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
本题有意思，是hash表的灵活运用。
思路：
1 计算好总字符数，和使用hash表A[26]记录好各个字符出现的次数
2 判断总字符是否可以被k整除，如果不可以，那么就不能分成k个子字符了
3 计算各个字符出现的次数是否能被k整除，如果不能，那么就不能分成k个子字符
4 根据字符出现的次数逐个打印
#include <string>
#include <iostream>
using namespace std;
void KString()
{
int k, len = 0;
cin>>k;
int A[26] = {0};
char a;
while (cin>>a)
{
A[a-‘a‘]++;
len++;
}
if (len == 0 || len % k)
{
cout<<-1;
return;
}
bool ks = true;
for (unsigned i = 0; i < 26; i++)
{
if (A[i] % k != 0) ks = false;
}
if (!ks) cout<<-1;
else
{
for (int d = 0; d < k; d++)
{
for (unsigned i = 0; i < 26; i++)
{
if (A[i])
for (unsigned j = 0; j < A[i]/k; j++)
{
cout<<char(i + ‘a‘);
}
}
}
}
}
codeforces A. k-String 题解