10. The sequence $a_{0}, a_{1}, \cdots, a_{n}$ satisfies
$$
a_{0}=\sqrt{3}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}} \text {, }
$$

where, $[x]$ denotes the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then $a_{2016}=$ $\qquad$
10. $3024+\sqrt{3}$.

From the given, we have
$$
\begin{array}{l}
a_{0}=1+(\sqrt{3}-1), \\
a_{1}=1+\frac{1}{\sqrt{3}-1}=1+\frac{\sqrt{3}+1}{2}=2+\frac{\sqrt{3}-1}{2}, \\
a_{2}=2+\frac{2}{\sqrt{3}-1}=2+(\sqrt{3}+1)=4+(\sqrt{3}-1), \\
a_{3}=4+\frac{1}{\sqrt{3}-1}=4+\frac{\sqrt{3}+1}{2}=5+\frac{\sqrt{3}-1}{2} .
\end{array}
$$

Therefore, $a_{2 k}=3 k+1+(\sqrt{3}-1)$,
$$
a_{2 k+1}=3 k+2+\frac{\sqrt{3}-1}{2} \text { . }
$$

Thus, $a_{2016}=3024+\sqrt{3}$.