5. Two players, A and B, are playing a game on a $2^{100} \times 100$ grid. They take turns filling in symbols in the empty cells of the first row, with A starting first. In each turn, A selects an empty cell in the first row and fills it with “$\times$”, then B selects an empty cell in the first row and fills it with “○”. Once all cells in the first row are filled, they start filling in the empty cells of the second row, and so on, until all cells are filled.

A's goal is to maximize the number of distinct rows in the grid, while B's goal is to minimize the number of distinct rows. If both players use their best strategies, how many distinct rows will there be in the grid?
$5.2^{50}$.
First prove that player B has a strategy to ensure that there are at most $2^{50}$ different rows in the grid.

Each row has 100 cells, which can be evenly divided into 50 $1 \times 2$ rectangles. After player A makes a move, player B selects the other cell in the same $1 \times 2$ rectangle and fills it with an O. Thus, there will be at most $2^{50}$ different rows in the grid.

Next, prove that player A has a strategy to ensure that there are at least $2^{50}$ different rows in the grid.

Player A can make an arbitrary move in the first row. Assume the first $k (k \geqslant 1)$ rows have been filled. Let there be $m$ different rows among the first $k$ rows, and $m < 2^{50}$. These $m$ different rows are called "good rows".
Now consider the $(k+1)$-th row.
The $m$ good rows contain a total of $50m$ $\times$ symbols. There must be a column with at least $\frac{50m}{100} = \frac{m}{2}$ $\times$ symbols. Player A fills the cell at the intersection of this column and the $(k+1)$-th row with a $\times$. Suppose in the $(k+1)$-th row, $i (1 \leqslant i < 50)$ $\times$ symbols and $i$ O symbols have already been filled. Let $x_i$ be the number of good rows among the $m$ good rows that contain $i$ $\times$ symbols and $i$ O symbols in the same columns as the $i$ $\times$ symbols and $i$ O symbols already filled. For these $x_i$ good rows, excluding the $i$ $\times$ symbols and $i$ O symbols in the columns, they form a grid of $x_i$ rows and $100 - 2i$ columns, containing $(50 - i)x_i$ $\times$ symbols. There must be a column with at least $\frac{(50 - i)x_i}{100 - 2i} = \frac{x_i}{2}$ $\times$ symbols. Player A fills the cell at the intersection of this column and the $(k+1)$-th row with a $\times$ (this is the $(i+1)$-th $\times$ in the $(k+1)$-th row). Player A continues to fill the $(k+1)$-th row in this manner.
By $x_1 \leqslant \frac{m}{2}, x_{i+1} \leqslant \frac{x_i}{2} (i = 1, 2, \cdots, 49)$, we have $x_{50} \leqslant \frac{x_1}{2^{49}} \leqslant \frac{m}{2^{50}} < 1$.
Thus, the $(k+1)$-th row is different from the $m$ good rows.
By following the above strategy, player A can ensure that there are at least $2^{50}$ different rows in the grid.
In conclusion, the answer is $2^{50}$.