To solve this problem, we need to print a pattern based on the given input. The input consists of two integers, let's call them N and M. 1. Read the values of N and M from the input. 2. Start a loop from 1 to N (inclusive) to iterate over each row of the pattern. 3. Inside the loop, start another loop from 1 to M (inclusive) to iterate over each column of the pattern. 4. Check if the current row or column is the first or last row/column. If it is, print "X" (capital letter X) as the element of the pattern. Otherwise, print "0" (digit zero). 5. After printing all the elements of a row, move to the next line. 6. Repeat steps 4 and 5 until all rows and columns are processed. 7. Print the final pattern. Here is the code in C to implement the above steps: ```c #include int main() { int N, M; scanf("%d %d", &N, &M); for (int i = 1; i

Question

To solve this problem, we need to print a pattern based on the given input. The input consists of two integers, let's call them N and M.

1. Read the values of N and M from the input.

2. Start a loop from 1 to N (inclusive) to iterate over each row of the pattern.

3. Inside the loop, start another loop from 1 to M (inclusive) to iterate over each column of the pattern.

4. Check if the current row or column is the first or last row/column. If it is, print "X" (capital letter X) as the element of the pattern. Otherwise, print "0" (digit zero).

5. After printing all the elements of a row, move to the next line.

6. Repeat steps 4 and 5 until all rows and columns are processed.

7. Print the final pattern.

Here is the code in C to implement the above steps:

```c
#include <stdio.h>

int main() {
    int N, M;
    scanf("%d %d", &N, &M);

for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= M; j++) {
            if (i == 1 || i == N || j == 1 || j == M) {
                printf("X ");
            } else {
                printf("0 ");
            }
        }
        printf("
");
    }

return 0;
}
```

This code reads the values of N and M from the input, then uses nested loops to iterate over each row and column of the pattern. It checks if the current element is on the first or last row/column and prints "X" in that case, otherwise it prints "0". Finally, it moves to the next line after printing all elements of a row.

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