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>
#include <vector>
#include <algorithm>
using namespace std;


void solve(long long A, long long B, long long C){
    vector<long long> abc = {A,B,C};
    sort(abc.begin(), abc.end());
    long long d1l = abc[2]-abc[1], d2l=abc[2]-abc[0];
    double d1 = (double)d1l, d2 = (double)d2l, ans;

    if(d1l%2==0 && d2l%2==0) ans=min(d1/2+d2/2, d1+(d2-d1)/2);
    else if(d1l%2==0 && d2l%2==1) ans=min(d1/2+ceil(d2/2)+1, d1+ceil((d2-d1)/2)+1);
    else if(d1l%2==1 && d2l%2==0) ans=min(d2/2+ceil(d1/2)+1, d1+ceil((d2-d1)/2)+1);
    else ans=min(floor(d1/2)+floor(d2/2)+1, d1+(d2-d1)/2);
    cout << ans << endl;
}

int main(){
    long long A;
    scanf("%lld",&A);
    long long B;
    scanf("%lld",&B);
    long long C;
    scanf("%lld",&C);
    solve(A, B, C);
    return 0;
}