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, B, C;
  scanf("%d%d%d", &A, &B, &C);

  int array[3] = {A, B, C};
  int array_len = sizeof(array) / sizeof(array[0]);
  sort(array, array + array_len);

  int is_even_count = 0;
  for (auto i : array) {
    if (i % 2 == 0) is_even_count++;
  }

  int ans = 0;
  if (is_even_count == 3 || is_even_count == 0) {
    ans += (array[2] - array[0]) / 2;
    ans += (array[2] - array[1]) / 2;
  } else if (is_even_count == 2) {
    for (int i = 0; i < 3; i++) {
      if (array[i] % 2 == 0) array[i]++;
    }
    ans++;
    ans += (array[2] - array[0]) / 2;
    ans += (array[2] - array[1]) / 2;
  } else if (is_even_count == 1) {
    for (int i = 0; i < 3; i++) {
      if (array[i] % 2 != 0) array[i]++;
    }
    ans++;
    ans += (array[2] - array[0]) / 2;
    ans += (array[2] - array[1]) / 2;
  }

  printf("%d", ans);
}