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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 6:01

#include <stdio.h>



int main()

{

    char* s[4] = { "%d\n", "Fizz\n", "Buzz\n", "FizzBuzz\n" };

    int i = 0;

    while(++i <= 100)

        printf(s[(i % 3 == 0) | ((i % 5 == 0) << 1)], i);

}

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