1. Let the function $f(x)=\log _{a} x(a>0, a \neq 1)$. If $f\left(x_{1} x_{2} \cdots x_{2003}\right)=8$, then the value of $f\left(x_{1}^{2}\right)+f\left(x_{2}^{2}\right)+\cdots+$ $f\left(x_{2003}^{2}\right)$ is ( ).
A. 4
B. 8
C. 16
D. $2 \log _{a} 8$
\begin{array}{l}\text { 1. } f\left(x_{1}^{2}\right)+f\left(x_{2}^{2}\right)+\cdots+f\left(x_{2003}^{2}\right)=\log a x_{1}^{2}+\log _{2} x_{2}^{2}+\cdots+\log _{a} x_{2003}^{2}=2\left(\log a x_{1}+\log _{a} x_{2}+\cdots+\right. \\ \left.\log _{a} x_{2003}\right)=2 \log _{a} x_{1} x_{2} \cdots x_{2003}=2 f\left(x_{1} x_{2} \cdots x_{2003}\right)=16 \text {. Therefore, the answer is } \mathrm{C} \text {. }\end{array}