1. The function $f(x)$ satisfies $f(x+3)=-\frac{1}{f(x)}$ for any real number $x$.
If $f(0)=2$, then $f(2013)=(\quad)$.
(A) $-\frac{1}{2}$
(B) $\frac{1}{2}$
(C) 2
(D) 2013
-1. A.
From the problem, we know
$$
f(x+6)=-\frac{1}{f(x+3)}=f(x) \text {. }
$$

Thus, $f(x)$ is a periodic function with a period of 6. Therefore, $f(2013)=f(3)=-\frac{1}{f(0)}=-\frac{1}{2}$.