哈希--#103. 子串查找

问文本串在模式串中出现了几次

双哈希，预处理出$b$的$i$次方，哈希转化公式：$sum_i$=($sum_{i-1}$*$b^{len}$+$s_i$+1)%$mod$

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <cstring>
#define ull long long
using namespace std;
const ull b1=131,P1=1000429,maxn=1e6+100,b2=133,P2=19491001;
ull power1[maxn],sum1[maxn],power2[maxn],sum2[maxn];
char s1[maxn],s2[maxn];
int main() {
    power1[0]=1;power2[0]=1;
    for (int i = 1; i < 1000000; i++) 
        power1[i]=(power1[i-1]*b1)%P1,power2[i]=(power2[i-1]*b2)%P2;
    scanf ("%s%s",s2+1,s1+1);
    int n=strlen(s1+1),m=strlen(s2+1);
    sum1[0]=sum2[0]=0;
    for (int i = 1; i <= m; i++) 
        sum1[i]=(sum1[i-1]*b1+(ull)(s2[i]+1))%P1,sum2[i]=(sum2[i-1]*b2+(ull)(s2[i]+1))%P2;
    ull ss1=0,ss2=0;
    for (int i = 1; i <= n; i++) 
        ss1=(ss1*b1+(ull)(s1[i]+1)) %P1,ss2=(ss2*b2+(ull)(s1[i]+1)) %P2;
    int ans=0;
    for (int i = 0; i <= m-n; i++) {
        if (ss1%P1==((sum1[i+n]-sum1[i]*power1[n])%P1+P1)%P1) 
        if (ss2%P2==((sum2[i+n]-sum2[i]*power2[n])%P2+P2)%P2) ++ans;
    }
    printf("%d\n",ans);
    return 0;
}