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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace test
{
    class Program
    {
        static void Main(string[] args)
        {
            var input = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
            var A = input[0];
            var B = input[1];
            var C = input[2];
            var max = Math.Max(A, Math.Max(B, C));
            var min = Math.Min(A, Math.Max(B, C));
            var mid = A + B + C - max - min;
            var even = new bool[] {
                max % 2 == 0 ? true : false,
                min % 2 == 0 ? true : false,
                mid % 2 == 0 ? true : false
            };
            var ans = 0;

            if (even[0] == even[1] && even[1] == even[2])
            {
                ans += (2 * max - min - mid) / 2;
            }
            else if (even[1] == even[2])
            {
                ans++;
                ans += (2 * max - min - mid - 2) / 2;
            }
            else
            {
                ans++;
                ans += (2 * max - min - mid + 1) / 2;
            }

            Console.WriteLine(ans);
        }
    }
}