4 ms·

Awesomely ingenious use of math above. 59 if newlines count as 1. i=1 while i<101:print'FizzBuzz'[ii%34:8--i**4%5]or i;i+=1

Awesomely ingenious use of math above. 59 if newlines count as 1. i=1 while i<101:print'FizzBuzz'[i*i%3*4:8--i**4%5]or i;i+=1

twiceaday 14y ago
Much simpler solution in 59 chars n=1 while n<101:print'Fizz'*(n%3<1)+'Buzz'*(n%5<1)or n;n+=1
twiceaday 14y ago
I came up with a shorter way to write the first index! n%-3&4 unbelievably it works
- glomph 14y ago
  haha, clever!