A - Two Abbreviations

最新推荐文章于 2024-12-16 18:16:05 发布

原创最新推荐文章于 2024-12-16 18:16:05 发布 · 229 阅读

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本文探讨了一个复杂的字符串匹配问题，旨在判断是否存在满足特定条件的好字符串，并找出其最短长度。条件包括字符串长度的整除性及通过特定方式组合字符得到目标字符串。文章提供了算法思路和代码实现，涉及数学、算法设计与字符串操作。

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 300300 points

Problem Statement

You are given a string SS of length NN and another string TT of length MM. These strings consist of lowercase English letters.

A string XX is called a good string when the following conditions are all met:

Let LL be the length of XX. LL is divisible by both NN and MM.
Concatenating the 11-st, (LN+1)(LN+1)-th, (2×LN+1)(2×LN+1)-th, ......, ((N−1)×LN+1)((N−1)×LN+1)-th characters of XX, without changing the order, results in SS.
Concatenating the 11-st, (LM+1)(LM+1)-th, (2×LM+1)(2×LM+1)-th, ......, ((M−1)×LM+1)((M−1)×LM+1)-th characters of XX, without changing the order, results in TT.

Determine if there exists a good string. If it exists, find the length of the shortest such string.

Constraints

1≤N,M≤1051≤N,M≤105
SS and TT consist of lowercase English letters.
|S|=N|S|=N
|T|=M|T|=M

Input

Input is given from Standard Input in the following format:

NN MM
SS
TT

Output

If a good string does not exist, print -1; if it exists, print the length of the shortest such string.

Sample Input 1 Copy

Copy

3 2
acp
ae

Sample Output 1 Copy

Copy

For example, the string accept is a good string. There is no good string shorter than this, so the answer is 66.

Sample Input 2 Copy

Copy

6 3
abcdef
abc

Sample Output 2 Copy

Copy

-1

Sample Input 3 Copy

Copy

15 9
dnsusrayukuaiia
dujrunuma

Sample Output 3 Copy

Copy

重要的坑说三遍

long long

long long
以后没啥事直接用long long

#include<bits/stdc++.h>
using namespace std;

long long gcd(long long a,long long b)
{
	 return (!b)?a:gcd(b,a%b);
}
int main()
{
	long long n,m;
	 map<long long ,long long>q;
	 cin>>n>>m;
	 int flag=0;
	 string s,t;
	 cin>>s;
	 cin>>t;
	/* if(n==n*m/gcd(n,m)||m==n*m/gcd(n,m))
	 cout<<"-1"<<endl;
	 else
	 cout<<n*m/gcd(n,m)<<endl;*/
	 long long l=n*m/gcd(n,m);
	 for(long long i=0;i<n;i++)
	 {
	 	 
	 	 	q[i*l/n+1]=s[i]-'a'+1;
		  
	 	 
	 }
	// cout<<endl;
	 
	 for(long long i=0;i<m;i++)
	 {
	 //	 cout<<i*l/m+1<<endl;
	 	 
	 	 	if(q[i*l/m+1]==t[i]-'a'+1||q[i*l/m+1]==0)
	 	 	continue;
	 	 	else
	 	 	{
	 	 		flag=1;
	 	 		break;
			  }
		  
	 }
	 if(flag)printf("-1");
	 else printf("%lld",l);
 }