【题解】codeforces1029A[Codeforces Round #506 (Div. 3)]A.Many Equal Substrings KMP

题目链接

Description

You are given a string $t$ consisting of $n$ lowercase Latin letters and an integer number $k$.

Let’s define a substring of some string $s$ with indices from $l$ to $r$ as $s[l…r]$.

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

It is guaranteed that the answer is always unique.

Input

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

The second line of the input contains the string $t$ consisting of exactly $n$ lowercase Latin letters.

Output

Print such string $s$ of minimum possible length that there are exactly $k$ substrings of $s$ equal to $t$.

It is guaranteed that the answer is always unique.

Examples

jnput

3 4
aba

Output

ababababa

Input

3 2
cat

Output

catcat

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类似 $KMP$ 算法，我们求出一个最长的相同前缀后缀长度，即 $next$ 数组，然后输出 $k-1$ 个 $S[i],i∈\big[0,s.size()-next[n-1]\big]$，最后再输出 $S$。

#include<cstdio>
#include<iostream>
using namespace std;
int nx[110];
int main()
{
	//freopen("in.txt","r",stdin);
	int n,k;
	string s;
	cin>>n>>k>>s;
	for(int i=1,j=0;i<s.size();i++)
	{
		while(j&&s[i]!=s[j])j=nx[j-1];
		if(s[i]==s[j])j++;
		nx[i]=j;
	}
	int len=s.size()-nx[n-1];
	for(int i=1;i<k;i++)
	    cout<<s.substr(0,len);
	cout<<s;
	return 0;
}

总结

字符串算法没怎么写过题，很生疏。