Rectangle shape of numerics

Question

for(int width=1; width<=5; width++) {
    if(width <= 1) {
        for(int width=1; width<=5; width++) {
            cout<<" "<<width<<" ";
        }
    } else if(width<5) {
        cout<< endl;
        for(int width2=5; width2<=9; width2++) {
            if(width2==5 || width2==9)
                cout<<" "<<width2<<" ";
            else
                cout<< " ";
        }
    } else {
        cout<< endl;
        for(int width3=13; width3>=9; width3--) {
            cout<<" "<<width3<<" ";
        }
    }
}

this code which I have posted above draws this shape

1 2 3 4 5
5          9
5          9
5          9
13 12 11 10 9

but I actually want my code to print it like this, I have tried a lot changing things but all in vain. so, I'm looking forward to you guys.

1 2 3 4 5
16          6
15          7
14          8
13 12 11 10 9

Answer 1

Here is solution

  int  width =6;
  int total =  (width-1)*4;
  for(int row=1; row <=width; row++)
  {
    if(row == 1 )
    {
        for(int pr=1; pr<=width; pr++)
        {
            cout<<" "<<pr<<" ";
        }
        cout<<"\n";
    }
    else if( row == width)
    {
        for(int pr=1; pr<=width; pr++)
        {
            cout<<" "<<(total-row-pr+3)<<" ";
        }
    }
    else 
    {
         for(int pr=1; pr<=width; pr++)
        {

            if(pr ==1 )
                cout<<" "<<(total-row+2)<<" ";
            else if(pr ==width)
                cout<<" "<<(width+row-1)<<" ";
            else
                cout<<" "<<" "<<" ";
        }
         cout<<"\n";
    }
  }

Answer 2

It helps to avoid magic numbers in your code using #defines or const variables. This makes it more readable and more extensible. For example if you wanted to make a square that was 20x20, your code would require a complete rewrite!

Start from this working solution to implement this principle into your coding.

#include <iostream>

using namespace std;

#define SIDE 4

int main(){

   int perimeter = SIDE * 4;

   for(int width=0; width<=SIDE; width++)
   {
      if(width < 1) {
         for(int width=0; width <= SIDE; width++) {
            cout<<" "<<width + 1<<" ";
         }
         cout<< endl;
      }
      else if(width < SIDE)
      {
         cout<<" "<<perimeter - width + 1 << "\t\t" << (SIDE + width) + 1;
         cout<< endl;
      }
      else
      {
         for(int width3 = perimeter - SIDE; width3 >= perimeter - 2 * SIDE; width3--) {
            cout<<" "<<width3 + 1<<" ";
         }
         cout<< endl;
      }
   }
   return 0;
}

Answer 3

If you print something on the console, going back in lines and carriage returns will be very messy.

The trick is to seperate the problem in 3 stages:

stage1: print the top line, simple enough

stage2: print the largest number wrapping around, then print some empty space and finish with the number at the end, make sure to increment and decrement the numbers accordingly.

stage3: print the last line.

Here is the code for the algorithm I just described:

#include <iostream>

using namespace std;


int main()
{
    const int width=6;

    const int height=6;

    int numberInFront=(height-1)*2 + (width-1)*2;
    int numberAtTheEnd= width;
    for(int i=1; i<width; ++i) cout<<i<<"\t";  //print top line
    cout<<endl;

    for(int i=0; i<height-1; ++i)
    {
       cout<<numberInFront<<"\t";
       for(int j=0; j<width-3; j++) cout<<"\t"; //print inner space
       cout<<numberAtTheEnd<<endl;
       numberInFront--;
       numberAtTheEnd++;
    }


    //print last line:
    int counter = numberInFront;
    while(counter!=numberAtTheEnd-1)
    {
        cout<<counter<<"\t";
        counter--;

    }
    return 0;
}