1. In the arithmetic sequence $\left\{a_{n}\right\}$, $S_{n}$ is the sum of the first $n$ terms of $a_{n}$, $S_{4}=1, S_{8}=4$. Then
$$
a_{2012}+a_{2003}+a_{2004}+a_{2005}
$$

is ( ).
(A) $\frac{2003}{2}$
(B) $\frac{2005}{2}$
(C) $\frac{2007}{2}$
(D) $\frac{2001}{2}$
$$
\begin{array}{l}
-1 . \mathrm{A} . \\
S_{8}-2 S_{4} \\
=\left(a_{8}-a_{4}\right)+\left(a_{1}-a_{3}\right)+\left(a_{6}-a_{2}\right)+\left(a_{5}-a_{1}\right) \\
=16 d=2 .
\end{array}
$$
$$
\begin{array}{l}
\text { and } a_{2002}+a_{2003}+a_{2004}+a_{2005}-S_{4} \\
=4 \times 2001 d=\frac{2001}{2} .
\end{array}
$$

Therefore, $a_{2002}+a_{2003}+a_{2004}+a_{2005}=\frac{2003}{2}$.