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>
int main()

{
	int N;
	scanf("%d", &N);
	
	double ma = N / 4;
	int wa;
	int c = 0;
	if(N % 4 == 0) printf("Yes");
  else
  {
	for(wa = 0; wa <= ma -1; wa++)
	{
		if(N % 7 == 0){
          c++;
		}
        N = N - 4;
	}
	
	if(c > 0) 	printf("Yes");	
	else printf("No");
  }
		return 0;
}