思路

打表。

提前算好。

时间复杂度:

$O(1)$

空间复杂度:

$O(1)$

Code

#include<bits/stdc++.h>
using namespace std;
string s[20]={"YES","NO","NO","NO","NO",
"YES","NO","YES","YES","YES",
"NO","YES","NO","YES","YES",
"YES","NO","YES","NO","YES"};
int main()
{
	int n;
	cin>>n;
	cout<<s[n]<<endl;
	return 0;
}