Step by Step solution about How to resolve the algorithm Mian-Chowla sequence step by step in the C++ programming language

The provided C++ code generates and displays the first 100 terms of the Mian-Chowla sequence. The Mian-Chowla sequence is a sequence of positive integers such that the sum of any two distinct terms is not equal to the value of any other term in the sequence. The code uses two main arrays:

• mc saves the Mian-Chowla sequence
• sums saves the sums of available pairs from the mc array The Contains function is used to check if an item exists in an array. The MianChowla function generates the Mian-Chowla sequence. It iterates from 1 to n to fill the mc array. For each value of i, it finds the smallest integer j such that the sum of any two distinct terms from mc with indices less than or equal to i is not equal to j.
• The outer loop iterates from 1 to n, updating the index of the current term being generated in the Mian-Chowla sequence.
• The inner loop iterates through potential values for the current term, starting from the previous term plus 1.
• For each potential value, the code checks if the sum of the current term and any previous term is already in the sums array. If it is, the code skips the current value and moves on to the next one.
• If the sum is not in the sums array, the code adds it and updates the length of the sums array.
• Once a valid value for the current term is found, it is stored in the mc array, and the code moves on to the next term. The main function calls the MianChowla function to generate the sequence. It then prints out the first 30 terms and terms 91 to 100 of the sequence, along with the time taken to generate the sequence.

using namespace std;

#include <iostream>
#include <ctime>

#define n 100
#define nn ((n * (n + 1)) >> 1)

bool Contains(int lst[], int item, int size) {
	for (int i = 0; i < size; i++) if (item == lst[i]) return true;
	return false;
}

int * MianChowla()
{
	static int mc[n]; mc[0] = 1;
	int sums[nn];	sums[0] = 2;
	int sum, le, ss = 1;
	for (int i = 1; i < n; i++) {
		le = ss;
		for (int j = mc[i - 1] + 1; ; j++) {
			mc[i] = j;
			for (int k = 0; k <= i; k++) {
				sum = mc[k] + j;
				if (Contains(sums, sum, ss)) {
					ss = le; goto nxtJ;
				}
				sums[ss++] = sum;
			}
			break;
		nxtJ:;
		}
	}
	return mc;
}

int main() {
	clock_t st = clock(); int * mc; mc = MianChowla();
	double et = ((double)(clock() - st)) / CLOCKS_PER_SEC;
	cout << "The first 30 terms of the Mian-Chowla sequence are:\n";
	for (int i = 0; i < 30; i++) { cout << mc[i] << ' '; }
	cout << "\n\nTerms 91 to 100 of the Mian-Chowla sequence are:\n";
	for (int i = 90; i < 100; i++) { cout << mc[i] << ' '; }
	cout << "\n\nComputation time was " << et << " seconds.";
}