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;
}