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 ll = long long;
using namespace std;
#define rep(i,n) for(int i=0, i##_len=(int)(n); i<i##_len; i++)
#define reps(i,n) for(int i=1 , i##_len=(int)(n);i<=i##_len;i++)
#define rrep(i,n) for(int i=((int)(n)-1);i>=0;i--)
#define rreps(i,n) for(int i=((int)(n));i>0;i--)
#define repi(i,x) for(auto i=(x).begin(),i##_fin=(x).end();i!=i##_fin;i++)
#define all(x) (x).begin(), (x).end()
#define F first
#define S second
#define mp make_pair
#define pb push_back
#define solve(a) ((a)?"Yes":"No")
typedef vector<int> Vi;
typedef vector<Vi> VVi;
typedef pair<int , int> Pi;
typedef vector<Pi> VPi;
typedef vector<long long> V;
typedef vector<V> VV;
typedef pair<long long , long long> P;
typedef vector<P> VP;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1;} return 0;}
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1;} return 0;}
const long long INFLL = 1LL<<60;
const int INF = 1<<30;
const double PI=acos(-1);
int main(){
	int a,b,c;
	int ans=0;
	cin >>a>>b>>c;
	int head=a;
	int sum=0;
	chmax(head,b);
	chmax(head,c);
	a-=head;
	b-=head;
	c-=head;
	sum=a+b+c;
	clog<<sum<<endl;
	if(sum%2==-1){
		sum-=3;
	}
	cout << -sum/2 << endl;
}
/*
2  5  4
-3 0 -1
0 3 2

2 6 3
-4 0 -3
-5 -1 -4

31 41 5
-10 0 -36  46 
*/