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;
typedef long long ll;
typedef vector<int> vec;

int main(){
  vec t;
  for(int i=0;i<3;i++){
    int a;
    cin>>a;
    t.push_back(a);
  }
  queue<vec > Q;
  map < vec , int > d;

  d[t]=0;
  Q.push(t);
  
  while(!Q.empty()){
    vec p=Q.front();
    Q.pop();
    if(p[0]==p[1]&&p[1]==p[2]){
      cout<<d[p]<<endl;
      break;
    }
    
    for(int i=0;i<3;i++){
      vec np=p;
      np[i]+=2;
      if(d.count(np)==0){
        d[np]=d[p]+1;
        Q.push(np);
      }
    }
    
    vec tmp=p;
    tmp[0]++;
    tmp[1]++;
    tmp[2]++;
    
    for(int i=0;i<3;i++){
      vec np=tmp;
      np[i]--;
      if(d.count(np)==0){
        d[np]=d[p]+1;
        Q.push(np);
      }
    }
  }
  return 0;
}