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

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

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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 $$s=a_1+a_2+\dots +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+\dots +R(a_1).$$

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 $a_1,a_2,\dots ,a_{k+1}$, 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
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Sample Onput

YES
NO
YES
NO
YES
NO
NO
1
2
3
4
5
6
7

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 $a_1=dokidokiliteratureclub$.

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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。

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解题思路 _发现规律

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

下面讨论满足$n\ge 2k+1$的情况：

其实找几个例子就能发现，$s$由3部分组成：$ABC$。其中

1. $A$和$C$是对称的
2. $A$、$B$、$C$都能为空
3. $A的长度\ge k$

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样例分析

样例1

5 1
qwqwq
1
2

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

样例2

2 1
ab
1
2

$2<1\ast 2+1$，不满足条件3，故输出NO。

样例3

3 1
ioi
1
2

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

样例4

4 2
icpc
1
2

$4<2\ast 2+1$，不满足条件3，故输出NO。

样例5

22 0
dokidokiliteratureclub
1
2

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

样例6

19 8
imteamshanghaialice
1
2

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
1
2

$A$ : $q$
$B$ : $wqw$
$C$ : $$
$6<3\ast 2+1$，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;
}
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原创不易，转载请附上原文链接哦~
Tisfy：https://blog.csdn.net/Tisfy/article/details/114647291