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 <stdlib.h>
#include <stdio.h>
#include <math.h>
int main() {
	int n;
	int max1, max2;
	int x, y;//x:4ドルケーキ個数 y:7ドルケーキ個数
	scanf("%d", &n);
	max1 = n / 4;	//4ドルケーキのMAX個数
	max2 = n / 7;	//7ドルケーキのMAX個数
	for (x = 0; x <= max1; x++) {
		for (y = 0; y <= max2; y++) {
			if ((y * 7 + x * 4) == n) {
				printf("Yes\n");
				return 0;
			}
		}
	}
	printf("No\n");

	return 0;
}