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(void)
{
	//変数を定義
	int n, cake = 4, donut = 7,ans = 0,yes = 0;
	scanf("%d", &n);

	for (int i = 0; i <= 100 ;i++) {
		for (int j = 0; j <= 100 ; j++) {
			if (n == cake * i + donut * j) {
				yes++;
				break;
			}
			else if(n < cake * i + donut * j){
				ans++;
				break;
			}
		}
		if (n == cake * i) {
			yes++;
			break;
		}
		if (n == donut * i) {
			yes++;
			break;
		}
	}

	if (yes != 0) {
		printf("Yes\n");
	}
	else if(ans != 0){
		printf("No\n");
	}

	return 0;
}