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 <stdio.h>
#include <vector>
#include <queue>
#include <string>
#include <cmath>
#define ll long long
#define FOR(i,t,x) for(int i=t;i<=x;i++)

using namespace std;

int ans,a,b,c;

int main(int argc, char const *argv[])
{
	cin>>a>>b>>c;
	int maks=max(a,max(b,c));
	while(a<(maks-1)){
		a+=2;ans++;
	}
	while(c<(maks-1)){
		c+=2;ans++;
	}
	while(b<(maks-1)){
		b+=2;ans++;
	}
	if(a==b&&b==c){
		cout<<ans<<endl;
		return 0;
	}
	if((a==maks&&b==maks)||(a==maks&&c==maks)||(b==maks&&c==maks)){
		ans+=2;
	}else{
		ans++;
	}
	cout<<ans<<endl;

	return 0;
}