西电复试之——真题2013A 斐波那契数列

西电复试之——真题2013A 斐波那契数列

F(0)=7;
F(1)=11;
F(n)=F(n-1)+F(n-2);

#include<iostream>
using namespace std;
int F(int n) {
	if (n == 0)
		return 7;
	if (n == 1)
		return 11;
	return F(n - 1) + F(n - 2);
}

int main() {
	int n;
	cin >> n;
	int m, num;
	while (n--) {
		cin >> m;
		num = F(m);
		if (num % 3 == 0)
			cout << "Yes" << endl;
		else
			cout << "No" << endl;
	}
	return 0;
}