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<string>
#include<algorithm>
using namespace std;
#define INF 1e9
#define rep(i, n) for(int i = 0; i < n; i++)
#define abs(n) ( (n) < 0 ) ? -(n) : (n)

int main(void){

  //入力
  int N;
  cin >> N;
  
  //処理
  bool can = false;	//できるならtrue
  for(int i = 0; i * 4 <= N; i++){
    if( (N - i*4)%4 == 0 || (N - i*4)%7 == 0){
      can = true;
      break;
    }
  }
  
  if(can) cout << "Yes" << endl;
  else cout << "No" << endl;

  return 0;
}