Example 5 There are 1988 unit cubes, and they (all or part of them) are arranged into 3 "squares" (i.e., 3 one-layer rectangular prisms with dimensions $a \times a \times 1, b \times b \times 1, c \times c \times 1$ where $a \leqslant b \leqslant c$) $A, B$, and $C$. Now, place square $C$ in the first quadrant of the $x O y$ plane, with one of its vertices at the origin, then place $B$ on $C$ so that each small block of $B$ is exactly on a small block of $C$, with the sides of $B$ parallel to the corresponding sides of $C$, but the perimeter of $B$ does not align with the perimeter of $C$; then place $A$ in the same manner on $B$, thus forming a "three-story building". For what values of $a, b$, and $c$ will the number of different forms of the three-story building relative to the origin be maximized?
Solution: According to the problem, the condition for $B$ to be placed on $C$ is $b \leqslant c - 2$, and there are $(c-b-1)^{2}$ different ways to place it; the condition for $A$ to be placed on $B$ is $a \leqslant b-2$, and there are $(b-a-1)^{2}$ different ways to place it. Thus, the problem is transformed into:
Finding $a, b, c$ that satisfy the conditions
$$
1 \leqslant a \leqslant b-2 \leqslant c-4 \left(a, b, c \in \mathbf{N}_{+}\right)
$$

and $a^{2}+b^{2}+c^{2} \leqslant 1988$, such that the function
$$
f=(b-a-1)^{2}(c-b-1)^{2}
$$

reaches its maximum value.
Obviously, when $a=1$, $f$ reaches its maximum value.
Thus, the problem is further transformed into:
Under the conditions $3 \leqslant b \leqslant c-2$ and $b^{2}+c^{2} \leqslant 1987$, find the maximum value of $g=(b-2)^{2}(c-b-1)^{2}$, or the maximum value of $p=(b-2)(c-b-1)$.
When $c$ is fixed, since
$$
(b-2)+(c-b-1)=c-3,
$$

$p$ is maximized when $|(b-2)-(c-b-1)|=|c-2b+1|$ is minimized (refer to Example 2, Solution 2). 

Since $c^{2} \leqslant 1987-b^{2} \leqslant 1987-9=1978$, it follows that $c \leqslant 44$.
We will discuss different values of $c$:
(1) If $c=44$, then
$$
b^{2} \leqslant 1987-44^{2}=51, 3 \leqslant b \leqslant 7.
$$

At this time, $p=(b-2)(c-b-1)=(43-b)(b-2)$ reaches its maximum value 180 when $b=7$ (at this time $|c-2b+1|$ is minimized).
(2) If $c=43$, then $3 \leqslant b \leqslant 11$. At this time, $p$ reaches its maximum value 279 when $b=11$.
(3) If $c=42$, then $3 \leqslant b \leqslant 14$. At this time, $p$ reaches its maximum value 324 when $b=14$.
(4) If $c=41$, then $3 \leqslant b \leqslant 17$. At this time, $p$ reaches its maximum value 345 when $b=17$.
(5) If $c=40$, then $3 \leqslant b \leqslant 19$. At this time, $p$ reaches its maximum value 340 when $b=19$.
(6) If $c \leqslant 39$, then
$$
\begin{aligned}
p & =(b-2)(c-b-1) \\
& \leqslant\left(\frac{c-3}{2}\right)^{2} \leqslant 18^{2}=324.
\end{aligned}
$$

In summary, when $a=1, b=17, c=41$, the number of different forms of the "three-story building" is the maximum, which is 345.