题解 数学方法 26行秒杀

#include <iostream>
using namespace std;
int main() {
	int T,n,k,n1,n2,sum=0;
	cin >> T;
	for (int i = 1;i <= T;i ++) {
		cin >> n >> k;
		int ans = 1000,cnt=0;
		for (int j = 1;j <= n;j ++) {
			cin >> n1;
			if (k == 4) {
				if (n1 % 2 == 0) {
					cnt ++;
				}
			}
			ans = min(ans,(k-n1%k) % k);
		}
		if (k == 4) {
			ans = min(ans,2 - cnt);
			if (cnt >= 2) {
				cout << 0 << endl;
				continue;
			}
		}
		cout << ans << endl;
	}
	
	return 0;
}