From: Kelly Tanguay <kelly.tanguay@...>
Date: 2008-02-03T04:33:00+09:00
Subject: Re: Learn to Program, by Chris Pine

This is my take on the same problem:


def roman_num number
set1 = [ 1, 5, 10, 50, 100, 500, 1000 ]
set2 = [ 'I', 'V', 'X', 'L', 'C', 'D', 'M' ]
numeral = []
  while number > 0
    if (number/(set1.last)) >= 1
      roman = (number/(set1.last))
      numeral.push((set2.pop)*roman)
      number = (number%(set1.pop))
    else
      set2.pop
      set1.pop
    end
  end
  puts 'Old Roman Numeral is ' + numeral.join + '.'
end

puts 'Please enter a number to see what it is in old roman numerals.'
number = gets.chomp.to_i
while number < 1 || number > 3999
  puts 'Please enter a number between 1 and 3999'
  number = gets.chomp.to_i
end
roman_num number



Jan_K wrote:
> Chapter 9, exercise 2 (page 76)
> 
> Old-school Roman numerals. In the early days of Roman numerals,
> the Romans didn?t bother with any of this new-fangled subtraction
> IX nonsense. No sir, it was straight addition, biggest to littlest -
> so 9 was written VIIII, and so on. Write a method that, when
> passed an integer between 1 and 3000 (or so), returns a string
> containing the proper old-school Roman numeral. In other words,
> old_roman_numeral 4 should return 'IIII'. Make sure to test
> your method on a bunch of different numbers. Hint: Use the integer
> division and modulus methods on page 36.
> For reference, these are the values of the letters used:
> I = 1 V = 5 X = 10 L = 50
> C = 100 D = 500 M = 1000
> 
> 
> Solution:
> ----------------------------------------------------------------
> def old_roman_number input
> 
>   while input < 1 || input > 3999
>     puts 'Please enter a number between 1 and 3999'
>     input = gets.chomp.to_i
>   end
> 
>   m_mod = input%1000
>   d_mod = input%500
>   c_mod = input%100
>   l_mod = input%50
>   x_mod = input%10
>   v_mod = input%5
> 
>   m_div = input/1000
>   d_div = m_mod/500
>   c_div = d_mod/100
>   l_div = c_mod/50
>   x_div = l_mod/10
>   v_div = x_mod/5
>   i_div = v_mod/1
> 
>   m = 'M' * m_div
>   d = 'D' * d_div
>   c = 'C' * c_div
>   l = 'L' * l_div
>   x = 'X' * x_div
>   v = 'V' * v_div
>   i = 'I' * i_div
> 
>   puts m + d + c + l + x + v + i
> 
> end
> 
> number = gets.chomp.to_i
> old_roman_number(number)

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