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CodeforcesRound#135(Div.2)-A.k-String_html/css
2020-11-27 15:57:35 责编:小采
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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 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::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; jsecond/n; i++) printf("%c", it->first); } } printf("\n"); } } return 0;}

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