1. Given a function $f(x)$ defined on $\mathbf{R}$ whose graph is centrally symmetric about the point $\left(-\frac{3}{4}, 0\right)$, and satisfies $f(x)=-f\left(x+\frac{3}{2}\right), f(-1)=1, f(0)=-2$. Then, the value of $f(1)+f(2)+\cdots+f(2006)$ is ( ).
(A) 1
(B) 2
(C) -1
(D) -2
$-1 . B$.
From the graph of the function $f(x)$ being centrally symmetric about the point $\left(-\frac{3}{4}, 0\right)$, we know that $f(x)=-f\left(-x-\frac{3}{2}\right)$.
Also, $f(x)=-f\left(x+\frac{3}{2}\right)$, so $f\left(-x-\frac{3}{2}\right)=f\left(x+\frac{3}{2}\right)$.
Therefore, $f(x+3)=f\left(\left(x+\frac{3}{2}\right)+\frac{3}{2}\right)$
$$
=-f\left(x+\frac{3}{2}\right)=f(x) \text {. }
$$

Thus, $f(x)$ is an even function with a period of 3.
Hence, $f(1)=f(-1)=1$,
$$
\begin{array}{l}
f(2)=f(-1+3)=f(-1)=1, \\
f(3)=f(0)=-2 .
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore, } f(1)+f(2)+\cdots+f(2006) \\
=668(f(1)+f(2)+f(3))+f(2005)+f(2006) \\
=f(2005)+f(2006)=f(1)+f(2)=2 .
\end{array}
$$