solution

• B被A整除，B是A的倍数。B / A
• A整除B =A把B整除 =Ｂ被A整除 =B / A
  ==>n可以整除不同的四个因数之和
  等价于 (a + b + c + d) % n == 0

#include<iostream>
#include<cmath>
using namespace std;
int main(){
	int k, n, t;
	scanf("%d", &k);
	while(k--){
		int num = 0, a[1000] = {0}, ans = 0;
		scanf("%d", &n);
		t = n;
		for(int i = 1; i <= sqrt(t); i++){
			if(t % i == 0){
				a[num++] = i;
				if(i * i != t) a[num++] = t / i;
			}
		}
		for(int i = 0; i < num; i++){
			for(int j = i + 1; j < num; j++){
				for(int k = j + 1; k < num; k++){
					for(int g = k + 1; g < num; g++){
						if((a[i] + a[j] + a[k] + a[g]) % n == 0){
							ans = 1;
							break;
						}
					}
				}
			}
		}
		if(ans) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}