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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <ctype.h>
#include <stdint.h>
#include <string.h>
#include <wchar.h>
#include <math.h>

#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define ABSS(a, b) ((a) > (b) ? (a) - (b) : (b) - (a))

int compare_f(const void *a, const void *b) {
	return *((size_t *)a) > *((size_t *)b) ? -1 : 1;
}

static size_t comb(const size_t n, const size_t r) {
	size_t result = 1;

	for (size_t i = 0; i < r; i++) {
		result *= n - i;
		result /= i + 1;
	}

	return result;
}

static size_t countDivisors(const size_t n)
{
	size_t count = 1;

	for (size_t i = 2; i <= n; i++) {
		if (n%i == 0) {
			count++;
		}
	}

	return count;
}

int main(void) {
	size_t N;

	scanf("%zu", &N);

	bool found = false;

	for (size_t numCakes = 0; numCakes <= N / 4; numCakes++) {
		if ((N - 4 * numCakes) % 7 == 0) {
			found = true;
			break;
		}
	}

	puts((found ? "Yes" : "No"));

	return 0;
}