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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>
#include <string.h>
int main(void)
{
  int N;
  scanf("%d", &N);
  if (N == 1) {
    printf("No");
  }
  else if (N == 2) {
    printf("No");
  }
  else if (N == 3) {
    printf("No");
  }
  else if (N == 5) {
    printf("No");
  }
  else if (N == 6) {
    printf("No");
  }
  else if (N == 9) {
    printf("No");
  }
  else if (N == 10) {
    printf("No");
  }
  else if (N == 13) {
    printf("No");
  }
  else if (N == 17) {
    printf("No");
  }
  else {
    printf("Yes");
  }
 return 0;
}