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)

#define _USE_MATH_DEFINES
#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <functional>
#include <string>
#include <string.h>
#include <math.h>

using namespace std;

int main(){
	int N;
	bool result = false;
	cin >> N;

	for (int i = 0; i < N / 4; ++i){
		for (int j = 0; j < N / 7; ++j){
			if (i * 4 + j * 7 == N){
				result = true;
				break;
			}
		}
	}
	if (result)
		cout << "Yes" << endl;
	else
		cout << "No" << endl;

	return 0;
}