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.
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.
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).
2 aazz
azaz
3 abcabcabz
-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 "); 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(" "); } } return 0; }