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 <iostream>
#include <vector>
#include <map>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <numeric>
#include <tuple>
#include <stack>

#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(x) x.begin(),x.end()
#define print(s) cout << s << endl
typedef long long ll;
using namespace std;

int main() {
    int n; cin >> n;
    rep(i,n/4){
        rep(j,n/7){
            if (4*i + 7*j == n) {
                cout << "yes" << endl;
                return 0;
            } 
        }
    }
    print("No");
}