多校4 Another Meaning 5763

最新推荐文章于 2025-11-03 09:47:51 发布

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本文介绍了一种基于哈希的字符串匹配算法，用于解决给定两个字符串时可能存在的多种匹配方式问题。通过使用动态规划和哈希函数，算法能够高效地计算出所有可能的匹配情况，并考虑了匹配过程中可能出现的不同情形。

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

4
hehehe
hehe
woquxizaolehehe
woquxizaole
hehehehe
hehe
owoadiuhzgneninougur
iehiehieh

Sample Output

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 1000000007

using 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;
}