matrix using std::vector

Hi everyone..
My desired output for this code is:

0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0

where n will be the value for the n x n matrix, which in this case is n=5.

I have achieved to layout the matrix pattern but all the elements displayed are
0. How can I get it to display an alternate pattern of 0 and 1?

Thanks..

Code:

#include <iostream>
#include <vector>
using namespace std;

int main()
{
                int n = 0;
	int i = 0, j = 1;

	vector<int> matrix;
	
	cout << "Enter a value for n where n >= 2: ";
	cin >> n;
	cout << endl;

	matrix.resize(n);
	
	if (n % 2 != 0)
	{
		for (i = 0; i < n; ++i)
		{
			for (j = 0; j < (n/2); ++j)
				cout << matrix[i] << " " << matrix[j] << " ";
			cout << matrix[i] << endl;
		}
		cout << endl;
	}
	else
		for (i = 0; i < n; ++i)
		{
			for (j = 0; j < (n/2); ++j)
				cout << matrix[i] << " " << matrix[j] << " ";
			cout << endl;
		}
		cout << endl;
}

If that is all you want to print, wouldn't it be easier to use nested for loops and not bother with a container in the first place?

Hi laserlight..

Thanks for the quick response. This is what I came up with. It works the way I wanted it to but do I really need to use a matrix for this? I guess, what I was trying to say is what is the most efficient way to write this program? I will greatly apreciate your opinion.

Thanks again..

Code:

#include <iostream>
#include <vector>
using namespace std;

int main()
{
	int n = 0;
	int bit0 = 0, bit1 = 1;
	int i = 0, j = 0;

	vector<int> matrix;
	
	cout << "Enter a value for n where n >= 2: ";
	cin >> n;
	cout << endl;

	matrix.resize(n);
	
	if (n % 2 != 0)
	{
		for (i = 0; i < (n/2); ++i)
		{
			for (j = 0; j < (n/2); ++j)
				cout << bit0 << " " << bit1 << " ";
			cout << bit0 << endl;
			for (j = 0; j < (n/2); ++j)
				cout << bit1 << " " << bit0 << " ";
			cout << bit1 << endl;
		}
		for (i = 0; i < (n/2); ++i)
		{
			cout << bit0 << " " << bit1 << " ";
		}
		cout << bit0 << endl;
	}
	else
		for (i = 0; i < (n/2); ++i)
		{
			for (j = 0; j < (n/2); ++j)
				cout << bit0 << " " << bit1 << " ";
			cout << endl;
			for (j = 0; j < (n/2); ++j)
				cout << bit1 << " " << bit0 << " ";
			cout << endl;
		}
		cout << endl;
}

This is what I came up with. It works the way I wanted it to but do I really need to use a matrix for this?

Good to see you solved the problem yourself
As I stated, you do not need to use a container.

I guess, what I was trying to say is what is the most efficient way to write this program?

Nested loops with some simple arithmetic will do:

Code:

#include <iostream>

using namespace std;

int main()
{
    int n = 0;
    cout << "Enter a value for n where n >= 2: ";
    cin >> n;

    cout << '\n';
    for (int i = 0; i < n; ++i)
    {
        for (int j = 0; j < n; ++j)
        {
            cout << ((i + j) &#37; 2) << ' ';
        }
        cout << '\n';
    }
}

Incidentally, with the above solution n < 2 will still be okay, even if n is negative.