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 <string>
#include <algorithm>
#include <cstdlib>
#include <iomanip>

using namespace std;

int main(void){
	int n;
	cin >> n;
	if(n%4==0 || n%7==0){
		cout << "Yes" << endl;
		exit(0);
	}
	if(n<7){
		if(n==4){
			cout << "Yes" << endl;
			exit(0);
		}else{
			cout << "No" << endl;
			exit(0);
		}
	}
	/*
	if((n%7)%4 == 0){
		cout << "Yes" << endl;
	}else{
		cout << "No" << endl;
	}*/

	for(int i=7;i<=n;i*=7){
		if(!((n-i)%4 == 0)){
			cout << "No" << endl;
			exit(0);
		}
	}
	cout << "Yes" << endl;
	return 0;
}