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W b s j s n n . S n N n ϕ(n) = p 1 (1− p−1)−s −1 p (5) and thus ∞ 0 B (x)xs−1dx =0− 1 s ∞ 0 xs dB(x W(s) s s)>0 (6) Hence W(s) s is the Mellin transform of B(t)Themethodusedin4 to establish (4) is essentially that of inversion of the Mellin transform with the abscissa of integration moved to −σForlarget. N g u n g u n d # m i e t e n w a h n s i n n e i n k i e z i n a k t i o n.
O p j n n n . It's only (n 1) * n if you use a naive bubblesort You can get a significant savings if you notice the following After each compareandswap, the largest element you've encountered will be in the last spot you were at. I s q ` J ( = .
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O v j n n n . Q \# C b ì. 0, thus 2s2 = s2 A ⇒ s = √A share Share a link to this answer Copy link.
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