How do I make this pattern with number of rows and columns specified by the user?-CodePudding

I received this assignment and I spent the last hour trying to figure it out but I don't seem to know how to do it. I have tried many ideas but none of them worked. Here is the expected output:

This is my final version of the code. It works however for some reason it doubles the number of columns. I'm not sure how to fix it.

int rows;
int columns;
int k = 0;
int j;
int i;
string pattern1;
string pattern2;
cout << "Enter number of desired rows: ";
cin >> rows;
cout << "Enter number of desired columns: ";
cin >> columns;
pattern1 = " \\@/";
pattern2 = " /*\\";

for (i = 1; i <= rows; i  )
{
    for (j = 1; j <= columns; j  )
    {
        if (i % 2 == 0)
        {
            cout << " /@\\";
        }
        else
            cout << pattern1;
        if (k == 0)
        {
            if (i % 2 == 0)
            {
                cout << " \\*/";
            }
            else
                cout << pattern2;
        }
    } cout << endl;
}

CodePudding user response：

I found the solution. Before, for each iteration of j it added two columns because the if statement was executed anyways because k is always 0. Now, I added a j at the end of the if statement so that whenever the compiler reads the if statement, it counts it as another iteration of the loop (what I mean is, we essentially make it that it adds two columns for every one iteration of the column (j) loop). But then I had to make the loop stop if j is actually equal to columns, thus I replaced the k==0 with j!=column. Here is the correct code.

 #include <iostream>
using namespace std;
int main()
{
    int rows;
    int columns;
    int k = 0;
    int j;
    int i;
    string pattern1;
    string pattern2;
    cout << "Enter number of desired rows: ";
    cin >> rows;
    cout << "Enter number of desired columns: ";
    cin >> columns;
    int nc = (columns/2);
    int nr = rows/2;
    pattern1 = " \\@/";
    pattern2 = " /*\\";

    for (i = 1; i <= rows; i  )
    {
        for (j = 1; j <= columns; j  )
        {
            if (i % 2 == 0)
            {
                cout << " /@\\";
            }
            else
                cout << pattern1;

            if (columns!=j)
            {
                if (i % 2 == 0)
                {
                    cout << " \\*/";
                }
                else
                    cout << pattern2;
                j  ;
            }
        } cout << endl;
    }
}

CodePudding user response：

Here's a way:

Two for loops, one for the rows and one for the columns.
A 2x2 matrix with the 2 patterns (even column, odd column) for the even rows and the 2 patterns for the odd rows.
For every row and column, patterns[i % 2][j % 2] will give you the even or odd pattern for the even or odd row.
You can print a space after each pattern except for the last column.

[Demo]

#include <iostream>  // cout
#include <string_view>
#include <vector>

void print_pattern(size_t nrows, size_t ncols)
{
    std::string_view patterns[2][2]{
        { R"(\@/)", R"(/*\)" },
        { R"(/@\)", R"(\*/)" },
    };
    for (size_t i{0}; i < nrows;   i)
    {
        for (size_t j{0}; j < ncols;   j)
        {
            std::cout << patterns[i % 2][j % 2] << ((j < ncols - 1) ? " " : "");
        }
        std::cout << "\n";
    }
}

int main()
{
    print_pattern(4, 5);
}