16. Let
$$
f(x)=x^{3}+a x^{2}+b x+c(1<a<b<c) \text {, }
$$
$g(x)$ is a polynomial with the leading coefficient 1, and the three roots of $g(x)$ are the reciprocals of the three roots of $f(x)$. Then $g(1)$ expressed as an algebraic expression in terms of $a$, $b$, and $c$ is ( ).
(A) $\frac{1+a+b+c}{c}$
(B) $1+a+b+c$
(C) $\frac{1+a+b+c}{c^{2}}$
( D) $\frac{a+b+c}{c^{2}}$
(E) $\frac{1+a+b+c}{a+b+c}$
16. A.

Let the three roots of $f(x)$ be $p$, $q$, and $r$. Then
$$
f(x)=(x-p)(x-q)(x-r) \text {. }
$$

According to the problem,
$$
\begin{array}{l}
g(x)=\left(x-\frac{1}{p}\right)\left(x-\frac{1}{q}\right)\left(x-\frac{1}{r}\right) . \\
\text { Therefore, } g(1)=\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)\left(1-\frac{1}{r}\right) \\
=\frac{(p-1)(q-1)(r-1)}{p q r} \\
=\frac{(1-p)(1-q)(1-r)}{-p q r} \\
=\frac{f(1)}{f(0)}=\frac{1+a+b+c}{c} .
\end{array}
$$