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>
using namespace std;
using ll = long long;
using P = pair<int, int>;

const ll INF64 = 1LL << 60;
const int INF32 = 1 << 29;
const ll MOD = 1000000007;


int main(){
	vector<int> a(3);
    cin >> a[0] >> a[1] >> a[2];
    sort(a.begin(), a.end());
    int num=0, ans =0;
    for(int i = 0; i < 3; ++i){
if(a[i]%2==1){++num;}
}
if(num==1){++ans;
    for(int i = 0; i < 3; ++i){
if(a[i]%2==1){--a[i];}
}
}
if(num==2){++ans;
for(int i = 0; i < 3; ++i){
if(a[i]% 2==0){--a[i];}
}
}

    ans += (a[2]-a[0])/2;
    
    ans += (a[2]-a[1])/2;
    
	cout << ans << endl;
    return 0;
}