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;
  int MAX;
  int cont = 0;
  scanf("%d%d%d",&A,&B,&C);
  MAX = (A > B) ? A : ((B > C) ? B : C);
  if(MAX - A){
    if((MAX - A) % 2){
      cont += ((MAX - A) / 2);
       A = MAX - 1 ;
    }
    else {
       cont += ((MAX - A) / 2);
        A = MAX;
    }
  }
  if(MAX - B){
    if((MAX - B) % 2){
      cont += ((MAX - B) / 2);
        B = MAX - 1 ;
    }
    else {
       cont += ((MAX - B) / 2);
         B = MAX;
    }
  }if(MAX - C){
    if((MAX - C) % 2){
      cont += ((MAX - C) / 2);
       C = MAX - 1 ;
    }
    else {
      cont += ((MAX - C) / 2);
        C = MAX;
    }
  }
  if(MAX * 3 - (A + B + C) == 2){
    cont += 1;
    printf("%d",cont);
  }
  else if(MAX * 3 - (A + B + C) == 0){
     printf("%d",cont);
  }
  else if(MAX * 3 - (A + B + C) == 1){
    cont += 2;
     printf("%d",cont);
  }
    return 0;
}