[2018-4-20]BNUZ套题比赛div2 CodeForces 548A Mike and Fax【补】

A. Mike and Fax

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.

He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.

He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.

Input

The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).

The second line contains integer k (1 ≤ k ≤ 1000).

Output

Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.

Examples

input

Copy

saba
2

output

Copy

NO

input

Copy

saddastavvat
2

output

Copy

YES

Note

Palindrome is a string reading the same forward and backward.

In the second sample, the faxes in his back-bag can be "saddas" and "tavvat".

题意及题解：给个字符串 和 回文串数量， 长度为回文长度（字符串长度除以所给回文串数量可得）来分割这个字符串（等长分割）， 判断分割出来的回文串是不是都是回文串 且 回文长度 都要 相等。ps: 中间不存在多余 字符的存在， 判 字符串长度 不能整除 回文串数量 时，直接输出no 

AC代码：

#include <bits/stdc++.h>

using namespace std;

string a;
int b;

int pd(int sta, int endd) {//计算回文串 回文长度，不是回文串返回0
	int i = sta, j = endd, cnt = 0;
	while (i <= j) {
		if (a[i] == a[j]) {
			cnt++;
		} else {
			return 0;
		}
		i++;
		j--;
	}
	return cnt;
}

int main() {
	set<int> s;
	int cnt = 0;
	cin >> a >> b;
	if (a.size() % b != 0) //不能整除说明一定不是他的包
		printf("NO\n");
	else if (a.size() >= b) {
		int len = a.size() / b;//分割后每节字符串的长度
		for (int i = 0; i < b; i++) {
			int p = pd(i * len, (i + 1) * len - 1);
			if (p != 0) {//是回文的情况
				cnt++;
				s.insert(p);
				if (s.size() > 1) {//集合中不只存在一个元素，证明有不同长度的回文串，根据题意不是他的包
					printf("NO\n");
					return 0;
				}
			} else {//不是回文的情况 输出no
				printf("NO\n");
				return 0;
			}
		}
		if (cnt == b)
			printf("YES\n");
		else
			printf("NO\n");
	} else //字符串长度比b小证明一定不匹配
		printf("NO\n");
	return 0;
}