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>

int main(){
  int a[3] = {};
  for(int i = 0; i < 3; i++){
    scanf("%d", &a[i]);
  }

  int temp = 0;

  for(int i = 0; i < 2; i++){
    if(a[i] > a[i+1]){
      temp = a[i+1];
      a[i+1] = a[i];
      a[i] = temp;
    }
  }
  if(a[0] > a[1]){
    temp = a[1];
    a[1] = a[0];
    a[0] = temp;
  }

  for(int i = 0; i < 3; i++)printf("%d ", a[i]);

  int c = 0;

  while(a[1] < a[2]){
    a[0] += 1;
    a[1] += 1;
    c += 1;
    }
  printf("\n");
  for(int i = 0; i < 3; i++)printf("%d ", a[i]);
    printf("\n");
    printf("%d", c);

  while(a[0] < a[1]){
    a[0] += 2;
    c += 1;
  }
  printf("\n");
  for(int i = 0; i < 3; i++)printf("%d ", a[i]);
    printf("\n");
    printf("%d", c);

  if(a[0] > a[1]){
    a[1] += 1;
    a[2] += 1;
    c += 1;
    for(int i = 0; i < 3; i++)printf("%d ", a[i]);
      printf("\n");
      printf("%d", c);
  }

  return 0;
}