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 <cstdio>
#include <iostream>
#include <algorithm>
#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
using namespace std;

bool isOK(int n) {
  REP(i, n) {
    int rest = n - 4*i;
    if(rest >= 0 && rest % 7 == 0) {
      return true;
    }
  }
  return false;
}

int main(void) {
  int N;
  cin >> N;
  bool ok = isOK(N);
  cout << (ok ? "Yes" : "No") << endl;
  return 0;
}