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: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
int main(void)
{   int a,b,c,i=0,x,y,z,e;
    scanf(" %d %d %d",&a,&b,&c);
    while(1)
    {
        if(a==b && b==c && c==a)
          {
              break;}
        if((a==b && c<a && (a-c)==1)||(b==c && a<b && (b-a)==1)||(a==c && b<a && (a-b)==1))
        {
            i=i+2;break;

        }
        if(a==b && c>a && (c-a)>=1)
        {
            x=c-a;i=i+x;
            break;
        }
        if(b==c && a>b && (a-b)>=1)
        {
            y=a-b;i=i+y;
            break;
        }
        if(a==c && b>a && (b-a)>=1)
        {
            z=b-a;i=i+z;
            break;
        }
        if(a!=b && b!=c && c!=a && a>b && b>c && (b-c)%2==0)
        {
            e=b-c;c=c+e;i=i+(e/2);
        }
        if(a!=b && b!=c && c!=a && a>c && c>b && (c-b)%2==0)
        {
            e=c-b;b=b+e;i=i+(e/2);
        }
        if(a!=b && b!=c && c!=a && c>b && b>a && (b-a)%2==0)
        {
            e=b-a;a=a+e;i=i+(e/2);
        }
        if(a!=b && b!=c && c!=a && c>a && a>b && (a-b)%2==0)
        {
            e=a-b;b=b+e;i=i+(e/2);
        }
        if(a!=b && b!=c && c!=a && b>a && a>c && (a-c)%2==0)
        {
            e=a-c;c=c+e;i=i+(e/2);
        }
        if(a!=b && b!=c && c!=a && b>c && c>a && (c-a)%2==0)
        {
            e=c-a;a=a+e;i=i+(e/2);
        }

        if(a!=b && b!=c && c!=a && a>b && b>c && (b-c)%2!=0)
        {
            z=a-b;b=b+z;c=c+z;i=i+z;
        }
        if(a!=b && b!=c && c!=a && a>c && c>b && (c-b)%2!=0)
        {
            z=a-c;b=b+z;c=c+z;;i=i+z;
        }
        if(a!=b && b!=c && c!=a && c>b && b>a && (b-a)%2!=0)
        {
            z=c-b;b=b+z;a=a+z;i=i+z;
        }
        if(a!=b && b!=c && c!=a && c>a && a>b && (a-b)%2!=0)
        {
            z=c-a;b=b+z;a=a+z;i=i+z;
        }
        if(a!=b && b!=c && c!=a && b>a && a>c && (a-c)%2!=0)
        {
            z=b-a;a=a+z;c=c+z;i=i+z;
        }
        if(a!=b && b!=c && c!=a && b>c && c>a && (c-a)%2!=0)
        {
            z=b-c;a=a+z;c=c+z;i=i+z;
        }
    }
    printf("%d",i);
    return 0;
}