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 <stdio.h>

#define CAKE 4
#define DONUT 7
int main(){
  int n;

  scanf( "%d", &n );

  if ( n < CAKE ){
    puts( "No" ); //printf( "%d\n", n );
    return 0;
  }else if ( n % CAKE == 0 ){
    puts( "Yes" ); //printf( "1:%d\n", n % CAKE );
    return 0;
  }else if ( n % DONUT == 0 ){
    puts( "Yes" ); //printf( "2:%d\n", n % DONUT );
    return 0;
  }else if ( ( n % DONUT ) % CAKE == 0 ) {
    puts( "Yes" ); //printf( "3:%d\n", ( n % DONUT ) % CAKE );
    return 0;
  // }else if ( ( n % CAKE ) % DONUT == 0 ) {
  //   puts( "Yes" ); printf( "4:%d\n", ( n % CAKE ) % DONUT );
  //   return 0;
  }else {
    int n_temp = n - CAKE;
    while( n_temp < 0 ){
      if ( n_temp % DONUT == 0 ){
        puts( "Yes" ); //printf( "3:%d\n", ( n % DONUT ) % CAKE );
        return 0;
      }
      n_temp -= CAKE;
    }
    n_temp = n - DONUT;
    while( n_temp < 0 ){
      if ( n_temp % CAKE == 0 ){
        puts( "Yes" ); //printf( "3:%d\n", ( n % DONUT ) % CAKE );
        return 0;
      }
      n_temp -= CAKE;
    }
  }

  puts( "No" );
  return 0;
}