Raffael Cavallaro <raffaelcavall...@pas.despam.s.il.vous.plait.mac.com> writes:
> CL-USER 119 > (defun fib (n)
> (/ (loop for k from 1 to n by 2
> sum (/ (* (expt 5 (/ (1- k) 2)) (fact n))
> (fact k) (fact (- n k))))
> (expt 2 (1- n))))
> CL-USER 122 > (time (fib 1000))
> Timing the evaluation of (FIB 1000)
> 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875
I have no idea what that fancy algorithm is doing, but the number it
prints appears to be the 2000th Fibonacci number rather than the 1000th.
And it took 0.76 seconds even compiled. This Haskell code took no
perceptible time even in the ghci interpreter:
main = print (fib 0 1 !! 2000)
where
fib a b = a:(a+b):fib b (a+b)
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