II. (This question is worth 40 points) Let integers $a_{1}, a_{2}, \cdots, a_{2011}$ satisfy $1=a_{1} \leq a_{2} \leq \cdots \leq a_{2019}=99$. Let $f=\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{2019}^{2}\right)-\left(a_{1} a_{3}+a_{2} a_{4}+a_{3} a_{5}+\cdots+a_{2017} a_{2019}\right)$.
Find the minimum value $f_{0}$ of $f$. And determine the number of arrays $\left(a_{1}, a_{2}, \cdots, a_{2019}\right)$ that make $f=f_{0}$.
Solution: From the given conditions, we have
$$
2 f=a_{1}^{2}+a_{2}^{2}+a_{2018}^{2}+a_{2019}^{2}+\sum_{i=1}^{2017}\left(a_{i+2}-a_{i}\right)^{2} .
$$

Since $a_{1}, a_{2}$ and $a_{i+2}-a_{i}(i=1,2, \cdots, 2016)$ are all non-negative integers, we have $a_{1}^{2} \geq a_{1}, a_{2}^{2} \geq a_{2}$, and $\left(a_{i+2}-a_{i}\right)^{2} \geq a_{i+2}-a_{i}(i=1,2, \cdots, 2016)$. Therefore,
$$
a_{1}^{2}+a_{2}^{2}+\sum_{i=1}^{2016}\left(a_{i+2}-a_{i}\right)^{2} \geq a_{1}+a_{2}+\sum_{i=1}^{2016}\left(a_{i+2}-a_{i}\right)=a_{2017}+a_{2018} .
$$
$\qquad$
From (1) and (2), we get
$$
2 f \geq a_{2017}+a_{2018}+\left(a_{2019}-a_{2017}\right)^{2}+a_{2018}^{2}+a_{2019}^{2},
$$

Combining $a_{2019}=99$ and $a_{2018} \geq a_{2017}>0$, we know
$$
\begin{aligned}
f & \geq \frac{1}{2}\left(2 a_{2017}+\left(99-a_{2017}\right)^{2}+a_{2017}^{2}+99^{2}\right) \\
& =\left(a_{2017}-49\right)^{2}+7400 \geq 7400 .
\end{aligned}
$$
20 points
On the other hand, let
$$
a_{1}=a_{2}=\cdots=a_{1920}=1, a_{1920+2 k-1}=a_{1920+2 k}=k(k=1,2, \cdots, 49), a_{2019}=99 \text {, }
$$

At this point, it is verified that all the inequalities above are equal, thus the minimum value of $f$ is $f_{0}=7400$.
$\qquad$
Next, consider the equality conditions of (3). At this point, $a_{2017}=a_{2018}=49$, and all inequalities in (2) are equal, i.e., $a_{1}=a_{2}=1$, and $a_{i+2}-a_{i} \in\{0,1\}(i=1,2, \cdots, 2016)$.

Therefore, $1=a_{1} \leq a_{2} \leq \cdots \leq a_{2018}=49$, and for each $k(1 \leq k \leq 49)$, there are at least two terms in $a_{1}, a_{2}, \cdots, a_{2018}$ that are equal to $k$. It is easy to verify that this is also a sufficient condition for (3) to be equal.

For each $k(1 \leq k \leq 49)$, let the number of terms in $a_{1}, a_{2}, \cdots, a_{2018}$ that are equal to $k$ be $1+n_{k}$, then $n_{k}$ is a positive integer, and $\left(1+n_{1}\right)+\left(1+n_{2}\right)+\cdots+\left(1+n_{49}\right)=2018$, i.e.,
$$
n_{1}+n_{2}+\cdots+n_{49}=1969 \text {. }
$$

The number of positive integer solutions $\left(n_{1}, n_{2}, \cdots, n_{49}\right)$ of this equation is $\mathrm{C}_{1968}^{48}$, and each solution uniquely corresponds to an array $\left(a_{1}, a_{2}, \cdots, a_{2019}\right)$ that makes (4) equal. Therefore, the number of arrays $\left(a_{1}, a_{2}, \cdots, a_{2019}\right)$ that make $f=f_{0}$ is $\mathrm{C}_{1968}^{48}$.
.40 points