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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

int main() {
  int a[3], count = 0;
  for (int i = 0; i < 3; i++) {
    cin >> a[i];
  } 

  int o = 0, e = 0;
  for (int i = 0; i < 3; i++) {
    a[i] % 2 ? o++ : e++;
  }

  if (o == 3 || e == 3) {
    int m = max(a[0], max(a[1], a[2]));
    for (int i = 0; i < 3; i++) {
      count += (m - a[i]) / 2;
    }
  } else if (e < o) {
    for (int i = 0; i < 3; i++) {
      if (a[i] % 2) {
        a[i]++;
      }
    }
    count++;
    int m = max(a[0], max(a[1], a[2]));
    for (int i = 0; i < 3; i++) {
      count += (m - a[i]) / 2;
    }
  } else if (e > o) {
    for (int i = 0; i < 3; i++) {
      if (!(a[i] % 2)) {
        a[i]++;
      }
    }
    count++;
    int m = max(a[0], max(a[1], a[2]));
    for (int i = 0; i < 3; i++) {
      count += (m - a[i]) / 2;
    }
  }

  cout << count << '\n';
  return 0;
}