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 <iostream>
#include <algorithm>
#include <vector>
#include <string>

using namespace std;

inline int get_steps(int a, int b) {
	return abs(a - b) / 2;
}

int main() {
	ios::sync_with_stdio(false);

	vector<int> vec(3);

	cin >> vec[0] >> vec[1] >> vec[2];

	sort(vec.begin(), vec.end());

	bool o1_p = vec[0] % 2;
	bool o2_p = vec[1] % 2;
	bool max_p = vec[2] % 2;

	int ans = get_steps(vec[0], vec[2]) + get_steps(vec[1], vec[2]);

	if ((o1_p == max_p && o2_p != max_p) || (o1_p != max_p && o2_p == max_p)) ans += 2;
	else if (o1_p != max_p && o2_p != max_p) ++ans;

	cout << ans;

	//system("pause");
	return 0;
}