On Tue, Jun 15, 2010 at 6:21 AM, Alain Ketterlin
<al...@dpt-info.u-strasbg.fr> wrote:
> You compute i**2 too many times (7/5 times more than necessary) and
> twice too many modulos. I suggest:
>
> c = { 0:1, 1:1, 2:1, 3:-1, 4:-1 }
> #or, why not: c = lambda i : +1 if (i%5) < 3 else -1
>
> s = 0
> for i in range(1,2011):
> s += c[(i-1)%5]*(i**2)
> print s
In fact, most of them are unnecessary:
from itertools import izip, cycle
def squares(start, stop):
square = start * start
step = start * 2 + 1
for root in xrange(start, stop):
yield square
square += step
step += 2
print sum(sign * square for sign, square in izip(cycle([1,1,1,-1,-1]),
squares(1, 2011)))
Now, anybody know how to make that version a one-liner without making
it go quadratic in run-time?
Cheers,
Ian
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