From: superpollo on 15 Jun 2010 07:49 goal (from e.c.m.): evaluate 1^2+2^2+3^2-4^2-5^2+6^2+7^2+8^2-9^2-10^2+...-2010^2, where each three consecutive + must be followed by two - (^ meaning ** in this context) my solution: >>> s = 0 >>> for i in range(1, 2011): .... s += i**2 .... if not (i+1)%5: .... s -= 2*i**2 .... if not i%5: .... s -= 2*i**2 ....>>> print s 536926141>>> bye From: Ulrich Eckhardt on 15 Jun 2010 08:12 superpollo wrote:> ... s += i**2 > ... if not (i+1)%5: > ... s -= 2*i**2 > ... if not i%5: > ... s -= 2*i**2 if not (i % 5) in [1, 2]: s += i**2 else: s -= i**2 Untested code. Uli -- Sator Laser GmbH Geschäftsführer: Thorsten Föcking, Amtsgericht Hamburg HR B62 932 From: Alain Ketterlin on 15 Jun 2010 08:21 superpollo writes: > goal (from e.c.m.): evaluate > 1^2+2^2+3^2-4^2-5^2+6^2+7^2+8^2-9^2-10^2+...-2010^2, where each three > consecutive + must be followed by two - (^ meaning ** in this context) > > my solution: > >>>> s = 0 >>>> for i in range(1, 2011): > ... s += i**2 > ... if not (i+1)%5: > ... s -= 2*i**2 > ... if not i%5: > ... s -= 2*i**2 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 Or, as a one liner using a list comprehension: print sum( ( c[(i-1)%5]*i**2 for i in xrange(1,2011) ) ) I don't know which one is the fastest (between dict+loop, dict+compr, lambda+loop and lambda+compr). -- Alain. From: superpollo on 15 Jun 2010 08:23 Ulrich Eckhardt ha scritto:> superpollo wrote: >> ... s += i**2 >> ... if not (i+1)%5: >> ... s -= 2*i**2 >> ... if not i%5: >> ... s -= 2*i**2 > > if not (i % 5) in [1, 2]: > s += i**2 > else: > s -= i**2 > > Untested code. does not work: >>> s = 0 >>> for i in range(1, 2011): .... if not (i % 5) in [1, 2]: .... s += i**2 .... else: .... s -= i**2 ....>>> print s 546627205>>> but this does: >>> s = 0 >>> for i in range(1, 2011): .... if i % 5 in [1, 2, 3]: .... s += i**2 .... else: .... s -= i**2 ....>>> print s 536926141 bye From: Stefan Behnel on 15 Jun 2010 08:23 superpollo, 15.06.2010 13:49:> goal (from e.c.m.): evaluate > 1^2+2^2+3^2-4^2-5^2+6^2+7^2+8^2-9^2-10^2+...-2010^2, where each three > consecutive + must be followed by two - (^ meaning ** in this context) > > my solution: > > >>> s = 0 > >>> for i in range(1, 2011): > ... s += i**2 > ... if not (i+1)%5: > ... s -= 2*i**2 > ... if not i%5: > ... s -= 2*i**2 Pretty ugly, if you ask me. What about this: s = 0 for i in range(1, 2011): if i%5 in (1,2,3): s += i**2 else: s -= i**2 Here's the obvious one-liner: s = sum(i**2 if i%5 in (1,2,3) else -i**2 for i in range(1, 2011)) Runs in ~600 usecs for me according to timeit - pretty fast. Stefan  |  Next  |  Last Pages: 1 2 3 4 5 6 7 8 Prev: [ANN] Robot Framework 2.5Next: logging = logging.getLogger(__name__)