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FizzBuzz everyday

Name: Anonymous 2011-09-14 3:47

(use '[match.core :only (match)])



(doseq [n (range 1 101)]

  (println (match [(mod n 3) (mod n 5)]

                  [0 0] "FizzBuzz"

                  [0 _] "Fizz"

                  [_ 0] "Buzz"

                  :else n)))

Name: Anonymous 2011-09-14 4:31

>>3
0x0d,0x0a
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