2016 Multi-University Training Contest 4 Another Meaning

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Another Meaning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 678    Accepted Submission(s): 314

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”.

 

Author

FZU

 

Source

2016 Multi-University Training Contest 4

题意：有两个串A和B，其中B有两个意思，求A能组成多少种意思

思路，先用kmp把A串中相同的子串末尾标记起来，在来进行dp；

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=100005;
const int mod = 1e9+7;
int nexta[N],vis[N],dp[N];
char a[N],b[N];
int n,m;
void getnext()
{
    memset(nexta,0,sizeof nexta);
    int j=-1,i=0;
    nexta[0]=-1;
    while(i<n)
    {
        if(j==-1||b[j]==b[i]) i++,j++,nexta[i]=j;
        else
        j=nexta[j];
    }
}
void kmp()
{
    int i=0,j=0;
    while(i<n&&j<m)
    {
        if(j==-1||a[i]==b[j])
        i++,j++;
        else 
        j=nexta[j];
        if(j==m)
        {
            j=nexta[j];
            vis[i]=1;
        }
    }
}
int main()
{
    int t,ca=1;
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof vis);
    cin>>a>>b;
    n=strlen(a);
    m=strlen(b);
    getnext();
    kmp();
    dp[0]=1;
    for(int i=1;i<=n;i++)
    {
        dp[i]=dp[i-1];
        if(vis[i])
        dp[i]+=dp[i-m];
        dp[i]%=mod;
    }
    printf("Case #%d: %d\n",ca++,dp[n]);
    }
    return 0;
    
 }