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
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string>
#include<iostream>
#include<iostream>
#include<cctype>
#include<cstdio>
#include<vector>
#include<stack>
#include<queue>
#include <algorithm>
#include<math.h>
#include<set>
#include<map>
#include <sstream>
#include<iomanip>
#include <ctype.h>
using namespace std;

//#include<bits/stdc++.h>

int main() {

	int n;
	cin >> n;
	bool f[110] = {};
	f[0] = 1;
	for (int i = 0; i <= n; i++) {
		if (f[i]) {
			f[i + 4] = 1;
			f[i + 7] = 1;
		}
	}
	if (f[n])cout << "Yes" << endl;
	else cout << "No" << endl;

	return 0;
}