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>

using namespace std;

#define mp make_pair
#define pb push_back
#define repi(i, p, q) for (long i = p; i < q; i++)
#define repd(i, p, q) for (long i = p; i >= q; i--)
#define sz(x) x.size()
#define INF 2000000000
#define MOD 1000000007

typedef map<int, int> MI;
typedef pair<int, int> PI;
typedef long long int LLI;
typedef long int LI;
typedef int I;

int main() {
    int n;
    cin >> n;
    int num = n%7;
    num = num%4;
    if(num == 0) {
        cout << "Yes";
    } else {
        int number = n%4;
        number = number%7;
        if(number == 0) {
            cout << "Yes";
        } else {
            cout << "No";
        }
    }
}