zero] + MOD)%MOD;
out.println(ans);
}
}
}
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Task solver = new Task();
solver.solve(in, out);
out.close();
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import sys
r = sys.stdin.readlines()
M = 10 ** 9 + 7
n, q = r[0].split(' ')
n = int(n)
q = int(q)
s = r[1]
p = [0]
for v in range(n):
p.append(p[v] + int(s[v]))
ans = []
for k in range(q):
a, b = r[k + 2].split(' ')
a = int(a)
b = int(b)
l = b - a + 1
one = p[b] - p[a - 1]
v = (pow(2, l, M) - pow(2, l - one, M) + M) % M
ans.append(str(v))
sys.stdout.write("\n".join(ans)) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | def init_input():
import os
from sys import stdin
it = iter(os.read(stdin.fileno(), 10 ** 7).split())
return lambda: next(it).decode(), lambda: int(next(it)), lambda: float(next(it))
ns, ni, nf = init_input()
MOD = 10 ** 9 + 7
n, q = ni(), ni()
s = 'x' + ns()
c = [0] * (n + 1)
for i in range(1, n + 1):
c[i] = c[i - 1] + (s[i] == '1')
p2 = [1] * (2 * n + 1)
for i in range(1, 2 * n + 1):
p2[i] = p2[i - 1] * 2 % MOD
out = []
for qq in range(q):
l, r = ni(), ni()
o = c[r] - c[l - 1]
z = (r - l + 1) - o
ans = (p2[o + z] - 1 - p2[z] + 1) % MOD
out.append(ans)
print(*out, sep='\n')
| PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | def beki(a,b):
waru=10**9+7
ans=1
while(b>0):
if(1 & b):
ans= ans * a %waru
b >>= 1
a=a * a % waru
return ans
n,m=map(int,input().split())
s=input()
ans=[]
waru=10**9+7
ruiseki=[0]*(n+1)
bekij=[1]*(n+1)
for i in range(n):
ruiseki[i+1]=ruiseki[i]+int(s[i])
bekij[i+1]=(bekij[i]*2)%waru
for i in range(m):
l,r=map(int,input().split())
ko=ruiseki[r]-ruiseki[l-1]
ans.append(str((((bekij[ko] -1)) * ((bekij[r+1-l - ko]))) %waru))
print("\n".join(ans)) | PYTHON3 |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | from sys import stdin
out = []
n,q = map(int, raw_input().split())
s = raw_input()
pref = [0]
for i in s:
if i == '0':
pref.append(pref[-1])
else:
pref.append(pref[-1]+1)
MOD = pow(10,9)+7
pow2 = [1]*(1+n)
for i in range(n):
pow2[i+1] = (pow2[i] *2)%MOD
for line in stdin.readlines():
l,r = map(int, line.split())
n = pref[r] - pref[l-1]
k = r - l + 1 - n
out.append(str((pow2[n]-1 + (pow2[n]-1)*(pow2[k]-1))%MOD))
print('\n'.join(out))
| PYTHON |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int mul(int a, int b) { return (long long)a * b % mod; }
int power(int a, long long b) {
int res = 1;
while (b > 0) {
if (b & 1) {
res = mul(res, a);
}
a = mul(a, a);
b >>= 1;
}
return res;
}
void solve() {
int n, q;
cin >> n >> q;
string s;
cin >> s;
vector<int> ps(n + 1);
for (int i = 0; i < n; i++) {
ps[i + 1] = ps[i] + (s[i] == '1');
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
int one = ps[r] - ps[l - 1];
int zeo = r - l + 1 - one;
cout << (power(2, one) - 1) * 1ll * power(2, zeo) % mod << "\n";
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
solve();
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
int n, q, f[N];
char c;
int binPow(int x, int y) {
int ans = 1;
while (y > 0) {
if (y & 1) ans = (1LL * ans * x) % mod;
x = (1LL * x * x) % mod;
y >>= 1;
}
return ans;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> q;
for (int i = 1; i <= n; i++) cin >> c, f[i] = f[i - 1] + (c == '1');
while (q--) {
int l, r;
cin >> l >> r;
int x = f[r] - f[l - 1], y = r - l + 1 - x;
int ans = binPow(2, x) - 1;
ans = (1LL * ans * binPow(2, y)) % mod;
if (ans < 0) ans += mod;
cout << ans << "\n";
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
int n, q, x, y, z, h;
string s;
long long a[100005], b[100005];
void init() {
b[0] = 1;
for (int i = 1; i <= n; i++) {
b[i] = b[i - 1] * 2 % mod;
}
}
int main() {
cin >> n >> q;
cin >> s;
init();
for (int i = 0; i < n; i++) a[i] = a[i - 1] + s[i] - 48;
while (q--) {
cin >> x >> y;
if (x == 1)
z = a[y - 1];
else
z = a[y - 1] - a[x - 2];
h = y - x + 1;
long long ans = (b[h] - b[h - z] + mod) % mod;
cout << ans << endl;
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
#pragma GCC optimize("03")
using namespace std;
long long int const mod = 1e9 + 7;
long long int pre[100010][2];
void mul(long long int &x, long long int val) {
x = (x * val) % mod;
if (x < 0) x += mod;
}
void add(long long int &x, long long int val) {
x = (x + val) % mod;
if (x < 0) x += mod;
}
long long int power(long long int x, long long int y) {
long long int res = 1;
while (y > 0) {
if (y & 1ll) mul(res, x);
mul(x, x);
y >>= 1;
}
return res;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int n, q, l, r, one, zero, x, y, val, d;
string st;
cin >> n >> q;
cin >> st;
for (long long int i = 1; i <= n; i++) {
d = st[i - 1] - '0';
pre[i][d] += 1;
}
for (long long int i = 1; i <= n; i++) {
pre[i][0] += pre[i - 1][0];
pre[i][1] += pre[i - 1][1];
}
while (q--) {
cin >> l >> r;
one = pre[r][1] - pre[l - 1][1];
zero = pre[r][0] - pre[l - 1][0];
if (one) {
x = (power(2, one) - 1);
val = x;
y = (power(2, zero) - 1);
mul(y, val);
add(x, y);
cout << x << "\n";
} else {
y = 0;
cout << y << "\n";
}
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long a[1000000];
long long bp(long long a, int n) {
long long res = 1;
while (n) {
if (n % 2) res *= a;
a *= a;
a %= mod;
n /= 2;
res %= mod;
}
return res;
}
int main() {
int n, q;
cin >> n >> q;
for (int i = 1; i <= n; i++) {
char c;
cin >> c;
a[i] = a[i - 1] + (c == '0');
}
while (q--) {
int l, r;
cin >> l >> r;
cout << (bp(2, r - l + 1) + 10 * mod - 1 - bp(2, a[r] - a[l - 1]) + 1) % mod
<< endl;
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.util.*;
import java.io.*;
public class Main
{
public static void main(String[] args)
{
FastReader reader = new FastReader();
PrintWriter writer = new PrintWriter(System.out);
int n = reader.nextInt();
int q = reader.nextInt();
String input = reader.nextLine();
int[] zeros = new int[n+1];
int[] ones = new int[n+1];
for (int i=0; i<n; i++)
if (input.charAt(i) == '0')
zeros[i+1]++;
else
ones[i+1]++;
for (int i=0; i<n; i++)
{
zeros[i+1] += zeros[i];
ones[i+1] += ones[i];
}
long mod = 1000000007;
long[] pow = new long[n+1];
pow[0] = 1;
for (int i=1; i<=n; i++)
pow[i] = (pow[i-1]*2) % mod;
while (q > 0)
{
int l = reader.nextInt()-1;
int r = reader.nextInt();
int z = zeros[r] - zeros[l];
int o = ones[r] - ones[l];
writer.println((pow[z] * (pow[o]+mod-1)) % mod);
q--;
}
writer.close();
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
} | JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | import java.io.*;
import java.util.stream.Stream;
/**
* @author madi.sagimbekov
*/
public class C1062C {
private static BufferedReader in;
private static BufferedWriter out;
public static void main(String[] args) throws IOException {
open();
int[] nq = readInts();
int n = nq[0];
int q = nq[1];
char[] a = readString().toCharArray();
int[] z = new int[n];
int[] o = new int[n];
for (int i = 0; i < n; i++) {
if (i == 0) {
if (a[i] == '0') {
z[i]++;
} else {
o[i]++;
}
} else {
if (a[i] == '0') {
z[i] = z[i - 1] + 1;
o[i] = o[i - 1];
} else {
o[i] = o[i - 1] + 1;
z[i] = z[i - 1];
}
}
}
long mod = 1000000007;
for (int i = 0; i < q; i++) {
int[] lr = readInts();
int l = lr[0];
int r = lr[1];
int zeros = z[r - 1] - (l == 1 ? 0 : z[l - 2]);
int ones = o[r - 1] - (l == 1 ? 0 : o[l - 2]);
long ans1 = pow(ones, mod) - 1;
long ans2 = pow(zeros, mod) - 1;
long ans = (ans1 % mod + (ans1 * ans2) % mod) % mod;
out.write(ans + "\n");
}
close();
}
private static long pow(int k, long mod) {
if (k == 0) {
return 1;
}
if (k == 1) {
return 2;
}
long res = pow(k / 2, mod);
if (k % 2 == 0) {
return (res * res) % mod;
} else {
return (res * res * 2) % mod;
}
}
private static int[] readInts() throws IOException {
return Stream.of(in.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
}
private static int readInt() throws IOException {
return Integer.parseInt(in.readLine());
}
private static long[] readLongs() throws IOException {
return Stream.of(in.readLine().split(" ")).mapToLong(Long::parseLong).toArray();
}
private static long readLong() throws IOException {
return Long.parseLong(in.readLine());
}
private static double[] readDoubles() throws IOException {
return Stream.of(in.readLine().split(" ")).mapToDouble(Double::parseDouble).toArray();
}
private static double readDouble() throws IOException {
return Double.parseDouble(in.readLine());
}
private static String readString() throws IOException {
return in.readLine();
}
private static void open() {
in = new BufferedReader(new InputStreamReader(System.in));
out = new BufferedWriter(new OutputStreamWriter((System.out)));
}
private static void close() throws IOException {
out.flush();
out.close();
in.close();
}
}
| JAVA |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const int INFMEM = 63;
const int INF = 1061109567;
const long long LINF = 4557430888798830399LL;
const double DINF = numeric_limits<double>::infinity();
const long long MOD = 1000000007;
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
const double PI = 3.141592653589793;
inline void fastll(long long &input_number) {
input_number = 0;
int ch = getchar_unlocked();
int sign = 1;
while (ch < '0' || ch > '9') {
if (ch == '-') sign = -1;
ch = getchar_unlocked();
}
while (ch >= '0' && ch <= '9') {
input_number = (input_number << 3) + (input_number << 1) + ch - '0';
ch = getchar_unlocked();
}
input_number *= sign;
}
inline void open(string a) {
freopen((a + ".in").c_str(), "r", stdin);
freopen((a + ".out").c_str(), "w", stdout);
}
inline void fasterios() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
long long n, q;
long long power[100005];
long long isi[100005];
long long satu[100005];
long long nol[100005];
int main() {
fasterios();
cin >> n >> q;
power[0] = 1;
for (int i = 1; i <= 100001; i++) power[i] = (power[i - 1] * 2) % MOD;
for (int i = 1; i <= n; i++) {
char tmp;
cin >> tmp;
isi[i] = (tmp == '1');
if (isi[i])
satu[i]++;
else
nol[i]++;
}
for (int i = 1; i <= n; i++) {
nol[i] += nol[i - 1];
satu[i] += satu[i - 1];
}
while (q--) {
long long l, r;
cin >> l >> r;
long long curnol = nol[r] - nol[l - 1];
long long cursatu = satu[r] - satu[l - 1];
long long ans = power[cursatu] - 1;
ans *= power[curnol];
cout << ans % MOD << '\n';
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long sz = 2e5 + 5;
const long long MD = 1e9 + 7;
pair<long long, long long> st[2 * sz];
long long arr[sz];
long long n, q;
string s;
pair<long long, long long> ADD(pair<long long, long long> &A,
pair<long long, long long> &B) {
return {A.first + B.first, A.second + B.second};
}
void build() {
for (long long i = 0; i < n; ++i) {
arr[i] = s[i] == '1';
}
for (long long i = 0; i < n; ++i) {
st[i + n] = {arr[i] == 0, arr[i] == 1};
}
for (long long i = n - 1; i >= 0; --i) {
st[i] = ADD(st[i << 1], st[i << 1 | 1]);
}
}
pair<long long, long long> query(long long L, long long R) {
--L;
pair<long long, long long> tot = {0LL, 0LL};
for (L += n, R += n; L < R; L >>= 1, R >>= 1) {
if (L & 1) tot = ADD(tot, st[L++]);
if (R & 1) tot = ADD(tot, st[--R]);
}
return tot;
}
long long powMod(long long a, long long p) {
long long res = 1;
for (; p; p >>= 1, a = ((a) * (a)) % MD) {
if (p & 1) res *= a, res %= MD;
}
return res;
}
long long MUL(long long a, long long b) {
a %= MD;
b %= MD;
return (a * b) % MD;
}
int main() {
scanf("%lld", &n), scanf("%lld", &q);
cin >> s;
build();
while (q--) {
long long L, R;
scanf("%lld", &L), scanf("%lld", &R);
pair<long long, long long> res = query(L, R);
long long le = (powMod(2LL, res.second) + MD - 1) % MD;
long long ri = (powMod(2LL, res.first) + MD - 1) % MD;
printf("%lld\n", MUL(le, 1LL + ri));
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long int bs(long long int a, long long int b) {
long long int res = 1;
while (b) {
if (b & 1) res *= a;
res %= ((long long int)1e9 + 7);
b >>= 1;
a *= a;
a %= ((long long int)1e9 + 7);
}
return res;
}
signed main() {
long long int n, m;
cin >> n >> m;
string a;
cin >> a;
long long int p[n];
memset(p, 0, sizeof p);
for (long long int i = 0; i < n; i++) {
if (i) p[i] += p[i - 1];
if (a[i] == '1') p[i]++;
}
while (m--) {
long long int l, r;
cin >> l >> r;
--l, --r;
long long int nos = p[r] - (l ? p[l - 1] : 0);
long long int noo = (r - l + 1) - nos;
long long int ptf = bs(2, nos) - 1;
long long int ptt = (bs(2, noo) - 1) * ptf;
cout << (ptt % ((long long int)1e9 + 7) + ptf % ((long long int)1e9 + 7)) %
((long long int)1e9 + 7)
<< '\n';
}
return 0;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | from __future__ import division
from sys import stdin, stdout
# from fractions import gcd
# from math import *
# from operator import mul
# from functools import reduce
# from copy import copy
from collections import deque, defaultdict, Counter
rstr = lambda: stdin.readline().strip()
rstrs = lambda: [str(x) for x in stdin.readline().split()]
rint = lambda: int(stdin.readline())
rints = lambda: [int(x) for x in stdin.readline().split()]
rstr_2d = lambda n: [rstr() for _ in range(n)]
rint_2d = lambda n: [rint() for _ in range(n)]
rints_2d = lambda n: [rints() for _ in range(n)]
pr = lambda args, sep: stdout.write(sep.join(map(str, args)) + '\n')
out = []
add = lambda a, b: (a + b) % mod
def arr_sum():
tem0, tem1 = [0], [0]
for i in range(n):
tem0.append(tem0[i] + int(s[i] == '0'))
tem1.append(tem1[i] + int(s[i] == '1'))
return tem0, tem1
n, q = rints()
s, qur, mod = rstr(), rints_2d(q), 10 ** 9 + 7
zero, one = arr_sum()
for l, r in qur:
all0, all1 = zero[r] - zero[l - 1], one[r] - one[l - 1]
if not all1:
out.append(0)
else:
v1 = add(-1, add(-pow(2, all0, mod), 1))
out.append(add(pow(2, all1 + all0, mod), v1))
pr(out, '\n')
| PYTHON |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long gcd(long long n1, long long n2) {
if (!n1) return n2;
if (!n2) return n1;
if (n1 % n2 == 0) return n2;
return gcd(n2, n1 % n2);
}
long long powmod(long long base, long long exponent) {
base %= 1000000007;
long long ans = 1;
while (exponent) {
if (exponent & 1) ans = (ans * base) % 1000000007;
base = (base * base) % 1000000007;
exponent /= 2;
}
ans %= 1000000007;
return ans;
}
int arr[1000100 + 1];
long long dp[1000100 + 1];
int main() {
clock_t begin = clock();
;
int i, j, k, n, q, l, r;
scanf("%d", &n);
scanf("%d", &q);
string s;
cin >> s;
for (i = 1; i <= n; i++) {
arr[i] = arr[i - 1];
if (s[i - 1] == '1') arr[i]++;
}
while (q--) {
scanf("%d", &l);
scanf("%d", &r);
int ones = arr[r] - arr[l - 1];
int zeroes = ((r - l + 1) - ones);
long long answer = powmod(2, ones) - 1;
answer = (answer * powmod(2, zeroes)) % 1000000007;
printf("%lld\n", answer);
}
clock_t end = clock();
double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
fprintf(stderr, "\nTime elapsed : %.3f seconds\n", elapsed_secs);
return 0;
;
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int nn = 1e5 + 5;
const long long MOD = 1e9 + 7;
long long pangkat[nn], arr[nn], pre0[nn], pre1[nn];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n, q;
pangkat[0] = 1;
for (int i = 1; i <= 1e5; i++) {
pangkat[i] = (pangkat[i - 1] * 2) % MOD;
}
cin >> n >> q;
char a;
for (int i = 1; i <= n; i++) {
cin >> a;
arr[i] = a - '0';
}
pre0[1] = (arr[1] == 0);
pre1[1] = (arr[1] == 1);
for (int i = 2; i <= n; i++) {
pre0[i] = pre0[i - 1] + (arr[i] == 0);
pre1[i] = pre1[i - 1] + (arr[i] == 1);
}
while (q--) {
int l, r;
cin >> l >> r;
int nol = pre0[r] - pre0[l - 1];
int satu = pre1[r] - pre1[l - 1];
long long bil1 = pangkat[nol];
long long bil2 = pangkat[satu] - 1;
while (bil2 < 0) bil2 += MOD;
cout << (bil1 * bil2) % MOD << "\n";
}
}
| CPP |
1062_C. Banh-mi | JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.
First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.
For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.
However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.
All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.
Input
The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).
The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.
Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.
Output
Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.
Examples
Input
4 2
1011
1 4
3 4
Output
14
3
Input
3 2
111
1 2
3 3
Output
3
1
Note
In the first example:
* For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2.
* For query 2: Both 3, 4 and 4, 3 ordering give the same answer.
In the second example, any order of eating parts leads to the same answer. | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long n, q, l, r, sum[100010], tmp1, tmp2, ans;
string s;
long long Pow(long long a, long long b) {
long long ans = 1;
while (b) {
if (b & 1 == 1) ans = (ans * a) % 1000000007LL;
a = (a * a) % 1000000007LL, b >>= 1;
}
return ans % 1000000007LL;
}
int main() {
cin >> n >> q;
cin >> s;
for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + s[i - 1] - '0';
while (q--) {
cin >> l >> r;
tmp1 = sum[r] - sum[l - 1], tmp2 = r - l + 1 - tmp1;
ans = Pow(2, tmp1) - 1;
if (ans == -1) ans += 1000000007LL;
ans = (ans * Pow(2, tmp2)) % 1000000007LL;
cout << ans % 1000000007LL << endl;
}
}
| CPP |
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