ruby 解答例https://codeiq.jp/magazine/2014/03/6633/

Posted 2021-05-16

# https://codeiq.jp/magazine/2014/03/6633/
# 出題は上記URLから

# Rubyで一番簡単な解答例
1.upto(100) do |n|
  puts n.to_s(2).to_i(3)
end

# 無理やり再帰を使った、というかSchemeっぽく解いた例
def concrete_function(buff , pointer, offset, weight, limit)
  return buff[1, limit] if offset < pointer || offset >= limit

  buff[offset + pointer + 1] = buff[pointer] + weight

  concrete_function(buff, 0, offset + pointer + 1, weight * 3, limit) if offset == pointer
  concrete_function(buff, pointer + 1, offset, weight, limit)
end

def result(limit)
  return concrete_function([0], 0, 0, 1, limit)
end

puts result(100).join("\n")