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
Input
Copy
3 4 aba
Output
Copy
ababababa
Input
Copy
3 2 cat
Output
Copy
catcat
题意：给你一个长度为n得串 让你输出一个串 里面有 k个字串n
题解：CF上面的题好像时间卡的非常紧 。。每次都RE 。我们这里用next数组看他是否有循环饥节 然后可以推出上的代码结论（不太好说，不过你很容易能懂）
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
string s;
int nexxt[550];
int n,m;
void getnext()
{
int i=0,j=-1;
nexxt[0]=-1;
while(i<n)
{
if(j==-1||s[i]==s[j])
{
i++,j++;
nexxt[i]=j;
}
else
j=nexxt[j];
}
}
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m;
cin>>s;
getnext();
cout<<s;
for(int i=1;i<m;i++)
{
for(int j=nexxt[n];j<n;j++)
{
cout<<s[j];
}
}
cout<<endl;
}