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)

问题描述

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.

输入

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|

输出

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.

样例

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

题解

dp
用kmp处理出匹配成功的结尾的字母位置，然后就是考虑每个位置选和不选的两种情况，转移下就可以了。

代码

#include<map>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define M (l+(r-l)/2)
#define bug(a) cout<<#a<<" = "<<a<<endl using namespace std;typedef __int64 LL;const int maxn=1e5+10;
const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const int mod=1e9+7;char s1[maxn],s2[maxn];
int vis[maxn];
vector<int> pos;
LL dp[maxn];int f[maxn];
void getFail(char *P){int m=strlen(P);f[0]=0; f[1]=0;for(int i=1;i<m;i++){int j=f[i];while(j&&P[i]!=P[j]) j=f[j];f[i+1]=P[i]==P[j]?j+1:0;}
}void find(char* T,char *P){int n=strlen(T),m=strlen(P);getFail(P);int j=0;for(int i=0;i<n;i++){while(j&&P[j]!=T[i]) j=f[j];if(P[j]==T[i]) j++;if(j==m){pos.push_back(i+1);vis[i+1]=1;}}
}void init(){pos.clear();memset(vis,0,sizeof(vis));
}int main() {int tc,kase=0;scanf("%d",&tc);while(tc--){init();scanf("%s%s",s1,s2);find(s1,s2);memset(dp,0,sizeof(dp));dp[0]=1;int n=strlen(s1),m=strlen(s2);for(int i=1;i<=n;i++){//第i位不选 dp[i]=dp[i-1];//第i位选 if(vis[i]) dp[i]+=dp[i-m];dp[i]%=mod;}printf("Case #%d: %I64d\n",++kase,dp[n]);}return 0;
}