Problem:
Score : 200 points

Problem Statement

La Confiserie d'ABC sells cakes at 4 dollars each and doughnuts at 7 dollars each.
Determine if there is a way to buy some of them for exactly N dollars. You can buy two or more doughnuts and two or more cakes, and you can also choose to buy zero doughnuts or zero cakes.

Constraints

N is an integer between 1 and 100, inclusive.

Input

Input is given from Standard Input in the following format:

N

Output

If there is a way to buy some cakes and some doughnuts for exactly N dollars, print Yes; otherwise, print No.

Sample Input 1

11

Sample Output 1

Yes

If you buy one cake and one doughnut, the total will be 4 + 7 = 11 dollars.

Sample Input 2

40

Sample Output 2

Yes

If you buy ten cakes, the total will be 4 \times 10 = 40 dollars.

Sample Input 3

3

Sample Output 3

No

The prices of cakes (4 dollars) and doughnuts (7 dollars) are both higher than 3 dollars, so there is no such way.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
#define FOR(i, a, b) for (int i = a; i < b; i++)
#define mFOR(i, a, b) for (int i = a; i > b; i--)
#define MP make_pair
#define PB push_back
#define ALL(v) v.begin(), v.end()
#define N 100007
#define INF 1000000007
using namespace std;

typedef long long ll;

ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll fceil(ll a, ll b) { return (a % b == 0 ? a / b : a / b + 1); }
int main()
{
   int n;
   cin >> n;
   cout << (n * 7 % 3 ? "No" : "Yes") << endl;
   return 0;
}