In case we have an algorithm which is pseudo-polynomial and runs in $O(n^2C)$ for some $C$ that is encoded in binary. Is it correct to say that if $C=2^n$ then $O(n^2C)=O(n^22^n)$ and because $n=\log C$ we can write that $O((\log C)^22^n)=O(2^n)$ because $2^n$ grows more rapidly than $\log C$?

没有正确的解决方案

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