This code is a snippet that performs the following actions: ###...

This code is a snippet that performs the following actions:

Initialization and Matrix Filling:

1. arr[i] = new double[n];:

   • Creates an array of size n and assigns it to the i-th position of the arr 2D array (assuming arr is a 2D array already declared earlier with size t rows).

2. The for loop (for(int j = 0; j < n; j++)) iterates through each index j in the newly created array arr[i].

3. Inside this loop:

   • The value at arr[i][j] is computed as:

     a + b * (Math.pow(2, j+1) - 1)
     

     • Math.pow(2, j+1) computes 2 raised to the power of (j + 1).
     • Subtracting 1 gives Math.pow(2, j+1) - 1.
     • Multiplying this result by b, then adding a gives the value assigned to arr[i][j].

Outputting the Values:

4. The second set of nested loops (for(int i = 0; i < t; i++) and for(int j = 0; j < arr[i].length; j++)) iterates over the 2D array arr.

5. For each element of arr:

   • The value is cast to an integer ((int) arr[i][j]) and printed with a space in between (%d format specifier in System.out.printf).

6. At the end of each row (i.e., inner loop), a newline is printed using System.out.println().

Summary:

• The code fills a 2D array arr with computed values based on the formula a + b * (2^(j+1) - 1) for every i and j.
• It then outputs these values row by row, casting the double values to int.

Example:

For clarity, let's assume:

• t = 2 (2 rows in the array),
• n = 3 (3 columns per row),
• a = 1 and b = 2.

Computation:

For i = 0:

• j = 0: arr[0][0] = 1 + 2 * (2^(0+1) - 1) = 1 + 2 * (2 - 1) = 1 + 2 * 1 = 3.
• j = 1: arr[0][1] = 1 + 2 * (2^(1+1) - 1) = 1 + 2 * (4 - 1) = 1 + 2 * 3 = 7.
• j = 2: arr[0][2] = 1 + 2 * (2^(2+1) - 1) = 1 + 2 * (8 - 1) = 1 + 2 * 7 = 15.

For i = 1 (similar calculations as above):

• arr[1][0] = 3, arr[1][1] = 7, arr[1][2] = 15.

Output:

The printed result would be:

3 7 15 
3 7 15

This example demonstrates how the code initializes and computes the 2D array, then outputs it row by row.