Problem:
Score : 300 points

Problem Statement

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order:

Choose two among A, B and C, then increase both by 1.

Choose one among A, B and C, then increase it by 2.

It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

Constraints

0 \leq A,B,C \leq 50

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the minimum number of operations required to make A, B and C all equal.

Sample Input 1

2 5 4

Sample Output 1

2

We can make A, B and C all equal by the following operations:

Increase A and C by 1. Now, A, B, C are 3, 5, 5, respectively.

Increase A by 2. Now, A, B, C are 5, 5, 5, respectively.

Sample Input 2

2 6 3

Sample Output 2

5

Sample Input 3

31 41 5

Sample Output 3

23

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

#![allow(non_snake_case)]

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn read_tuple<T: std::str::FromStr + Copy>() -> (T, T, T) {
    let v: Vec<T> = read_vec();
    (v[0], v[1], v[2])
}

fn main() {
    let mut n: Vec<i32> = read_vec();
    n.sort();
    let mut ans = 0;
    if n[0] % 2 != n[1] % 2 && n[0] % 2 != n[2] % 2 {
        n[0] -= 1;
        ans += 1;
    } else if n[1] % 2 != n[0] % 2 && n[1] % 2 != n[2] % 2 {
        n[1] -= 1;
        ans += 1;
    } else if n[2] % 2 != n[0] % 2 && n[2] % 2 != n[1] % 2 {
        n[2] -= 1;
        ans += 1;
    }
    ans += (n[2] * 2 - n[0] - n[1]) / 2;
    println!("{}", ans);
}