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Jason Bailey wrote:

> for key, value in @data
>   count, num  um.divmod(value)
>   reply << (key * count)
> end


That's neat, I approached it solely as a parsing problem, not a 
mathematical one.
My solution is a tad longer, but it works (I think) :)

Tim


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#!/usr/bin/env ruby

module RomanNumerals
  BASICS   'I'
, 'V', 'X'
0, 'L'0, 'C'
00, 'D'00, 'M'
000 }
  def self.convert( n )
    (48..57)  n[0] ? arabic_to_roman( n ) : roman_to_arabic( n )
  end
  def self.roman_to_arabic( n )
    total, previous  , 0  
    n.to_s.split(//).reverse.each do |numeral|
      current  ASICS[ numeral.upcase ]
      total + revious
      total -  * current if current < previous
      previous  urrent
    end
    total + revious
    return total
  end
  def self.arabic_to_roman( n )
    numerals, tens, basic  ], 1, BASICS.invert  
    n.to_s.split(//).collect { |i| i.to_i }.reverse.each do |digit|
      numerals.unshift case digit
        when 2, 3     then basic[ tens ] * digit
        when 4        then basic[ tens ] + basic[ tens * 5 ]
        when 9        then basic[ tens ] + basic[ tens * 10 ]
        when 6, 7, 8  then basic[ tens * 5 ] + basic[ tens ] * ( digit - 5 )
        when 1, 5     then basic[ tens * digit ]
      end
      tens  ens * 10
    end
    return numerals.to_s
  end
end

if ARGV.empty?
  ARGV.push( $stdin.read.chomp )
else
  ARGV.collect! { |file| IO.read( file ) }
end

ARGV.collect! do |input|
  input.split("\n").collect { |line| RomanNumerals.convert( line ) }.join("\n")
end

$stdout.write ARGV.join("\n")+"\n"



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