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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

int main(void){

  int a=0, b=0, c=0;
  int x=0, k=0;
  int m=0, n=0, buffer=0;
  int num[3]={0};
  int max=0, num_sum=0;

  for(m=0; m<3; m++){
    scanf("%d", num+m);
    num_sum=num_sum+num[m];
  }

  for(m=0; m<2; m++){
    for(n=m+1; n<3; n++){
      if(num[m]<num[n]){
	buffer=num[m];
	num[m]=num[n];
	num[n]=buffer;
      }
    }
  }
  m=0;
  max=num[0];

  while((num_sum+2*m) % 3 != 0 || ((num_sum+2*m) / 3 ) < max){
    m++;
  }

  x=(num_sum+2*m) / 3;

  k=( 3*x-num_sum ) / 2;

  printf("%d\n", k);

  return 0;
}