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>
#include <stdlib.h>

int main()
{
int x,y,z;
int a,b,count=0;
scanf("%d %d %d",&x,&y,&z);
if(x%2==0 && y%2==0 && z%2!=0)
{
    x=x+1;
    y=y+1;
    count=count+1;
}
else if(x%2==0 && z%2==0 && y%2!=0)
{
    x=x+1;
    z=z+1;
    count=count+1;
}
else if(z%2==0 && y%2==0 && x%2!=0)
{
    z=z+1;
    y=y+1;
    count=count+1;
}
else if(x%2!=0 && z%2!=0 && y%2==0)
{
    x=x+1;
    z=z+1;
    count=count+1;
}
else if(x%2!=0 && y%2!=0 && z%2==0)
{
    x=x+1;
    y=y+1;
    count=count+1;
}

else if(z%2!=0 && y%2!=0 && x%2==0)
{
    z=x+1;
    y=y+1;
    count=count+1;
}
if((x%2==0 && y%2==0 && z%2==0)||(x%2!=0 && y%2!=0 && z%2!=0))
 {
 if(x>=y && x>=z){
   a= (x-y)/2;
   b= (x-z)/2;

  }
  else if(y>=x && y>=z){
   a= (y-x)/2;
   b= (y-z)/2;

  }
 else  if(z>=y && z>=x){
   a= (z-y)/2;
   b= (z-x)/2;

   }
  }

 if(count!=1)
  printf("%d\n",a+b);
 else if(count==1)
  printf("%d\n",a+b+1);

   return 0;
}