8. In the figure shown in Figure 3, on both sides of square $P$, there are $a$ and $b$ squares to the left and right, and $c$ and $d$ squares above and below, where $a$, $b$, $c$, and $d$ are positive integers, satisfying
$$
(a-b)(c-d)=0 \text {. }
$$

The shape formed by these squares is called a "cross star".
There is a grid table consisting of 2014 squares, forming a $38 \times 53$ grid. Find the number of cross stars in this grid table.
(Tao Pingsheng, provided)
8. For a cross, the cell $P$ referred to in the problem is called the "center block" of the cross.

When $a=b$, the cross is called "standing"; when $c=d$, it is called "lying" (some crosses are both standing and lying).

If the union of a row and a column of a rectangle $R$ is exactly a cross $S$, then $R$ is called the "source rectangle" of $S$. In the source rectangle $R$, the cross $S$ is uniquely determined by the position of the center block, and the crosses corresponding to different source rectangles must be different. Moreover, by the definition of the cross, the number of rows and columns of the source rectangle is at least 3; a source rectangle with an even number of rows does not contain a lying cross; a source rectangle with an even number of columns does not contain a standing cross.

Now consider the $m \times n$ grid $P_{m, n}$. The number of source rectangles of size $(2k+1) \times l$ is $(m-2k)(n-l+1)$, where $k, l$ satisfy
$$
3 \leqslant 2k+1 \leqslant m, 3 \leqslant l \leqslant n.
$$

For each $(2k+1) \times l$ source rectangle, the center block of a lying cross must be in the $(k+1)$-th row and in the 2nd to $(l-1)$-th columns, so this source rectangle corresponds to $l-2$ lying crosses.
$$
\begin{array}{l}
\text { Hence } A=\sum_{k=1}^{\left[\frac{m-1}{2}\right]} \sum_{l=3}^{n}(l-2)(m-2k)(n-l+1) \\
=\left[\sum_{k=1}^{\left[\frac{m-1}{2}\right]}(m-2k)\right]\left[\sum_{i=1}^{n-2} l(n-l-1)\right], \\
\text { where, } \sum_{k=1}^{\left[\frac{m-1}{2}\right]}(m-2k) \\
=\left[\frac{m-1}{2}\right] \frac{(m-2)+\left(m-2\left[\frac{m-1}{2}\right]\right)}{2} \\
\quad=\left[\frac{m-1}{2}\right]\left(m-\left[\frac{m+1}{2}\right]\right), \\
\sum_{i=1}^{n-2} l(n-l-1)=\left(\sum_{l=1}^{n-2} l\right)(n-1)-\sum_{l=1}^{n-2} l^{2} \\
=\frac{(n-2)(n-1)^{2}}{2}-\frac{(n-2)(n-1)(2n-3)}{6} \\
=\frac{n(n-1)(n-2)}{6} .
\end{array}
$$

Thus, $A=\left[\frac{m-1}{2}\right]\left(n-\left[\frac{m+1}{2}\right]\right) \frac{n(n-1)(n-2)}{6}$.
Similarly, the number of source rectangles of size $k \times (2l+1)$ in the grid $P_{m, n}$ is $(m-k+1)(n-2l)$, and each $k \times (2l+1)$ source rectangle corresponds to $k-2$ standing crosses. Therefore,
$$
\begin{aligned}
B & =\sum_{k=3}^{m}\left[\sum_{k=1}^{\left.n-\frac{n-1}{2}\right]}(k-2)(m-k+1)(n-2l)\right. \\
& =\left[\frac{n-1}{2}\right]\left(n-\left[\frac{n+1}{2}\right]\right) \frac{m(m-1)(m-2)}{6} .
\end{aligned}
$$

Each $(2k+1) \times (2l+1)$ source rectangle corresponds to exactly one cross that is both standing and lying (the center block is at the center of the source rectangle), and the grid $P_{m, n}$ contains $(m-2k)(n-2l)$ such source rectangles, so similarly we get
$$
\begin{aligned}
C & \left.=\sum_{k=1}^{\left[\frac{m-1}{2}\right]}\right]\left[\frac{n-1}{2}\right] \\
& =\left[\frac{m-1}{2}\right]\left(m-\left[\frac{m+1}{2}\right]\right)\left[\frac{n-1}{2}\right]\left(n-\left[\frac{n+1}{2}\right]\right), \text { (3) }
\end{aligned}
$$

which is the number of crosses that are counted twice in $A+B$.
In particular, for the case $m=38, n=53$, we have
$$
\begin{array}{l}
{\left[\frac{m-1}{2}\right]\left(m-\left[\frac{m+1}{2}\right]\right)=18 \times 19=342,} \\
\frac{m(m-1)(m-2)}{6}=8436, \\
{\left[\frac{n-1}{2}\right]\left(n-\left[\frac{n+1}{2}\right]\right)=26 \times 26=676,} \\
\frac{n(n-1)(n-2)}{6}=23426 .
\end{array}
$$

Substituting into equations (1), (2), and (3), we get
$$
\begin{array}{l}
A=342 \times 23426=8011692, \\
B=676 \times 8436=5702736, \\
C=342 \times 676=231192 .
\end{array}
$$

Thus, the number of crosses in the grid $P_{38,53}$ is
$$
A+B-C=13483236
$$