Rubyで英数字を作って文字数を数える Rubyでオイラープロジェクトを解こう！Problem17
If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of “and” when writing out numbers is in compliance with British usage.
1から5の数は英単語で one, two, three, four, five と書かれるが、これには全部で3 + 3 + 5 + 4 + 4 = 19 の文字が使われている。
1から1000(one thousand)までの全ての数を英単語で書いた場合、何文字が使われるか。
注記：スペースとハイフンは数えない。例えば、342 (three hundred and forty-two) は23文字、115 (one hundred and fifteen)は20文字からなる。数字を書くときの”and”の使用は英国式に従う。
方針：
- 1から1000までの数字を英単語に変換するnum_to_wordメソッドを作る
- 入力範囲の数字の文字数の合計をカウントするcount_lettersメソッドを作る
WORDS = {0 => "", 1 => "one", 2 => "two", 3 => "three", 4 => "four", 5 => "five", 6 => "six", 7 => "seven", 8 => "eight", 9 => "nine", 10 => "ten", 11 => "eleven", 12 => "twelve", 13 => "thirteen", 14 => "fourteen", 15 => "fifteen", 16 => "sixteen", 17 => "seventeen", 18 => "eighteen", 19 => "nineteen", 20 => "twenty", 30 => "thirty", 40 => "forty", 50 => "fifty", 60 => "sixty", 70 => "seventy", 80 => "eighty", 90 => "ninety"}
def count_letters(range)
ans = 0
range.each do |n|
word = num_to_word(n)
ans += word.gsub(/[-\s]/, "").length
end
ans
end
def num_to_word(n)
word = ""
hund, ten = n.divmod(100)
case hund
when 1..9
if ten.zero?
word << "#{WORDS[hund]} hundred"
else
word << "#{WORDS[hund]} hundred and "
end
when 10
word << "one thousand"
end
case ten
when 10..19
word << WORDS[ten]
when 1..9, 20..99
ten, one = ten.divmod(10)
if one.zero?
word << "#{WORDS[ten*10]}"
elsif ten.zero?
word << "#{WORDS[one]}"
else
word << "#{WORDS[ten*10]}-#{WORDS[one]}"
end
end
word
end
count_letters(1..1000) # => 21124
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