Vance A Heron
3/7/2005 8:27:00 AM
I think the additional stipulation for "subtractors"
(other than the order of magnitude one) is that the
"subtractors" must be a power of 10 (I, X or C) *not*
V, L, or D.
That's what I coded.
Puzzled over the subtractor vs the repeats, until
I figured out one need to do both.
Here's my solution - now I'll be able to look
at everyone elses :)
Vance
--- roman.rb ---
#! /usr/bin/env ruby
def is_arabic?(n)
return n =~ /^[0-9]*$/
end
def to_roman(n)
r = ''
rd = %w{M D C L X V I}
val = [1000, 500, 100, 50, 10, 5, 1]
val.each_with_index { |v, i|
c = (n - (n % v))/v
if c < 4 && c > 0
c.times { r << rd[i] }
n -= (c * v)
end
s = (i % 2 == 0) ? i+2 : i+1
if (s < 7) && (n >= (val[i] - val[s]))
r << rd[s] << rd[i]
n -= (val[i] - val[s])
end
}
return r
end
def to_arabic(n)
val = { 'I' => 1, 'V' => 5, 'X' => 10, 'L' => 50,
'C' => 100, 'D' => 500, 'M' => 1000 }
sum = 0
lastval = 1000
n.each_byte {|c|
sum += val[c.chr]
sum -= (lastval * 2) if val[c.chr] > lastval
lastval = val[c.chr]
}
return sum
end
f = File.open(ARGV[0])
while line = f.gets
line.chomp!
val = is_arabic?(line) ? to_roman(line.to_i) : to_arabic(line)
puts "#{line} -> #{val}"
end
----