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: Time Limit Exceeded)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Diagnostics.CodeAnalysis;
using System.Linq;
using System.Runtime.CompilerServices;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp6
{
    class Program
    {
        static void Main(string[] args)
        {
            var source = Console.ReadLine().Split(' ').Select(s => int.Parse(s));

            var a = source.ElementAt(0);
            var b = source.ElementAt(1);
            var c = source.ElementAt(2);

            var seedList = new List<int> { a, b, c };

            for (int i = 0; i < Int32.MaxValue; i++)
            {
                if (a == b && b == c)
                {
                    Console.WriteLine(i);
                    break;
                }

                
                // 同じ値が２つある時はそれぞれに１を足す

                

                if (seedList.Count(i1 => seedList.Min() == i1) == 1)
                {
                    //一番小さい値が存在するときは小さいやつに２を足す
                    seedList[seedList.IndexOf(seedList.Min())] += 2;
                }
                else
                {
                    for (int j = 0; j < 3; j++)
                    {
                        var minValue = seedList.Min();
                        if (seedList[j] == minValue)
                        {
                            seedList[j] += 1;
                        }
                    }
                }
            }
        }
    }
}