Example 2: There are 1988 unit cubes, and they (all or part) are used to form three right square prisms $A$, $B$, and $C$ with heights of 1 and base side lengths of $a$, $b$, and $c$ ($a<b<c$). Now, place $A$, $B$, and $C$ in the first quadrant, with each base parallel to the coordinate axes, one vertex of $C$ at the origin, $B$ on top of $C$, and every unit cube of $B$ on top of a unit cube of $C$, but the boundaries of $B$ do not align with any boundaries of $C$. Similarly, $A$ is on top of $B$, and every unit cube of $A$ is on top of a unit cube of $B$, but the boundaries of $A$ do not align with any boundaries of $B$. This forms a three-story building. Question: For what values of $a$, $b$, and $c$ can the number of three-story buildings formed be maximized? (11th Austrian-Polish Mathematics Competition)
Analysis and Solution: From "boundary misalignment", we have $a \leqslant b-2 \leqslant c-4$. Thus, placing $A$ on $B$ has $(b-a-1)^{2}$ ways, and placing $B$ on $C$ has $(c-b-1)^{2}$ ways. Therefore, there are $P=(b-a-1)^{2}(c-b-1)^{2}$ different three-story buildings. Thus, the problem is equivalent to: for all positive integers $a, b, c$ satisfying $1 \leqslant a \leqslant b-2 \leqslant c-4, a^{2}+b^{2}+c^{2} \leqslant 1988$, find the maximum value of $P=(b-a-1)^{2}(c-b-1)^{2}$.

Clearly, $P \leqslant(b-2)^{2}(c-b-1)^{2}$, with equality when $a=1$. Therefore, we only need to find the maximum value of $Q=(b-2)(c-b-1)$ for all positive integers $b, c$ satisfying $3 \leqslant b \leqslant c-2, b^{2}+c^{2} \leqslant 1987$.

It is easy to see that
$$Q \leqslant \frac{[(b-2)+(c-b-1)]^{2}}{4}=\frac{(c-3)^{2}}{4}$$

Thus, we only need to determine the range of $c$ to find the maximum value of $\frac{(c-3)^{2}}{4}$. From the condition, we have $c^{2} \leqslant 1987-b^{2} \leqslant 1987, c \leqslant 44$, so $Q \leqslant \frac{(44-3)^{2}}{4}<421, Q \leqslant 420$.

Unfortunately, equality cannot hold, so we need to find the maximum value by considering successive maxima.
Fixing $c$, we have
$$\begin{aligned}
Q & =-b^{2}+(1+c) b+2-2 c \\
& =-\left(b-\frac{1+c}{2}\right)^{2}+\frac{(1+c)^{2}}{4}+2-2 c
\end{aligned}$$

From the graph of the quadratic function, we know that $Q \leqslant \frac{(1+c)^{2}}{4}+2-2 c=A(c)$ (when $c$ is odd, $b=\frac{1+c}{2}$); or $Q \leqslant \frac{(1+c)^{2}}{4}-\frac{1}{4}+2-2 c=B(c)$ (when $c$ is even, $b=\frac{c}{2}$).

Now, we need to find the maximum values of $A(c)$ and $B(c)$.
From $b^{2}+c^{2} \leqslant 1987$, we have $c \leqslant 44$. Thus, when $c$ is odd,
$$Q \leqslant A(c) \leqslant A(43)=\frac{(1+43)^{2}}{4}+2-86=400.$$

When $c$ is even,
$$Q \leqslant B(c) \leqslant B(44)=\frac{(1+44)^{2}}{4}-\frac{1}{4}+2-88=420.$$

Although $Q \leqslant 420$ always holds, equality cannot be achieved. In fact, for equality in (2), we need $c=44$ and $b=\frac{c}{2}=22$. However, $(44,22)$ does not satisfy $b^{2}+c^{2} \leqslant 1987$. Therefore, we cannot uniformly fix $c$ to find the maximum value of $Q=(b-2)(c-b-1)$; we need to classify the values of $c$ and obtain different forms of the maximum function.

When $c=44$, $Q=-b^{2}+45 b-86$. Since $b^{2} \leqslant 1987-c^{2}=51, 3 \leqslant b \leqslant 7$, so when $b=7$, $Q$ reaches its maximum value of 180;

When $c=43$, $3 \leqslant b \leqslant 11$, at this time, $Q \leqslant 279$, equality holds when $b=11$;
When $c=42$, $3 \leqslant b \leqslant 14$, at this time, $Q \leqslant 324$, equality holds when $b=14$;
When $c=41$, $3 \leqslant b \leqslant 17$, at this time, $Q \leqslant 345$, equality holds when $b=17$;
When $c=40$, $3 \leqslant b \leqslant 19$, at this time, $Q \leqslant 340$, equality holds when $b=19$;
When $c \leqslant 39$, $Q \leqslant \frac{[(b-2)+(c-b-1)]^{2}}{4}=\frac{(c-3)^{2}}{4} \leqslant 18^{2}=324$.
From the above, we know that $Q \leqslant 345$ always holds. And when $(a, b, c)=(1,17,41)$, equality holds. Therefore, when $(a, b, c)=(1,17,41)$, $P$ reaches its maximum value of $345^{2}$.