hdu5763 Another Meaning(kmp+dp)

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

Another Meaning

[align=center]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 999    Accepted Submission(s): 476
[/align]

[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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|

 

[align=left]Output[/align]
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.
 

[align=left]Sample Input[/align]

4
hehehe
hehe
woquxizaolehehe
woquxizaole
hehehehe
hehe
owoadiuhzgneninougur
iehiehieh

[align=left]Sample Output[/align]

Case #1: 3
Case #2: 2
Case #3: 5
Case #4: 1

HintIn 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”.
题目大意：给你两串字符串，问第一串哪些部分可以用第二串字符来代替，问组合起来总共会有多少个

思路：kmp+dp

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
char st[111111],temp[111111];
int num[111111];
int Next[111111];
long long mod = 1e9 + 7;
int dp[111111];
void getNext()
{
memset(Next,0,sizeof(Next));
Next[0]=-1;
int j=0;
int k=-1;
while(temp[j])
{
if(k == -1 || temp[j] == temp[k])
{
j++;
k++;
Next[j] = k;
}
else
{
k = Next[k];
}
}
}
int main()
{
int t;
scanf("%d",&t);
for (int cas = 1 ; cas <= t ; cas ++)
{
memset(num,0,sizeof(num));
scanf("%s",st);
scanf("%s",temp);
getNext();
int n = strlen(st);
int m = strlen(temp);
dp[0] = 1;
int j = 0;
for (int i = 0 ; i < n; i++)
{
while (j && st[i] != temp[j])
j = Next[j];
if (st[i] == temp[j])
j++;
dp[i + 1] = dp[i];
if (j == m)
{
dp[i + 1] = (dp[i + 1] + dp[i - m + 1]) % mod;
j = Next[j];
}
}
printf("Case #%d: %lld\n",cas,dp
);
}
}