`%`演算子を使わずとも算数の知識だけでFizzBuzzができる例 - Qiita

Help us understand the problem. What are the problem?

[kiyo_hiko]

modなしならループ使うかな。Rubyではないが (defun fizzbuzz-to (n) (do ((i 1 (1+ i)) (f 1 (if (<= 2 f) 0 (1+ f))) (b 1 (if (<= 4 b) 0 (1+ b))) (acc nil)) ((< n i) (nreverse acc)) (push (cond ((= f b 0) 'FizzBuzz) ((= f 0) 'Fizz) ((= b 0) 'Buzz) (t i)) acc)))

[tbpg]

(ฅ・ω・)ฅ { FizzBuzzer だ }