题目链接：

Another Meaning

Time Limit: 2000/1000 MS (Java/Others)    

Memory Limit: 65536/65536 K (Java/Others)


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
 
 
题意：
 
给了一个单词它有两种意思,现在有一个字符串.问你这个字符串有多少种不同的意思;
 
思路：
 
设dp[i]表示[1,i]这段字符串的意思总数,考虑以i位结尾的的这一小段是否和单词匹配;
如果匹配,dp[i]=dp[i-1]+dp[i-strlen(B)];不匹配的话dp[i]=dp[i-1];
然后就是kmp找出所有的匹配位置的结点flag[i]标记下来;
 
AC代码:
/************************************************ 
┆  ┏┓　　　┏┓ ┆    
┆┏┛┻━━━┛┻┓ ┆ 
┆┃　　　　　　　┃ ┆ 
┆┃　　　━　　　┃ ┆ 
┆┃　┳┛　┗┳　┃ ┆ 
┆┃　　　　　　　┃ ┆  
┆┃　　　┻　　　┃ ┆ 
┆┗━┓　  　┏━┛ ┆ 
┆　　┃　  　┃　　┆　　　　　　 
┆　　┃　  　┗━━━┓ ┆ 
┆　　┃　 AC代马 　　┣┓┆ 
┆　　┃　        　　┏┛┆ 
┆　　┗┓┓┏━┳┓┏┛ ┆ 
┆　　　┃┫┫　┃┫┫ ┆ 
┆　　　┗┻┛　┗┻┛ ┆       
************************************************ */  


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+100;
const int maxn=(1<<8);
const double eps=1e-8;

char A[N],B[N];
int nex[N],flag[N];
LL dp[N];

inline void makenext()
{
    int k = -1,j = 0,len = strlen(B);
    nex[0] = -1;
    while(j < len)
    {
        if(k == -1||B[j] == B[k])
        {
            j++;
            k++;
            nex[j] = k;//相等的话就往后继续;
        }
        else k = nex[k];//不等的话就相当于kmp一样,把模式串的这个子串用已经求出来的next跳转;
    }
}
inline void kmp()
{
    mst(flag,0);
    makenext();
    int posP = 0,posT = 0;
    int lenP = strlen(B),lenT = strlen(A);
    while(posP < lenP&&posT < lenT)
    {
        if(posP == -1||B[posP] == A[posT])
        {
            posP++;
            posT++;
        }
        else posP = nex[posP];
        if(posP == lenP)
        {
            flag[posT-1]=1;
            posP=nex[posP];
        }
    }
}

int main()
{       
        int t,Case=0;
        read(t);
        while(t--)
        {
            scanf("%s",A);
            scanf("%s",B);
            kmp();
            int len=strlen(A),l=strlen(B);
            dp[0]=1;
            For(i,1,len)
            {
                if(!flag[i-1])dp[i]=dp[i-1];
                else dp[i]=(dp[i-1]+dp[i-l])%mod;
            }
            printf("Case #%d: %lld\n",++Case,dp[len]);
        }
        return 0;
}