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(unused_macros)]
macro_rules! get {
    ([$t:ty] ; $n:expr) => {
        (0..$n)
        .map(|_| get!([$t])).collect::<Vec<_>>()
    };
    ($($t:ty),+ ; $n:expr) => {
        (0..$n).map(|_| get!($($t),+)).collect::<Vec<_>>()
    };
    ([$t:ty]) => {
        rl().split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect::<Vec<$t>>()
    };
    ($t:ty) => {
        rl().parse::<$t>().unwrap()
    };
    ($($t:ty),*) => {
        {
            let line =rl();
            let mut iter = line.split_whitespace();
            ($(iter.next().unwrap().parse::<$t>().unwrap()),*)
        }
    };
}
fn rl() -> String {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).unwrap();
    line.trim().to_string()
}

fn main() {
    let mut v: Vec<usize> = get!([usize]);
    let mut odd_num = 0;
    for i in 0..v.len() {
        if v[i] & 1 == 1 {
            odd_num += 1;
        }
    }
    let mut ans = 0;
    if odd_num == 1 {
        ans += 1;
        for i in 0..v.len() {
            if v[i] & 1 == 0 {
                v[i] += 1;
            }
        }
    } else if odd_num == 2 {
        ans += 1;
        for i in 0..v.len() {
            if v[i] & 1 == 1 {
                v[i] += 1;
            }
        }
    }
    v.sort();
    let biggest = v[2];
    let middle = v[1];
    let smallest = v[0];
    ans += biggest - middle;
    ans += (biggest - ((biggest - middle) + smallest)) / 2;
    println!("{}", ans);
}