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

最新推荐文章于 2019-03-26 17:24:53 发布

ZXM_ZhanX

最新推荐文章于 2019-03-26 17:24:53 发布

阅读量351

点赞数

CC 4.0 BY-SA版权

分类专栏：题解 codeforces 文章标签： c acm codeforces 题解

本文链接：https://blog.youkuaiyun.com/qq_40731186/article/details/80041629

题解同时被 2 个专栏收录

29 篇文章

订阅专栏

codeforces

15 篇文章

订阅专栏

本文介绍了一道算法题目，需要判断给定的字符串是否能由相同长度的多个回文子串组成。通过等长分割字符串，并验证每个子串是否为回文串，最终确定是否符合题目要求。

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.

$\text{[math]}$

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
inputCopy
saba
2
outputCopy
NO
inputCopy
saddastavvat
2
outputCopy
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;
}