Java中的棋盘格模式，带有嵌套的for循环和2x2分组

Question

I'm totally stumped on this introductory java homework question. We have to use nested for loops to make an m by n checkerboard composed of X's and O's. M is the number of rows and N is the number of columns. I can make a general checkerboard pattern with the code below but the thing I'm having trouble wrapping my head around is they want the characters to be grouped in to 2x2 groupings. So for the code posted below the end result should look like this:

XXOO
XXOO
OOXX

I'm sure it's not that difficult but I've tried everything I can think of for several hours and still can't seem to figure it out. I'm getting super frustrated which isn't helping things either :/ Thanks in advance for any and all help!

public class Homework
{
    public static void main (String[] args)
    {
        int m = 3;
        int n = 4;
        for(int rows = 0; rows<m; rows++)
        {
            for(int cols = 0; cols<n; cols++)
            {
                if((rows+cols)%2 ==0) System.out.print("X");
                else System.out.print("O");
            }
            System.out.println();
        }
    }
}

Answer 1

For each value of rows and cols , you'll print X if they are both 0 or 1 OR if they are both 2 or 3. The second-to-last bit in the value contains that information; it's 0 if it's 0 or 1, and 1 if it's 2 or 3. You'll need to know that particular bit on both values. Use value & 2 to extract the second-to-last bit.

Print X if both are the same (both 0 or both 1), and O if they aren't the same.

XX  |  OO
    |
XX  |  OO
(1) |  (2)
----+----
(3) |  (4)
OO  |  XX
    |
OO  |  XX

• In scenario (1), both second-to-last bits are 0s.
• In scenario (2), the row's second-to-last bit is a 0, but the col's is a 1.
• In scenario (3), the row's second-to-last bit is a 1, but the col's is a 0.
• In scenario (4), both second-to-last bits are 1s.

Print the X if both second-to-last bits are the same, else print O .