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>
using namespace std;
int main(){
  int ans=1<<30;
  int A,B,C,a,b,c,bitf,cnt;
  cin >> A >> B >> C;
  //AB,BC,CA++をするか、しないかを決めて揃える。
  for(bitf=0;bitf<(1<<3);bitf++){
    a=A;b=B;c=C;cnt=0;
    if(bitf & 1<<0 ){a++;b++;cnt++;}
    if(bitf & 1<<1 ){b++;c++;cnt++;}
    if(bitf & 1<<2 ){c++;a++;cnt++;}
    //偶奇パリティチェック
    if(a%2==b%2 && a%2==c%2){
    int max_abc = max(a,max(b,c));
    cnt+=(3*max_abc-a-b-c)/2;
    ans = min(ans,cnt);
    }
  }
  cout << ans << endl;
  return 0;
}