January 2019 1 16 Report
Czy dobrze rozwiązałam?
T(1) = 1, T(n) = 2T( \frac{n}{4}) + 2n --\ \textgreater \ O(n^{log_{4}8} )
ponieważ a = 2, b = 4, f(n) = 2n. Zatem a = 2 < f(b) = 8 i podstawiam do "wzorów" dla rekurencji o postaci T(1) = 1, T(n) = aT(n/b) + f(n)
More Questions From This User See All

Recommend Questions



Life Enjoy

" Life is not a problem to be solved but a reality to be experienced! "

Get in touch

Social

© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.