HDU 5763 Another Meaning

Problem Description

As is known to all, in many cases, a word has two meanings. Such as “hehe”, which not only means “hehe”, but also means “excuse me”. 
Today, ?? is chating with MeiZi online, MeiZi sends a sentence A to ??. ?? is so smart that he knows the word B in the sentence has two meanings. He wants to know how many kinds of meanings MeiZi can express.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each test case contains two strings A and B, A means the sentence MeiZi sends to ??, B means the word B which has two menaings. string only contains lowercase letters.

Limits
T <= 30
|A| <= 100000
|B| <= |A|

 

Output

For each test case, output one line containing “Case #x: y” (without quotes) , where x is the test case number (starting from 1) and y is the number of the different meaning of this sentence may be. Since this number may be quite large, you should output the answer modulo 1000000007.

 

Sample Input

4
hehehe
hehe
woquxizaolehehe
woquxizaole
hehehehe
hehe
owoadiuhzgneninougur
iehiehieh

 

Sample Output

Case #1: 3
Case #2: 2
Case #3: 5
Case #4: 1

Hint
In the first case, “ hehehe” can have 3 meaings: “*he”, “he*”, “hehehe”.
In the third case, “hehehehe” can have 5 meaings: “*hehe”, “he*he”, “hehe*”, “**”, “hehehehe”.
 

用kmp确定每个匹配的位置，用dp统计答案。
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
const int INF = 0x7FFFFFFF;
int T, n, f[N], nt[N], cas = 0;
char S[N], s[N];

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%s%s", S, s);
        nt[0] = -1;    n = strlen(S);
        for (int i = 0; s[i]; i++)
        {
            int k = nt[i];
            while (k >= 0 && s[i] != s[k]) k = nt[k];
            nt[i + 1] = k + 1;
        }
        f[0] = 1;
        for (int i = 0, j = nt[1]; S[i];)
        {
            if (j < 0 || S[i] == s[j])
            {
                f[i + 1] = f[i];
                i++, j++;
                if (!s[j]) f[i] = (f[i] + f[i - j]) % mod;
            }
            else j = nt[j];
        }
        printf("Case #%d: %d\n", ++cas, f[n]);
    }
    return 0;
}