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

传送门 Time Limit: 1 second Memory Limit: 256 megabytes

Problem Description

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 $s$ and a parameter $k$ , you need to check if there exist $k+1$ non-empty strings $a_1,a_2...,a_{k+1}$ , such that $s=a_1+a_2+\ldots +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+\ldots+R(a_{1}).$

Here $+$ represents concatenation. We define $R(x)$ as a reversed string $x$ . For example $R(abcd) = dcba$ . Note that in the formula above the part $R(a_{k+1})$ is intentionally skipped.

Input

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

The first line of each test case description contains two integers $n$ , $k$ ( $1\le n\le 100$ , $0\le k\le \lfloor \frac{n}{2} \rfloor$ ) — the length of the string $s$ and the parameter $k$ .

The second line of each test case description contains a single string $s$ of length $n$ , consisting of lowercase English letters.

Output

For each test case, print "YES" (without quotes), if it is possible to find $a_1,a_2,\ldots,a_{k+1}$ , and "NO" (without quotes) otherwise.

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

Sample Input

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

Sample Onput

YES
NO
YES
NO
YES
NO
NO

Note

In the first test case, one possible solution is $a_1=qw$ and $a_2=q$ .

In the third test case, one possible solution is $a_1=i$ and $a_2=o$ .

In the fifth test case, one possible solution is $a_1=dokidokiliteratureclub$ .

题目大意 _{看懂题意的话可以直接跳过}

把一个长度为n的字符串，分成 $2k+1$ 份小字符串。其中第 $i$ 个小字符串与第 $2k+1-i$ 个字符串对称（ $abc$ 与 $cba$ 对称），问能不能做到。

解题思路 _发现规律

首先长度为 $n$ 要分成 $2k+1$ 份，就要满足 $n\ge2n+1$ 。若不满足直接 $NO$ 。

下面讨论满足 $n\ge2n+1$ 的情况：

其实找几个例子就能发现， $s$ 由3部分组成： $A B C$ 。其中

$A$ 和 $C$ 是对称的

$A$ 、 $B$ 、 $C$ 都能为空

$A的长度\ge k$

样例分析

样例1

5 1
qwqwq

$A$ : $q$ $B$ : $wqw$ $C$ : $q$ 满足上述3个条件，故输出YES。

样例2

2 1
ab

$2<1*2+1$ ，不满足条件3，故输出NO。

样例3

3 1
ioi

$A$ : $i$ $B$ : $o$ $C$ : $i$ 满足上述3个条件，故输出YES。

样例4

4 2
icpc

$4<2*2+1$ ，不满足条件3，故输出NO。

样例5

22 0
dokidokiliteratureclub

$A$ : $B$ : $dokidokiliteratureclub$ $C$ : 满足上述3个条件，故输出YES。

样例6

19 8
imteamshanghaialice

无法找到两个个长度 $\ge 8$ 的 $A$ 和 $B$ 使得 $A$ 、 $B$ 对称，故输出NO。

样例7

6 3
aaaaaa

$A$ : $q$ $B$ : $wqw$ $C$ : $$ $6<3*2+1$ ，不满足条件3，故输出NO。

AC代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;

int main()
{
    int N;
    cin >> N;
    while (N--) //N组样例
    {
        int n, k;
        cin >> n >> k;
        string s;
        cin >> s;
        if (k * 2 >= n) //不满足条件3
        {
            puts("NO");
            continue;
        }
        for (int i = 0; i < k; i++) //前面k个和后面k个相对称吗
        {
            if (s[i] != s[n - i - 1]) //不对称
            {
                puts("NO");
                goto loop; //跳出
            }
        }
        puts("YES");
    loop:; //跳出到这里
    }
    return 0;
}

同步发文于我的CSDN,原创不易，转载请附上原文链接哦~
Tisfy：blog.csdn.net/Tisfy/artic…

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