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Comment by nappy-doo | Hacker News Reader

My best C solution (scores a 96):

#define d printf(

main(i){i<101?(!(i%3)?d"Fizz"):0)|(!(i%5)?d"Buzz"):0)?d"\n"):d"%d\n",i),main(i+1):0;}

I was working on the same lines, and managed to bump it up to a score of 106:

  #define P printf(
  main(n){(n%3?0:P"Fizz"))+(n%5?0:P"Buzz"))?:P"%d",n);P"\n");n>99?:main(n+1);}

you can get rid of the !() if you reverse the ?:

  #define d printf(
  main(i){i<101?(i%3?0:d"Fizz"))|(i%5?0:d"Buzz"))?d"\n"):d"%d\n",i),main(i+1):0;}

excellent, missed that. :)