【字符串哈希】 BKDR算法

給定一個字符串，問你子串S1[l1...r1]  與 子串S2[l2...r2] 是否相等。

字符串哈希就是將字符串映射到整數，然後就和普通的哈希就一樣了。

bkdr： hash[i] = (hash[i-1] * p + s[i]  -  'a'  +  1)  % mod

一般取值：P = 13,  mod = 101

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1000000+8;
const int p = 13;
const int mod = 1e9 + 7;
char s[maxn];
int bkdr(int l, int r)
{
	int hash = s[l] - 'a' + 1;
	int len  = r-l+1;
	for(int i = l+1; i <= r; i++) hash = (hash*p + s[i] -'a' + 1) % mod;
    return hash;
}
int main()
{
    int t;
    while(scanf("%s", s) == 1){
        scanf("%d", &t);
        int l1, r1, l2, r2;
		for(int i = 0; i < t; i++){
			scanf("%d %d %d %d",&l1, &r1, &l2, &r2);
			int hash1 = bkdr(l1-1, r1-1);
			int hash2 = bkdr(l2-1, r2-1);
			if(hash1 == hash2) printf("Yes\n");
			else printf("No\n");
		}
	}
    return 0;
}