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 Python (3.4.3) (judge verdict: Accepted)

A,B,C = map(int,input().split())
L = [A,B,C]
t = 0
L.sort()
if L[0] == L[1] == L[2]:
  print(0)
else:
  for i in range(3):
    if L[i]%2 == 0:
      t += 1
  if t == 3 or t == 0:
    print((L[2]-L[0])//2+(L[2]-L[1])//2)
  elif t == 2:
    if L[0]%2 != 0:
      L[1] += 1
      L[2] += 1
      print(1+(L[2]-L[0])//2+(L[2]-L[1])//2)
    elif L[1]%2 != 0:
      L[0] += 1
      L[2] += 1
      print(1+(L[2]-L[0])//2+(L[2]-L[1])//2)
    else:
      L[0] += 1
      L[1] += 1
      print(1+(L[2]-L[0])//2+(L[2]-L[1])//2)
  else:
    if L[0]%2 == 0:
      L[1] += 1
      L[2] += 1
      print(1+(L[2]-L[0])//2+(L[2]-L[1])//2)
    elif L[1]%2 == 0:
      L[0] += 1
      L[2] += 1
      print(1+(L[2]-L[0])//2+(L[2]-L[1])//2)
    else:
      L[0] += 1
      L[1] += 1
      print(1+(L[2]-L[0])//2+(L[2]-L[1])//2)