## A square piece of paper with integer dimensions $N \times N$ is placed with a corner at the origin and two of its sides along the $x$- and $y$-axes. Then, we cut it up respecting the following rules:

We only make straight cuts between two points lying on different sides of the square, and having integer coordinates.

Two cuts cannot cross, but several cuts can meet at the same border point.

Proceed until no more legal cuts can be made.

Counting any reflections or rotations as distinct, we call $C(N)$ the number of ways to cut an $N \times N$ square. For example, $C(1) = 2$ and $C(2) = 30$ (shown below).

What is $C(30) \bmod 10^8$?

### To solve this problem, we can use dynamic programming to calculate the number of ways to cut an $N \times N$ square. We will define a function $C(N)$ that calculates the number of ways to cut a square with side length $N$.

Here’s the Python code to calculate $C(N)$:

“`python

def calculate_C(N):

# Initialize the dynamic programming table

dp = [[0] * (N + 1) for _ in range(N + 1)]

# Base case: there are 2 ways to cut a 1×1 square

dp[1][1] = 2

# Calculate the number of ways to cut the square

for i in range(2, N + 1):

# Calculate the number of ways to cut a square with side length i

# by considering the previous cuts of squares with smaller side lengths

dp[i][i] = dp[i – 1][i – 1] * 2 + (i – 1)

for j in range(1, i):

dp[i][i] += dp[i][j] * 2 # Add the number of ways to cut a square of side length j on the other side

# Calculate the number of ways to cut squares of side lengths less than i

for j in range(i – 1, 0, -1):

dp[i][j] = dp[i][j + 1] + (dp[i – j][j] – dp[i – j][j + 1])

return dp[N][N] % (10**8)

# Calculate C(30) modulo 10^8

result = calculate_C(30)

print(result)

“`

The code uses a 2-dimensional list `dp` to store the number of ways to cut a square of each side length up to `N`. The base case is initialized with `dp[1][1] = 2`, representing the two ways to cut a 1×1 square.

The outer loop iterates from 2 to `N` to calculate the number of ways to cut a square with side length `i` by considering the previous cuts. The inner loop calculates the number of ways to cut a square of side length `j` on the other side. Finally, the code calculates the number of ways to cut squares of side lengths less than `i`.

The result is then printed out modulo `10^8`, as required in the problem statement.

Running this code will give you the value of $C(30) \bmod 10^8$.

##### More Answers:

Pseudo Square RootBillionaire

At Least Four Distinct Prime Factors Less Than 100