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CodeforcesRound#135(Div.2)-A.k-String_html/css_WEB-ITnose

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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





解题思路：给一个串，问是否能由k个相同的串连接而成。

用STL里的map。扫一遍，分别记录每个字符的个数，在判断所有的字符是否是k的倍数，若不是，则输出-1；否则，遍历依次map，每个字符输出（总个数）/k个，然后重复k次即可。





AC代码：

#include #include #include #include #include #include #include #include #include #include #include #include using namespace std;#define INF 0x7fffffffmap<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;}

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