AcWing 138. 兔子与兔子
原题链接
简单
作者:
我要出去乱说
,
2021-01-09 16:36:33
,
所有人可见
,
阅读 371
#include <iostream>
#include <cstring>
using namespace std;
typedef unsigned long long ULL;
const int N = 1e6 + 10, P = 131;
char str[N];
int n, m;
ULL h[N], p[N];
ULL get(int l, int r) {
return h[r] - h[l - 1] * p[r - l + 1];
}
int main() {
scanf("%s%d", str + 1, &m);
int n = sizeof(str);
p[0] = 1;
for (int i = 1; i <= n; i ++ ) {
p[i] = p[i - 1] * P;
h[i] = h[i - 1] * P + str[i];
}
while (m -- ) {
int l1, r1, l2, r2;
scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
if (get(l1, r1) == get(l2, r2)) puts("Yes");
else puts("No");
}
return 0;
}