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)

use std::io::*;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::BTreeSet;
use std::collections::HashSet;
use std::cmp;
use std::f64::consts;

fn main() {
    let mut vec: Vec<i32> = (0..3).map(|_| read()).collect();
    vec.sort();

    let max = vec[2];
    let middle = vec[1];
    let mut min = vec[0];

    let mut count = 0;
    count += max-middle;

    min+=(max-middle);

    if (max-min) % 2 == 0 {
        count += (max-min) / 2;
    } else {
        count += (max-min)/2 + 2;
    }

    println!("{}", count);
}

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")
}

// 最大公約数
fn gcd(a: i32, b: i32) -> i32 {
    match b {
        0 => a,
        _ => gcd(b, a % b)
    }
}

// 最小公倍数
fn lcm(a: i32, b: i32) -> i32 {
    a * b / gcd(a, b)
}

// 階乗
fn kaijou(n: i32)->i32 {
    if n == 1 {
        return n;
    }
    return n * kaijou(n-1);
}