16. In $\triangle A B C$, the three interior angles are $\angle A$, $\angle B$, and $\angle C$, and $2 \angle C - \angle B = 180^{\circ}$. Additionally, the ratio of the perimeter of $\triangle A B C$ to its longest side is $m$. Therefore, the maximum value of $m$ is
16. $\frac{9}{4}$.

From $2 \angle C-\angle B=180^{\circ}$, we know $120^{\circ}>\angle C>90^{\circ}$, so $\angle C$ is the largest. Its longest side is $c$, and we have $-\frac{1}{2}<\cos C<0$.
$$
\begin{array}{l}
\text { Also, } \angle A=180^{\circ}-\angle B-\angle C=360^{\circ}-3 \angle C, \text { then } \\
m=\frac{a+b+c}{c}=\frac{-\sin 3 C-\sin 2 C+\sin C}{\sin C} \\
=4 \sin ^{2} C-2 \cos C-2 \\
=-4\left(\cos C+\frac{1}{4}\right)^{2}+\frac{9}{4} .
\end{array}
$$

When $\cos C=-\frac{1}{4}$, $m$ has the maximum value $\frac{9}{4}$.