Codeforces - 219A - k-String

题目：http://codeforces.com/problemset/problem/219/A

A. k-String

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A string is called a k-string if it can be represented as k concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.

You are given a string s, consisting of lowercase English letters and a positive integer k. Your task is to reorder the letters in the string s in such a way that the resulting string is a k-string.

Input

The first input line contains integer k (1 ≤ k ≤ 1000). The second line contains s, all characters in s are lowercase English letters. The string length s satisfies the inequality 1 ≤ |s| ≤ 1000, where |s| is the length of string s.

Output

Rearrange the letters in string s in such a way that the result is a k-string. Print the result on a single output line. If there are multiple solutions, print any of them.

If the solution doesn't exist, print "-1" (without quotes).

Sample test(s)

input

2
aazz

output

azaz

input

3
abcabcabz

output

-1

分析：

用数组来统计每个字母出现的次数，如果某个字母出现的次数不是k的整数倍，则返回-1。

代码：

#include <iostream>
#include <string>

int main()
{
	int k;
	std::string s;
	int c[26] = {0};

	std::cin >> k >> s;

	if(s.size() % k != 0)
	{
		std::cout << -1 << std::endl;
		return 0;
	}

	for(int i = 0; i < s.size(); ++i)
	{
		++c[s[i] - 'a'];
	}

	for(int i = 0; i < 26; ++i)
	{
		if(c[i] % k != 0)
		{
			std::cout << -1 << std::endl;
			return 0;
		}
	}

	std::string res;
	for(int i = 0; i < 26; ++i)
	{
		for(int j = 0; j < c[i] / k; ++j)
		{
			res += char('a' + i);
		}
	}

	for(int i = 0; i < k; ++i)
	{
		std::cout << res;
	}
	std::cout << std::endl;

	return 0;
}

        感谢阅读！