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

A. k-Stringtime 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：感謝莫莫的解題報告

這個題目主要是需要理解題意，靜下心來看下，其實很簡單。
題意：
     題目給一個K，給定一個字串，裡麵包含若干字母。如果字串可以用裡面所有的字母重新組合為K-string，輸出重組以後的字串，否則輸出-1。K-string即為一個字串可以分為一個子串複製K倍然後得到字串。
解題思路：
     先判斷是不是K-string，用一個數組儲存從a~z出現的次數。如果某一個次數不是K的倍數，那麼便直接輸出-1即可。否則則將裡面的次數全部除以K，用來輸出。

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<algorithm>#include<map>using namespace std;char a[1004],c[1004];             //a是輸入的字串，c則是需要連續輸出K倍的子串int b[26];                       //儲存a~z字母出現額次數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)                      //不滿足次數是K的倍數，跳出，輸出-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;                  //次數除以K                 for(j=0;j<x;j++)                     c[++ll]=i+'a';                 //用c來儲存需要輸出的子串              }          c[++ll]='\0';          for(i=0;i<k;i++)                   //輸出重組以後的K-string              cout<<c;          cout<<endl;       }    }    return 0;}