ive got the below code to print a pattern (attached below). However i'd like to just use one loop

#include<iostream>
using namespace std;
int main(){
    int n;
    cin>>n;
    for(int i=1;i<=n;i  ){
        for(int j=1;j<=i;j  ){
            cout<<"*";
        }
        for(int j=1;j<=n-i;j  ){
            if(j%2!=0){
                cout<<"_";
            }else{
                cout<<".";
            }
        }
        cout<<endl;
    }
    for(int i=1;i<n;i  ){
        for(int j=1;j<=n-i;j  ){
            cout<<"*";
        }
        for(int j=1;j<=i;j  ){
            if(j%2==0){
                cout<<".";
            }else{
                cout<<"_";
            }
        }
        cout<<endl;
    }
}
    

when n = 5, heres the output.

*_._.
**_._
***_.
****_
*****
****_
***_.
**_._
*_._.

how do i just make this into one single loop

CodePudding user response：

Try this and see how it does what you want to understand the step you did not find on your own:

#include<iostream>
using namespace std;
int main() {
    int n;
    cin >> n;
    for (int i = 1; i <= n*2-1; i  ) {
        if (i <= n)
        {
            for (int j = 1; j <= i; j  ) {
                cout << "*";
            }
            for (int j = 1; j <= n - i; j  ) {
                if (j % 2 != 0) {
                    cout << "_";
                }
                else {
                    cout << ".";
                }
            }
            cout << endl;
        }
        else
        {
            for (int j = 1; j <= n*2 - i; j  ) {
                cout << "*";
            }
            for (int j = 1; j <= i-n; j  ) {
                if (j % 2 == 0) {
                    cout << ".";
                }
                else {
                    cout << "_";
                }
            }
            cout << endl;
        }
    }
}

CodePudding user response：

I'd like to just use one loop.

I'll take it literally and show a starting point for a possible solution.

// Let's start by figuring out some dimensions.
int n;
std::cin >> n;                  
int height = 2 * n - 1;
int area = n * height;

// Now we'll print the "rectangle", one piece at a time.
for (int i = 0; i < area;   i)
{ //            ^^^^^^^^
  // Extract the coordinates of the char to be printed.
  int x = i % n;
  int y = i / n;
  
  // Assign a symbol, based on such coordinates.
  if ( x <= y  and  x <= height - y - 1 )
  { // ^^^^^^       ^^^^^^^^^^^^^^^^^^^    Those are the diagonals.  
    std::cout << '*'; // This prints correctly the triangle on the left...
  }
  else
  {
    std::cout << '_'; // <---  But of course, something else should done here. 
  }
  
  // End of row.
  if ( x == n - 1 )
    std::cout << '\n';
}

CodePudding user response：

If you look at the pattern, then you can see a sort of "triangles". And this already gives a hint for the solution. Use a triangle function.

Please read about it here.

Then you will notice that always the "aboslute"-function, in C std::abs, is involved.

But first of all, it is easily visible that the number rows to print is always the width of a triangle * 2.

And the number of charcters in the pattern, can be calculated by applying the triangle function. Example for width 5:

     Number of stars        number of dashdot
Row  width-abs(row-width)   abs(row-width)

 1             1                  4 
 2             2                  3
 3             3                  2
 4             4                  1
 5             5                  0
 6             4                  1
 7             3                  2 
 8             2                  3
 9             1                  4  

And this can be implemented easily now.

#include <iostream>
#include <cmath>
#include <algorithm>

using namespace std::string_literals;

int main() {
    // Get the max width of the pattern and perform a short input validation
    int maxWidth{};
    if ((std::cin >> maxWidth) and (maxWidth > 0)) {

        // The number of rows for the pattern is dependent on the width. It is a simple relation
        const int numberOfRows = 2 * maxWidth;

        // Show all rows
        for (int row = 1; row < numberOfRows;   row) {

            // Use triangle formular to create star pattern
            std::string starPattern(maxWidth - std::abs(row - maxWidth), '*');

            // Create dashDot pattern
            std::string ddp(std::abs(row - maxWidth), '\0');
            std::generate(ddp.begin(), ddp.end(), [i = 0]() mutable { return i   % 2 ? '.' : '_'; });

            // Show output
            std::cout << (starPattern ddp) << '\n';
        }
    }
    else std::cout << "\n*** Error: Invalid input\n\n";
}

Of course you can also create the whole pattern wit std::generate.

Maybe it is too complex for now.

And less understandable.

See:

#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>

int main() {
    // Get the max width of the pattern and perform a short input validation
    if (int width{}; (std::cin >> width) and (width > 0)) {

        std::vector<std::string> row(2 * width - 1);

        std::for_each(row.begin(), row.end(), [&, r = 1](std::string& s) mutable {
        s = std::string(width - std::abs(r - width), '*');
        std::string ddp(std::abs(r   - width),'\0');
        std::generate(ddp.begin(), ddp.end(), [&, i = 0]()  mutable{ return i   % 2 ? '.' : '_'; });
        s  = ddp; std::cout << s << '\n'; });
    }
    else std::cout << "\n*** Error: Invalid input\n\n";
}