多校4 Another Meaning 5763

Another Meaning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1266    Accepted Submission(s): 595

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

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”.
 

 

题意：给一a串给一b串，问有几种匹配方式？（可匹配可不匹配）

思路：第一个字符匹配上了那后面Nb-1个字符都必须匹配，第一个字符没匹配上，后面的字符可匹配可不匹配，所以前后是有关系的，用DP，分两种情况匹配上没匹配上，两种情况又分别分为后面匹配上和后面没匹配上，列递推式，普通匹配极限数据会超时（虽然当时没超...）所以匹配加个kmp或哈希

代码：

#include <cstdio>#include <cstring>#include <string>#include <iostream>#include <cmath>#include <map>#define mod 1000000007using namespace std;const unsigned long long B = 1e8+7;    /*mod*/const int MAXN = 100000+100;char a[MAXN],b[MAXN],tmp[MAXN];long long dp[MAXN];bool vis[MAXN];void hashfind(){    int al=strlen(a),bl=strlen(b);    if(al>bl)    {        strcpy(tmp,a);        strcpy(a,b);        strcpy(b,tmp);    }    unsigned long long t=1,ah=0,bh=0;    for(int i=0; i<al; i++)    {        t*=B;        ah=ah*B+a[i];        bh=bh*B+b[i];    }    for(int i=0; i+al<=bl; i++)    {        if(ah==bh)vis[i]=1;                       //匹配到的坐标        if(i+al<bl)bh=bh*B+b[i+al]-b[i]*t;    }//    return ans;}int main(){    int t;    cin>>t;    int cas = t;    while(t--)    {        scanf("%s%s",b,a);        memset(vis,0,sizeof vis);        hashfind();        int m = strlen(b),n = strlen(a);        for(int i = 0; i <= m; i++)            dp[i] = 1;        for(int i = m-n; i >=0; i--)        {            if(vis[i])                dp[i] = (dp[i+n] + dp[i+1])%mod;                      else                dp[i] =  dp[i+1]%mod;        }        cout<<"Case #"<<cas-t<<": "<<dp[0]<<endl;    }    return 0;}