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: Wrong Answer)

#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <limits.h>
#define min(x, y) (x < y ? x : y)
#define ll long long int
#define Lff long double

int main()
{
  int N;
  scanf("%d", &N);

  if (N % 4 == 0)
  {
    printf("Yes");
    return 0;
  }

  else if (N % 7 == 0)
  {
    printf("Yes");
    return 0;
  }
  
  else if (N % 11 == 0)
  {
    printf("Yes");
    return 0;
  }

  else
  {
    int a = N % 4;
    int b = a % 7;

    if (b == 0)
    {
      printf("Yes");
      return 0;
    }
  }

  printf("No");

  return 0;
}