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: Runtime Error)

#include <stdio.h>

int main(void){
	int N,i,j,x=4,y=7,a=0;
	scanf("%d",&N);
	for(i = 0; i < N; i++){
		for(j = 0; j < N; j++){
			a= (x * i) + (y * j);
			//
			printf("i= %d j= %d a= %d n= %d ",i,j,a,N);
			printf("\n");
			
			if(N % 4 == 0){
				printf("Yes\n");
				return 0;
			}
			else if(N % 7 == 0){
				printf("Yes\n");
				return 0;
			}
			else if(N % a == 0){
				printf("Yes\n");
				return 0;
			}
			else{
				printf("No\n");
				return 0;
			}
		}
	}
	return 0;
}