matrix using std::vector

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

Quote:

---

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;
}

---

Quote:

---

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.

Quote:

---

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.