$7 \cdot 100$ A square piece of paper is cut into two parts along a straight line, one of which is then cut into two parts along a straight line, and then one of the 3 pieces is cut into two parts, and so on. To obtain 73 polygons with 30 sides, what is the minimum number of cuts needed?
[Solution] Each cut increases the number of polygons by 1. Therefore, after $k$ cuts, $k+1$ polygons are obtained. On the other hand, each cut increases the sum of the interior angles of the polygons by $2 \pi$. After $k$ cuts, the sum of the interior angles of the $k+1$ polygons is $2 \pi(k+1)$.

Among all the 30-sided polygons, the sum of the interior angles is $73 \cdot 28 \pi$. The number of the remaining polygons is $k+1-73=k-72$, and their sum of interior angles is no less than $(k-72) \pi$. This leads to
$$
73 \cdot 28 \pi + (k-72) \pi \leqslant (k+1) 2 \pi,
$$

Solving this, we get $k \geqslant 73 \cdot 27 - 1 = 1970$, meaning at least 1970 cuts are required.
These 1970 cuts can be made as follows: first, use 72 cuts to divide the square into 73 rectangles. Then, use the corner-cutting method to cut each rectangle 26 times to get a 30-sided polygon, resulting in a total of $72 + 73 \times 26 = 1970$ cuts to obtain 73 30-sided polygons.
In summary, at least 1970 cuts are needed.