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 (1.42.0) (judge verdict: Accepted)

#![allow(non_snake_case)]
#![allow(unused_imports)]

use proconio::{input, fastout};
use proconio::marker::*;
use whiteread::parse_line;
use std::collections::*;
use num::*;
use superslice::*;

use itertools::Itertools;
use itertools_num::ItertoolsNum;

#[fastout]
fn main() {
    input!{
        a: usize, b: usize, c: usize,
    }

    if a == b && b == c && c == a {
        println!("0")
    } else {
        let mut abc_vec = vec![a, b, c];
        abc_vec.sort();

        let diff = vec![abc_vec[2]-abc_vec[0], abc_vec[2]-abc_vec[1]];

        // both even
        if diff.iter().all(|x| x % 2 == 0) {
            println!("{}", diff.iter().sum::<usize>() / 2)
        } else if diff.iter().all(|x| x % 2 == 1) {
            // both odd
            println!("{}", diff.iter().sum::<usize>() / 2)
        } else {
            println!("{}", (diff.iter().sum::<usize>()+1) / 2 + 1)
        }
    }
}