January 2019 1 11 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)
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