Codeforces Round #135 (Div. 2)---A. k-String

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 sin 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

解题思路：给一个串。问能否由k个同样的串连接而成。

用STL里的map。扫一遍。分别记录每一个字符的个数。在推断全部的字符是否是k的倍数，若不是，则输出-1；否则，遍历依次map。每一个字符输出（总个数）/k个，然后反复k次就可以。

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

map<char, int> m;

int main()
{
#ifdef sxk
freopen("in.txt","r",stdin);
#endif
int n;
string s;
while(scanf("%d",&n)!=EOF)
{
cin>>s;
int len = s.size();
for(int i=0; i<len; i++)
m[s[i]] ++;
map<char, int>::iterator it;
int flag = 1;
for(it=m.begin(); it!=m.end(); it++){
if(it->second % n){
flag = 0;
break;
}
}
if(!flag)  printf("-1\n");
else{
for(int j=0; j<n; j++){
for(it=m.begin(); it!=m.end(); it++){
for(int i=1; i<=it->second/n; i++)
printf("%c", it->first);
}
}
printf("\n");
}
}
return 0;
}