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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
    int a, b, c, ans=0, i, j, k, max, max2, max3;
    scanf("%d %d %d", &a, &b, &c);
    if (a == b && b == c)
        printf("0");
    else if((a == b) || (b==c) || (a==c))
        printf("2");
    else
    {
        if(a>b)
        {
            if(b>c)
            {
                max = c;
                max2 = b;
                max3 = a;
            }
            else if(b<c)
            {
                max = a;
                max2 = c;
                max3 = b;
            }
        }
        else if(a<b)
        {
            if(b<c)
            {
                max = a;
                max2 = b;
                max3 = c;
            }
            else if(b>c)
            {
                max = b;
                max2 = c;
                max3 = a;
            }
        }
    }
    ans = max - max2;
    max3 += ans;
    if ((max - max3)%2 != 0)
    {
        ans = ans +((max-max3)/2)+2;
   //     printf("Here 1\n");
    }
    else
    {
        ans = ans + ((max-max3)/2);
   //     printf("Here 2\n");
    }
    printf("%d", ans);
    return 0;

}