hoppity-hop.rb

# http://www.facebook.com/jobs_puzzles/index.php?puzzle_id=7
cap = ARGV[0] ? File.read(ARGV[0]).to_i : 189

=begin
  A simplistic, brute force solution:
=end

mod = lambda {|n, m| n%m==0}
word = lambda {|k| return 'Hop' if mod[k,15]; return 'Hophop' if mod[k,5]; return 'Hoppity' if mod[k,3]}
(1..cap).map {|i| s = word[i]; s ? puts(s) : next}
puts "—————————————————"

=begin
  Lets try and be a bit more clever.
  Exploiting that the modulo operation will repeat itself consistently, we can find the 
  quotient and remainder of the inputted number (here read into the variable cap). Along
  with the sequence of output for the first 15 integers, we solve the problem by printing:
  
  The base output repeated `quotient` times and then
  The first n words in the base output, where n is the sum of the 
  quotients of the remainder when divided by 3 and 5.
=end

words = %w[Hoppity Hophop Hoppity Hoppity Hophop Hoppity Hop]
quo, rem = cap.divmod(15)
puts words*quo + words.first(rem/3+rem/5)
puts "—————————————————"

=begin
  But this leads us to a somewhat simpler solution.
  With any given input n, the number of printed terms will be n/3+n/5-n/15.
    (Please don't make me give a formal proof, but I'm using 
    the principle of inclusion/exclusion or something like that.)
  So we can go from the 1st term up to the (n/3+n/5-n/15)st and consecutively print
  the words as they appear in the base case (1..15) output.
=end

(1..(cap/3+cap/5-cap/15)).map {|n| puts words[(n-1)%7]}