HDU - 5763 Another Meaning

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.InputThe 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|

OutputFor 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”.

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

char s1[100005],s2[100005],nex[100005];
long long  vis[100005],dp[100005];

int main(){
    long long t,i,j,cas=1;
    scanf("%lld",&t);
    while(t--){
        scanf("%s%s",s1,s2);// s2 是匹配串 
        memset(nex,-1,sizeof(nex)); //初始化为-1 
        memset(vis,0,sizeof(vis));
        memset(dp,0,sizeof(dp));
        long long n=strlen(s1); 
        long long m=strlen(s2);
        j=-1;
        for(i=1;i<m;i++){
            while(j>=0&&s2[j+1]!=s2[i])j=nex[j];
            if(s2[j+1]==s2[i])j++;
            nex[i]=j;
        }    
        j=-1;
        for(i=-1;i<n-1;i++){
            while(j>=0&&s2[j+1]!=s1[i+1])j=nex[j];
            if(s2[j+1]==s1[i+1])j++;
            if(j==m-1){
                //j=next[j]; //重叠匹配 
                //j=-1; //从头匹配 
                j=nex[j];
                vis[i+1]=1;
            }
        }
        long long num=0;
        for(i=0;i<m;i++){
            dp[i]=vis[i];
        }
        for(;i<n;i++){
            if(vis[i]){
                dp[i]=(dp[i-1]+dp[i-m]+1)%1000000007;
            }
            else dp[i]=dp[i-1];
        }
        printf("Case #%lld: %lld\n",cas++,(dp[n-1]+1)%1000000007);
    }
    return 0;
}