Problem:
Score : 100 points

Problem Statement

Snuke has a favorite restaurant.

The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer.

So far, Snuke has ordered N meals at the restaurant.
Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen.
Find x-y.

Constraints

1 ≤ N ≤ 100

Input

The input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

20

Sample Output 1

15800

So far, Snuke has paid 16000 yen, and the restaurant has paid back 200 yen. Thus, the answer is 15800.

Sample Input 2

60

Sample Output 2

47200

Snuke has paid 48000 yen for 60 meals, and the restaurant has paid back 800 yen.

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn main() {
    let n: i32 = read();
    let mut ans = n * 800;
    ans -= n / 15 * 200;
    println!("{}", ans);
}

// 以下関数

use std::io::*;
use std::str::FromStr;
pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

pub const MOD: u64 = 10e9 as u64 + 7;

pub fn is_prime(n: i32) -> bool {
    if n < 2 {
        return false;
    } else if n == 2 {
        return true;
    }
    let mut i = 2;
    while i * i < n {
        if n % i == 0 {
            return false;
        } else {
            i += 1;
            continue;
        }
    }
    true
}

pub fn lcm(mut n: i32, mut m: i32) -> i32 {
    n = if n > m { m } else { n };
    m = if n > m { n } else { m };
    let mut i = n;
    while i % m != 0 {
        i += n;
    }
    i
}

pub fn abs(n: i32) -> i32 {
    if n >= 0 {
        n
    } else {
        -n
    }
}

pub fn max(n: i32, m: i32) -> i32 {
    if n >= m {
        n
    } else {
        m
    }
}

pub fn min(n: i32, m: i32) -> i32 {
    if n <= m {
        n
    } else {
        m
    }
}