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: Accepted)

#define _USE_MATH_DEFINES
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <cctype>
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <iomanip>
#include <cstdio>
#include <numeric>
#include <functional>
#include <cstdlib>
#include <regex>
#include <list>
#include <cstring>
int main() {
	int n;
	std::cin >> n;
	int a = n;
	for (int i = 1;i < (n / 4) + 1;++i) {
		if ((n - 4 * i) % 3 == 0&&(n-4*i)/3<=i) {
			std::cout << "Yes" << std::endl;
			return 0;
		}
	}
	std::cout << "No" << std::endl;
		return 0;
}