One. (This question is worth 40 points) Let real numbers $a_{1}, a_{2}, \cdots, a_{2016}$ satisfy $9 a_{i}>11 a_{i+1}^{2}(i=1,2, \cdots, 2015)$. Find the maximum value of $\left(a_{1}-a_{2}^{2}\right) \cdot\left(a_{2}-a_{3}^{2}\right) \cdots \cdot\left(a_{2015}-a_{2016}^{2}\right) \cdot\left(a_{2016}-a_{1}^{2}\right)$.
Let $P=\left(a_{1}-a_{2}^{2}\right) \cdot\left(a_{2}-a_{3}^{2}\right) \cdots \cdots\left(a_{2015}-a_{2016}^{2}\right) \cdot\left(a_{2016}-a_{1}^{2}\right)$.
From the given, for $i=1,2, \cdots, 2015$, we have $a_{i}-a_{i+1}^{2}>\frac{11}{9} a_{i+1}^{2}-a_{i+1}^{2} \geq 0$.

If $a_{2016}-a_{1}^{2} \leq 0$, then $S \leq 0$.
10 points
Now consider the case where $a_{2016}-a_{1}^{2}>0$. Let $a_{2017}=a_{1}$. By the AM-GM inequality, we have
$$
\begin{array}{l}
P^{\frac{1}{2016}} \leq \frac{1}{2016} \sum_{i=1}^{2016}\left(a_{i}-a_{i+1}^{2}\right)=\frac{1}{2016}\left(\sum_{i=1}^{2016} a_{i}-\sum_{i=1}^{2016} a_{i+1}^{2}\right) \\
=\frac{1}{2016}\left(\sum_{i=1}^{2016} a_{i}-\sum_{i=1}^{2016} a_{i}^{2}\right)=\frac{1}{2016} \sum_{i=1}^{2016} a_{i}\left(1-a_{i}\right) \\
\leq \frac{1}{2016} \sum_{i=1}^{2016}\left(\frac{a_{i}+\left(1-a_{i}\right)}{2}\right)^{2}=\frac{1}{2016} \cdot 2016 \cdot \frac{1}{4}=\frac{1}{4} \text {, } \\
P \leq \frac{1}{4^{2016}} \text {. } \\
\end{array}
$$

Therefore,
when $a_{1}=a_{2}=\cdots=a_{2016}=\frac{1}{2}$, the equality holds, and we have $9 a_{i}>11 a_{i+1}^{2}$ $(i=1,2, \cdots, 2015)$, at this point $P=\frac{1}{4^{2016}}$.
In conclusion, the maximum value sought is $\frac{1}{4^{2016}}$.
40 points