HDU 6153 A Secret 套路，求解前缀在本串中出现的次数

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 2e6 + 20;
int nextliu[maxn], dp[maxn];
char str[maxn], sub[maxn];
void get_next (char str[], int nextliu[], int lenstr) {
    int i = 1, j = 0;
    nextliu[1] = 0;
    while (i <= lenstr) {
        if (j == 0 || str[i] == str[j]) {
            nextliu[++i] = ++j;
        } else j = nextliu[j];
    }
    return ;

}
const int MOD = 1e9 + 7;
int dp2[maxn];
void work() {
    scanf("%s%s", str + 1, sub + 1);
    int lenstr = strlen(str + 1), lensub = strlen(sub + 1);
    reverse(str + 1, str + 1 + lenstr);
    reverse(sub + 1, sub + 1 + lensub);
    for (int i = 1; i <= lensub; ++i) dp2[i] = 1;
    get_next(sub, nextliu, lensub);
    for (int i = lensub; i >= 1; --i) {
        int t = nextliu[i + 1] - 1;
        dp2[t] += dp2[i];
        dp2[t] %= MOD;
    }
    int to = lensub + 1;
    sub[to++] = ‘$‘;
    for (int i = 1; i <= lenstr; ++i) sub[to++] = str[i];
    sub[to] = ‘\0‘;
    to--;

    for (int i = 1; i <= to; ++i) dp[i] = 1;
    get_next(sub, nextliu, to);
    for (int i = to; i >= 1; --i) {
        int t = nextliu[i + 1] - 1;
        dp[t] += dp[i];
        dp[t] %= MOD;
    }
//    printf("%d\n", dp[2] - dp2[2]);
    LL ans = 0;
    for (int i = 1; i <= lensub; ++i) {
        ans += 1LL * i * ((dp[i] + MOD - dp2[i]) % MOD);
        ans %= MOD;
    }
    printf("%I64d\n", ans);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}