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 D4 4
#define D7 7

int main(void){
  int money,amari,check,check1,amariA,checkA,check1A;
  //printf("%d\n",D7);
  //for(money=1;money<100;money++){
  scanf("%d",&money );
  //printf("*********************\n");
  //printf("money %d\n", money);
  amari = money % D4;
  check1 = (money - amari) / D4;
  //printf("amari %d\n",amari );
  //printf("check1 %d\n",check1 );
  check = amari % 3;
  //printf("check %d\n",check );
/////////////////////////////////
  //printf("money %d\n", money);
  amariA = money % D7;
  //printf("%d\n",money %D7 );
  check1A = (money - amariA) / D7;
  //printf("amari %d\n",amariA );
  //printf("check1 %d\n",check1A );
  checkA = amari % 3;
  //printf("check %d\n",checkA );

  //printf("%d",money );
  if(check == 0&&check1 != 0){
    printf("Yes\n");
  }else if(checkA == 0&&check1A != 0){
    printf("Yes\n");
  }else if(amari == 0 && check1 != 0 || amariA == 0 && check1A != 0){
    printf("Yes\n");
  }else{
    printf("No\n");

  }
//}
  return 0;
}