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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <string>
using namespace std;
//変数デバッグ
#define DEB(variable) cout << #variable << '=' << variable << endl

//for簡易表記(引数ミス防止)
#define FOR(LoopVariable,numberOFbegin,numberOFend) for (int LoopVariable = (numberOFbegin); (LoopVariable) < (numberOFend); (LoopVariable)++)
#define REP(LoopVariable,numberOFend) for(int LoopVariable = 0;(LoopVariable)<(numberOFend);LoopVariable++)

int main(){

    bool firstope = 0;
    vector<int> num(3);
    vector<bool> pari(3);
    
    REP(i,3){
        cin >> num[i];
    }

    
    {
        int tmp;

        if(num[0]>num[1]){
            tmp = num[1];
            num[1] = num[0];
            num[0] = tmp;
        }
        if(num[1]>num[2]){
            tmp = num[2];
            num[2] = num[1];
            num[1] = tmp;
        }
        if(num[0]>num[1]){
            tmp = num[1];
            num[1] = num[0];
            num[0] = tmp;
        }
    }

    REP(i,3){
        pari[i] = num[i] % 2;
    }
    

    if(pari[0]!=pari[1]){
        if(pari[0]==pari[2]){
            firstope=1;
            num[0]++;
            num[2]++;
        }else if(pari[1]==pari[2]){
            firstope=1;
            num[1]++;
            num[2]++;
        }
    }
    

    cout<<firstope+(num[2]-num[1])+((num[1]-num[0])/2)<<endl;

    return 0;
}