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: Wrong Answer)

#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<vector>
#include<string> 
#include<queue>
#include<algorithm>
using namespace std;
int main()
{
	int a[3],b[2],i,m,ans,k;
	while(scanf("%d%d%d",&a[0],&a[1],&a[2])!=EOF)
	{
		k=0;
		m=max(a[0],max(a[1],a[2]));
		for(i=0;i<3;i++)
		{
			if((m-a[i])!=0)b[k++]=m-a[i];
			else;
		}
		if(k==0)ans=0;
		else if(k==1)
		{
			if(b[0]%2)ans=b[0]/2+2;
			else ans=b[0]/2;
		}
		else {
			if(b[0]%2+b[1]%2==1)
		ans=max(b[0]+1,b[1]+1);
		else ans=(b[0]+b[1])/2;
		}
		printf("%d\n",ans);
	}
	return 0;
}