HDU 5763

传送门

给你串A和串B，|B| <= |A|

已知串B有两种可能含义，求串A的总可能含义数

dp[i]表示串A的从头开始长度为i的串的可能意义的数目

若该长度为i的串的后缀与模式串B匹配，则该后缀可以选择替换或者不替换

dp[i] = dp[i - 1] + dp[i - Bl]   否则dp[i] = dp[i - 1]

#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e5 + 10;
char s1[maxn], s2[maxn];
int dp[maxn];
const ULL B = 100000007;

void contain(char *a, char *b) {
    int al = strlen(a), bl = strlen(b);
    ULL t = 1;
    for (int i = 0; i < al; i++) t *= B;
    ULL ah = 0, bh = 0;
    for (int i = 0; i < al; i++) ah = ah * B + a[i];
    for (int i = 0; i < al; i++) bh = bh * B + b[i];
    
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < al; i++) dp[i] = 1;
    
    for (int i = 0; i + al <= bl; i++) {
        dp[i + al] = dp[i + al - 1];
        if (ah == bh) {
            dp[i + al] = (dp[i + al] + dp[i]) % MOD;
        }
        if (i + al < bl) bh = bh * B + b[i + al] - b[i] * t;
    }
}


int main(int argc, const char * argv[]) {
    int T;
    scanf("%d", &T);
    for (int t = 1; t <= T; t++) {
        scanf("%s%s", s1, s2);
        contain(s2, s1);
        int l1 = strlen(s1);
        printf("Case #%d: ", t);
        printf("%d
", dp[l1]);
        
    }
    return 0;
}