Pattern 366

#include <stdio.h>

int main()
{
  int n = 5;
  int i,j;

  for(i=n; i >= 1; i--)
  {
    for(j=1; j <= n; j++)
    {
      if (j == n - i + 1 || j == n || i == n)
      {
        printf("%d ",i);
      }
      else
      {
        printf("  "); // 2ws
      }
    }
    printf("\n");
  }
  return 0;
}

#include <iostream.h>

int main()
{
  int n = 5;
  

  for(int i =n; i >= 1; i--)
  {
    for(int j =1; j <= n; j++)
    {
      if (j == n - i + 1 || j == n || i == n)
      {
        cout<<i<<" ";
      }
      else
      {
        cout<<"  "; // 2ws 
      }
    }
    cout<<endl;
  }
  return 0;
}

class PatternProg
{
	public static void main(String args[])
	{
	  int n = 5;


	  for (int i = n; i >= 1; i--)
	  {
		for (int j = 1; j <= n; j++)
		{
		  if (j == n - i + 1 || j == n || i == n)
		  {
			System.out.print(i+" ");
		  }
		  else
		  {
			System.out.print("  "); //2ws
		  }
		}
		System.out.println();
	  }
	  
	}
}

using System;

class PatternProg
{
  public static void Main()
  {
    int n = 5;


    for (int i = n; i >= 1; i--)
    {
      for (int j = 1; j <= n; j++)
      {
        if (j == n - i + 1 || j == n || i == n)
        {
          Console.Write(i + " ");
        }
        else
        {
          Console.Write("  "); //2ws
        }
      }
      Console.WriteLine();
    }

    Console.ReadKey(true);

  }
}

n = 5

for x in range(n, 0, -1):
  for y in range(1, n + 1):
    if y == n - x + 1 or y == n or x == n:
       print(str(x)+" ", end="")
    else:
       print("  ", end="")  # 2ws
  print()