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 procon_read<T: std::str::FromStr>() ->  Vec<T> {
	let mut buf = String::new();
	io::stdin().read_line(&mut buf).expect("");
	let iter = buf.split_whitespace();
	let mut v: Vec<T> = Vec::new();
	for i in iter {
		v.push(i.parse().ok().unwrap());
	}
	v
}

fn main() {
	let v:Vec<i32> = procon_read();
	let n = v[0];
	let up4 = n/4;
	let up7 = n/7;
	let mut flag = false;
	for i in 0..(up4+1) {
		for j in 0..(up7+1) {
			if 4*i + 7*j == n {
				flag = true;
				break;
			}
		}
		if flag {
			break;
		}
	}
	if flag {
		println!("Yes");
	}else{
		println!("No");
	}
}