Step by Step solution about How to resolve the algorithm Look-and-say sequence step by step in the C programming language

The provided C code generates and prints the look-and-say sequence. The look-and-say sequence is a sequence of numbers where each term is the number of times the previous term repeats itself, followed by the term itself.

Here's how the code works:

1. Initialize three character pointers: a, b, and x. a is initially allocated memory for two characters (to store the first number, "1"). b is initially a null pointer, and x is used as a temporary variable.

2. Enter a for loop that continues until b can't be reallocated to a larger size.

3. Inside the loop:

   • Print the value of x, which is a copy of the current string in a.
   • Count the number of consecutive occurrences of each character in a and build the next term of the look-and-say sequence in b. The code uses len to keep track of the current length of b and cnt to count consecutive occurrences of a character.
   • Update a to point to the newly allocated memory for b and set x to a (so x points to the current string).

4. Continue the loop until realloc fails, indicating that no more memory can be allocated.

This code will print the look-and-say sequence, starting with "1" and continuing indefinitely. The first few terms of the sequence are:

1
11
21
1211
111221
312211
13112221
1113213211
...

#include <stdio.h>
#include <stdlib.h>

int main()
{
	char *a = malloc(2), *b = 0, *x, c;
	int cnt, len = 1;

	for (sprintf(a, "1"); (b = realloc(b, len * 2 + 1)); a = b, b = x) {
		puts(x = a);
		for (len = 0, cnt = 1; (c = *a); ) {
			if (c == *++a)
				cnt++;
			else if (c) {
				len += sprintf(b + len, "%d%c", cnt, c);
				cnt = 1;
			}
		}
	}

	return 0;
}