Marko Rauhamaa <ma...@pacujo.net> writes: > And in fact, the sqrt optimization now makes the original version 20% > faster: ... > bound = int(math.sqrt(n))
That could conceivably fail because of floating point roundoff or overflow, e.g. fac(3**1000). A fancier approach to finding the integer square root might be worthwhile though. -- https://mail.python.org/mailman/listinfo/python-list
