2016 Multi-University Training Contest 4 Another Meaning

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.LimitsT <= 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
4hehehehehewoquxizaolehehewoquxizaoleheheheheheheowoadiuhzgneninouguriehiehieh
 

Sample Output
Case #1: 3Case #2: 2Case #3: 5Case #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;     }