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 Rust (1.15.1) (judge verdict: Accepted)

use std::io;

fn main() {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    let mut iter = buf.trim().split_whitespace();
    let n: i64 = iter.next().unwrap().parse().unwrap();
    let s = n / 7 + 1;
    let f = n / 4 + 1;
    let mut ans = 0;
    for number_s in 0..s {
        for number_f in 0..f {
            if 7 * number_s + 4 * number_f == n {
                ans += 1;
            }
        }
    }

    if ans >= 1 {
        println!("Yes");
    } else {
        println!("No");
    }
}