A. Split it! Codeforces Round #706 (Div. 2)

A. Split it!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Kawashiro Nitori is a girl who loves competitive programming.

One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.

Given a string ss and a parameter kk, you need to check if there exist k+1k+1 non-empty strings a1,a2...,ak+1a1,a2...,ak+1, such that

s=a1+a2+…+ak+ak+1+R(ak)+R(ak−1)+…+R(a1).s=a1+a2+…+ak+ak+1+R(ak)+R(ak−1)+…+R(a1).

Here ++ represents concatenation. We define R(x)R(x) as a reversed string xx. For example R(abcd)=dcbaR(abcd)=dcba. Note that in the formula above the part R(ak+1)R(ak+1) is intentionally skipped.

Input

The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1001≤t≤100)  — the number of test cases. The description of the test cases follows.

The first line of each test case description contains two integers nn, kk (1≤n≤1001≤n≤100, 0≤k≤⌊n2⌋0≤k≤⌊n2⌋)  — the length of the string ss and the parameter kk.

The second line of each test case description contains a single string ss of length nn, consisting of lowercase English letters.

Output

For each test case, print "YES" (without quotes), if it is possible to find a1,a2,…,ak+1a1,a2,…,ak+1, and "NO" (without quotes) otherwise.

You can print letters in any case (upper or lower).

Example

input

Copy

7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa

output

Copy

YES
NO
YES
NO
YES
NO
NO

Note

In the first test case, one possible solution is a1=qwa1=qw and a2=qa2=q.

In the third test case, one possible solution is a1=ia1=i and a2=oa2=o.

In the fifth test case, one possible solution is a1=dokidokiliteraturecluba1=dokidokiliteratureclub.

AC代码：

#include <iostream>
#include <algorithm>
#include <map>
#include <string>
 
using namespace std;
 
const int MAX = 1e5+5;
int a[MAX];
 
int main(){
	//freopen("a.txt", "r", stdin);
	int t;
	cin >> t;
	while (t--){
		bool f = true;
		int n, k;
		string s;
		cin >> n >> k >> s;
		if (!k) cout << "YES" << '\n';
		else if (n / k > 1 && n != k *2){
			for (int i = 0; i < k; i++) {
				if (s[i] != s[s.size() - i - 1]){
					f = false;
					break;
				}
			}
			if (f) cout << "YES\n";
			else cout << "NO\n";
		}else cout << "NO\n";
	}
 
	return 0;
}

标签：case,string,706,a1,a2,Split,test,Div,ak
来源： https://blog.csdn.net/weixin_53484451/article/details/114648889