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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>

int main()
{
  int a,b,c;
  int max=-1;
  int p;
  int ans;
  scanf("%d%d%d",&a,&b,&c);
  if(max<a){
    max=a;
    p=1;
  }
  if(max<b){
    max=b;
    p=2;
  }
  if(max<c){
    max=c;
    p=3;
  }
  if((3*max)%2==(a+b+c)%2){
    if(p==1){
      ans=(2*max-b-c)/2;
    }
    else if(p==2){
      ans=(2*max-a-c)/2;
    }
    else if(p==3){
      ans=(2*max-a-b)/2;
    }
  }
  else{
    if(p==1){
      ans=(2*max+3-b-c)/2;
    }
    else if(p==2){
      ans=(2*max+3-a-c)/2;
    }
    else if(p==3){
      ans=(2*max+3-a-b)/2;
    }
  }
  printf("%d\n",ans);
  return 0;
}