hdu 5763 Another Meaning（2016 Multi-University Training Contest 4——dp+kmp）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5763

Another Meaning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 802    Accepted Submission(s): 372

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

 

题目大意：给t组测试数据，每组数据有两个字符串，第一个是母串，第二个是子串；子串有两种含义，问根据子串可以计算出母串有多少种含义。

解题思路：

首先要想根据母串和子串找到有两种含义的串，再根据next数组，用dp判断这里可不可以变成两个含义，可以就加上，否则就等于前一状态。

详见代码。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N=100000+10;
#define Mod 1000000007

char ch[N],str[N];
int Next[N],a[N],dp[N];
int lenc,lens;

int get_next()
{
    int i=0,j=-1;
    Next[0]=-1;
    while (i<lens)
    {
        if (j==-1||str[i]==str[j])
        {
            i++;
            j++;
            Next[i]=j;
        }
        else
            j=Next[j];
    }
}

void kmp()
{
    int i=0,j=0;
    get_next();
    while (i<lenc)
    {
        if(j==-1||str[j]==ch[i])
        {
            i++;
            j++;
        }
        else
            j=Next[j];
        if (j==lens)
        {
            a[i]=1;//表示这个匹配成功
            j=Next[j];
            //cout<<i<<endl;
            // return i-j+1;
        }
    }
}

int main()
{
    int t,Case=1;
    scanf("%d",&t);
    while (t--)
    {
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        scanf("%s%s",ch,str);//ch母串，str子串
        lenc=strlen(ch);
        lens=strlen(str);
        kmp();
        dp[0]=1;
        for (int i=1;i<=lenc;i++)
        {
            dp[i]=(dp[i]+dp[i-1])%Mod;
            if (a[i]==1)
                dp[i]=(dp[i]+dp[i-lens])%Mod;
        }
        printf ("Case #%d: %d\n",Case++,dp[lenc]);
    }
    return 0;
}