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)

#include <iostream>
#include <vector>
#include <string>
using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define Yes cout << "Yes" << endl
#define No cout << "No" << endl
#define REP(i,n) for(int i=0;i<n;i++)
#define pb push_back


// vectorの中身をcout << vs;で表示
template <class T>
ostream &operator<<(ostream &o, const vector<T> &v)
{
    o << "{";
    for (int i = 0; i < (int)v.size(); i++) o << (i > 0 ? ", " : "") << v[i];
    o << "}";
    return o;
}

int main(void) {
	// cin, coutの結びつきを解除し、速くする
	cin.tie(0);
   	ios::sync_with_stdio(false);

	int n; cin >> n;

	bool buyable = false;
	REP(i, n/7+1) {
		if ((n - 7 * i) % 4 == 0) {
			buyable = true;
			break;
		}
	}
	buyable?Yes:No;
}

// 0 % 4 == 0 ではない？
// そこでエラー。7の倍数の時、Noになっていた