If g(n) = sqrt(n)sqrt(n), does the complexity of g(n) = O(2n)?
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比较两个指数函数时的一个有用技巧是让它们具有相同的底数：
√n√n = (2lg √n)√n = 2√n lg √n
Now you're comparing 2√n lg √n against 2n, and hopefully from that it's easy to see that the former function does not grow as rapidly as the latter, so √n√n = O(2n) is indeed true.
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