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: Accepted)

#include <stdio.h>
int main(void)
{
  int a,b,c,result;
  scanf("%d %d %d",&a,&b,&c);
  int a_=a,b_=b,c_=c;
  if (a<b) ;
  else a=b_,b=a_;
  a_=a,b_=b,c_=c;
  if (b<c) ;
  else b=c_,c=b_;
  a_=a,b_=b,c_=c;
  if (a<b) ;
  else a=b_,b=a_;
  a_=a,b_=b,c_=c;
//  printf("%d %d %d",a,b,c);
  if (((b-a)%2==0)&&((c-b)%2==0)) result=(((c-a)/2)+((c-b)/2));
  else if (((b-a)%2==0)&&((c-b)%2==1))
  {
    a++;
    b++;
    result++;
    result+=((c-a)/2+(c-b)/2);
  }
  else if (((b-a)%2==1)&&((c-b)%2==0))
  {
    c++;
    b++;
    result++;
    result+=((c-a)/2+(c-b)/2);
  }
  else if (((b-a)%2==1)&&((c-b)%2==1))
  {
    c++;
    a++;
    result++;
    result+=((c-a)/2+(c-b)/2);
  }
  printf("%d",result);
  return 0;
}