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) {
    // n = 読み込む数値、f = フラグ、i = ケーキの個数、j = ドーナツの個数
	int n;
	int f = 0, i = 0, j = 0;
	
    // 数値読み込み
	scanf("%d", &n);
    // 4で割り切れるならフラグ1
	if(n % 4 == 0) f = 1;
    // 7で割り切れるならフラグ1
	if(n % 7 == 0) f = 1;
    // 4 * i + 7 * j で割り切れるならフラグ1 にするループ文
	for(i = 4; i < 101; i += 4) {
		if(f == 1) break;
		for(j = 7; j < 101; j += 7) {
			if(n % (i + j) == 0) f = 1;
		}
	}
    // 出力
	printf( f ? "Yes\n" : "No\n");
	return 0;
}