[F] Fix compilation error
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@@ -104,6 +104,7 @@ By definition of $g \in \cO(f)$, we know $g(n) \leq c_0 \cdot f(n)$.
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g(n) &\leq c_0 \cdot (\floor{f(n)} + \floor{f(n)}) \quad{\text{because } 1 \leq \floor{f(n)}} \\
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g(n) &\leq 2c_0 \cdot \floor{f(n)}\\
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g(n) &\leq c_1 \cdot \floor{f(n)}
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\end{align*}
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\end{proof}
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\end{enumerate}
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