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 <bits/stdc++.h>

using namespace std;
using llong = long long;
using P = pair<llong, llong>;

#define BE(x) x.begin(), x.end()

const llong inf = llong(1e18)+7;
const llong mod = 1e9+7;

int main(){
	int N;
	cin >> N;

	bool check = false;
	for(int i = 0; i <= 100; i++)
		if((N-i*4)%7 == 0 || (N-i*7)%4 == 0)
			check = true;

	if(check)
		cout << "Yes" << endl;
	else
		cout << "No" << endl;

	return 0;
}