Codeforces 506 Div3.A 寻找最长公共前后缀

最新推荐文章于 2025-06-28 11:48:06 发布

原创最新推荐文章于 2025-06-28 11:48:06 发布 · 291 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#kmp

kmp 专栏收录该内容

2 篇文章

订阅专栏

本文探讨了一个有趣的字符串构造问题，目标是构建最短的字符串，使其包含特定子串恰好k次。通过分析输入字符串的周期性，可以高效地解决此问题。示例代码展示了如何根据子串的重复模式来构建所需的字符串。

A. Many Equal Substrings

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string tt consisting of nn lowercase Latin letters and an integer number kk.

Let's define a substring of some string ss with indices from ll to rr as s[l…r]s[l…r].

Your task is to construct such string ss of minimum possible length that there are exactly kk positions ii such that s[i…i+n−1]=ts[i…i+n−1]=t. In other words, your task is to construct such string ss of minimum possible length that there are exactly kk substrings of ss equal to tt.

It is guaranteed that the answer is always unique.

Input

The first line of the input contains two integers nn and kk (1≤n,k≤501≤n,k≤50) — the length of the string tt and the number of substrings.

The second line of the input contains the string tt consisting of exactly nn lowercase Latin letters.

Output

Print such string ss of minimum possible length that there are exactly kk substrings of ss equal to tt.

It is guaranteed that the answer is always unique.

Examples

input

Copy

3 4
aba

output

Copy

ababababa

input

Copy

3 2
cat

output

Copy

catcat

这个题目虽然简单，但是揭示了kmp求next数组的原理

#include<iostream>
#include<cstring>
using namespace std;
int main(){
	int n,k;
	char s[101];
	cin>>n>>k;
	cin>>s;
	int d=strlen(s);
//	if(d==1){
//		cout<<s<<endl;
//	}
//	else{
		int len;
		for(int j=1,i=d-2;i>=0&&j<=d-1;j++,i--){
			for(int l=0,r=j;l<=i,r<d;l++,r++){
				if(s[l]!=s[r]){
					goto k;
				}
			}
			len=i+1;
			break;
			k:;
		}
//		if(len>d/2){
//			cout<<s<<endl;
//		}
//		else{
//		cout<<len<<endl;
			if(len==0){
				for(int i=0;i<k;i++){
					cout<<s;
				}
				cout<<endl;
			}
			else{
				cout<<s;
				for(int i=0;i<k-1;i++){
					for(int j=len;j<d;j++){
						cout<<s[j];
					}
				}
				cout<<endl;
			}	
//		}
//	}
}