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

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 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

2aazz

output

azaz

input

3abcabcabz

output

-1

ps：感谢莫莫的解题报告

这个题目主要是需要理解题意，静下心来看下，其实很简单。
题意：
     题目给一个K，给定一个字符串，里面包含若干字母。如果字符串可以用里面所有的字母重新组合为K-string，输出重组以后的字符串，否则输出-1。K-string即为一个字符串可以分为一个子串复制K倍然后得到字符串。
解题思路：
     先判断是不是K-string，用一个数组保存从a~z出现的次数。如果某一个次数不是K的倍数，那么便直接输出-1即可。否则则将里面的次数全部除以K，用来输出。

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<algorithm>#include<map>using namespace std;char a[1004],c[1004];             //a是输入的字符串，c则是需要连续输出K倍的子串int b[26];                       //保存a~z字母出现额次数int main(){    int i,j,k;    while(cin>>k)    {      memset(b,0,sizeof(b));      cin>>a;      for(i=0;i<strlen(a);i++)          b[a[i]-'a']++;      int flag=0;      for(i=0;i<26;i++)          if(b[i]%k!=0)                      //不满足次数是K的倍数，跳出，输出-1          {             flag=1;                           break;          }      if(flag)          cout<<"-1"<<endl;       else       {          int ll=-1;          for(i=0;i<26;i++)              if(b[i])              {                 int x=b[i]/k;                  //次数除以K                 for(j=0;j<x;j++)                     c[++ll]=i+'a';                 //用c来保存需要输出的子串              }          c[++ll]='\0';          for(i=0;i<k;i++)                   //输出重组以后的K-string              cout<<c;          cout<<endl;       }    }    return 0;}