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Codeforces 506 Div3.A 尋找最長公共前後綴

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

這個題目雖然簡單,但是揭示了kmp求next陣列的原理

#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;
			}	
//		}
//	}
}