How to print this pattern?

[obeeey]

Hi,
my aim is to print this pattern:

o----
-o---
--o--
---o-
----o

where n=5 is a side length of the square, x=3 is how many times this square has to be printed horizontally.

So the result for n=5, x=3 should be:

o----o----o----
-o----o----o---
--o----o----o--
---o----o----o-
----o----o----o

This is the code for the square, easy as it is:

int main()
{
int n=5;
    for(int row=0; row<n; ++row)
    {
        for(int col=0; col<n; ++col)
        {
            if(col==row)
                cout << "o";
            else
                cout << "-";
        }
    cout << endl;
    }
    return 0;
}

I know how to print this pattern a few times vertically, but my real problem is how to do this horizontally - what I see from the debugger is that the program is built from the left side of the line to the right side and then there is a new line, so I am confused with what kind of conditions I should print "o".

All I have, but it's not working fully:

int main()
{
int n=5;
int x=3;
int i=0;

    for(int row=0; row<n; ++row)
    {
        for(int col=0; col<x*n; ++col)
            {
            if(col==n*i || col==row)  //MISSING CONDITION?
            {
            cout << "o";
            ++i;
            }
            else
                cout << "-";
            }
    cout << endl;
    }
    return 0;
}

Please, is there a simple way that I'm missing? Could anyone help me?

Thanks so much.

[lastchance]

#include <iostream>
using namespace std;

int main()
{
   int n=5;
   int x=3;

   for( int row = 0; row < n; ++row)
   {
      for ( int col = 0; col  < x * n; ++col )
       {
            if ( ( n + col - row ) % n )  // NOT SO MISSING CONDITION
            {
               cout << "-";
            }
            else
            {
               cout << "o";
            }
        }
        cout << '\n';
    }
}

[obeeey]

Wow, thank you so much, you really saved my evening. I completely forgot about the usage modulo... Now it's all clear.

Have a great time!