1. Given $\triangle A B C$ is an acute triangle,
$$
\begin{array}{l}
f(x)=-x^{2}+2 x+m, \\
p=\sin A+\sin B+\sin C, \\
q=\cos A+\cos B+\cos C .
\end{array}
$$

Then ( ).
(A) $f(p)>f(q)$
(B) $f(p)=f(q)$
(C) $f(p)<f(q)$
(D) Any of the above three cases is possible
$-、 1 . C$.
From $\triangle A B C$ being an acute triangle, we get
$$
\angle A+\angle B>\frac{\pi}{2} \Rightarrow \angle A>\frac{\pi}{2}-\angle B>0 \text {. }
$$

Thus, $\sin A>\cos B$.
Similarly, $\sin B>\cos C, \sin C>\cos A$.
Therefore, $p>q$.
Since $\cos A+\cos B+\cos C$
$$
=1+4 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}>1 \text {, }
$$

we have $p>q>1$.
Since $f(x)$ is a decreasing function in $(1,+\infty)$, therefore, $f(p)<f(q)$.