Codeforces Round #666 (Div. 2) A. Juggling Letters

problem

solution

统计各个字符数，只要都是n的倍数就可以了

#include<iostream>
#include<string>
#include<cstring>
using namespace std;
int num[250];
int main(){
	int T;  cin>>T;
	while(T--){
		int n;  cin>>n;
		memset(num,0,sizeof(num));
		for(int i = 1; i <= n; i++){
			string s;  cin>>s;
			for(int i = 0; i < s.size(); i++)
				num[s[i]-'A']++;
		}
		int ok = 1;
		for(int i = 0; i <= 100; i++){
			if(num[i]%n!=0){
				ok = 0; break;
			}
		}
		if(ok)cout<<"YES\n";
		else cout<<"NO\n";
	}
	return 0;
}