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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<functional>
#include <random>

using namespace std;

int main(void) {
	int N;
	int a[101];
	scanf("%d", &N);
	

	for (int i = 0; i < 100; i++) {
		a[i] = 0;
	}
	for (int i = 0; 4*i <= 100; i++) {
		for (int j = 0; 4 * i + 7 * j <= 100; j++) {
			a[4 * i + 7 * j] = 1;
		}
	}
	if (a[N] == 1) {
		printf("Yes\n");
	}
	else {
		printf("No\n");
	}

}