Обсуждение @ 1081. Двоичная последовательность @ Timus Online Judge

AC with this formula

Послано Sergey 9 окт 2011 01:01

I've used this formula to solve this problem:
a(n) = 2^(b(n)-1) + a(n - c(1+b(n)))
b(n) = -1+floor(log(((n+0.2)*sqrt(5)))/log((1+sqrt(5))/2));
c(0) = 0, c(1) = 1, c(n) = c(n-1) + c(n-2)

Re: AC with this formula

Послано daftcoder [Yaroslavl SU] 16 окт 2011 17:46

INSANE!

Re: AC with this formula

Послано htzfun 1 май 2012 17:44

I don't want to live in this world anymore.

Re: AC with this formula

Послано Md. Taufique Hussain 3 июн 2013 13:32

wow!!!! you probably made an easy thing too complicated.

Re: AC with this formula

Послано Shubh Srivastava 24 май 2018 00:24

Looks like you're using golden ratio to calculate Fibonacci

Re: AC with this formula

Послано Hristo Nikolaev (B&W) 18 апр 2023 17:54

Way too complicated. I solved entirely using bitwise operations and an array with the first 64 Fibonacci numbers.

Обсуждение задачи 1081. Двоичная последовательность