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)

// Last Change: 03/21/2019 23:24:43.
#include <algorithm>
#include <array>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <memory>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

using namespace std;
using ll = long long;

namespace NS {
void LoopUntilZeroInpput() {
  int hogegegege = 0;
  while (cin >> hogegegege && hogegegege != 0) {
  }
}
} // namespace NS

int main() {
  vector<int> ipt(3);
  vector<int> evenInfo;
  vector<int> oddInfo;

  for (int i = 0; i < 3; ++i) {
    cin >> ipt.at(i);
    if (ipt.at(i) % 2 == 0) {
      evenInfo.push_back(i);
    } else {
      oddInfo.push_back(i);
    }
  }

  int ans = 0;
  //偶奇揃え
  if (evenInfo.size() == 2) {
    ++ipt.at(evenInfo.at(0));
    ++ipt.at(evenInfo.at(1));
    ans = 1;
  } else if (oddInfo.size() == 2) {
    ++ipt.at(oddInfo.at(0));
    ++ipt.at(oddInfo.at(1));
    ans = 1;
  }

  sort(ipt.begin(), ipt.end());
  ans += (ipt[2] - ipt[0]) / 2;
  ans += (ipt[2] - ipt[1]) / 2;

  cout << ans << endl;

  //NS::LoopUntilZeroInpput();
}