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 <string>
#include <functional> 
#include <vector>
#include <numeric>
#include <stack>
#include <queue>
#include <utility>


using namespace std;

int main(){

    vector<int> a(3);
    cin >> a[0] >> a[1] >> a[2];

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

    if(a[0]%2 == 0 && a[1]%2 == 0 && a[2]%2 == 0){
        cout << (a[2]-a[0])/2 + (a[2]-a[1])/2;
        return 0;
    }

    if(a[0]%2 == 1 && a[1]%2 == 1 && a[2]%2 == 1){
        cout << (a[2]-a[0])/2 + (a[2]-a[1])/2;
        return 0;
    }

    if((a[0]%2 == 1 && a[1]%2 == 0 && a[2]%2 == 1) || (a[0]%2 == 0 && a[1]%2 == 1 && a[2]%2 == 0)){
        a[0] += 1;
        a[2] += 1;
        sort(a.begin(),a.end());
        cout << (a[2]-a[0])/2 + (a[2]-a[1])/2 + 1;
        return 0;
    }


    if((a[0]%2 == 1 && a[1]%2 == 0 && a[2]%2 == 0) || (a[0]%2 == 0 && a[1]%2 == 1 && a[2]%2 == 1)){
        a[1] += 1;
        a[2] += 1;
        sort(a.begin(),a.end());
        cout << (a[2]-a[0])/2 + (a[2]-a[1])/2 + 1;
        return 0;
    }

    if((a[0]%2 == 0 && a[1]%2 == 0 && a[2]%2 == 1) || (a[0]%2 == 1 && a[1]%2 == 1 && a[2]%2 == 0)){
        cout << a[2] - a[1] + (a[1] - a[0])/2;
    }

    return 0;
}