# Your Problem:
# Given a non-negative int n, compute recursively (no loops) the count of the
# occurrences of 8 as a digit, except that an 8 with another 8 immediately to
# its left counts double, so 8818 yields 4. Note that mod (%) by 10 yields
# the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the
# rightmost digit (126 / 10 is 12).
#
#
# count8(8) → 1
# count8(818) → 2
# count8(8818) → 4
#
# Credit for this problem goes to Nick Parlante from CodingBat.com
# Do not modify anything outside of the function or the function itself.
# CODE START
def getCount8(num:int) -> int:
# CODE END
# Press Ctrl+E (Command+E on mac) to test your code