Pattern 51

#include <stdio.h>

int main()
{
  int n = 5;
  int i,j;
  int x = (n * (n + 1)) / 2;

  for(i = n; i >= 1; i--)
  {
    for (j = n; j >= i; j--)
    {
      printf("%2d ", x--);
    }
    printf("\n");
  }
  return 0;
}

#include <iostream.h>

int main()
{
  int n = 5;
  
  int x = (n * (n + 1)) / 2;

  for(int i = n; i >= 1; i--)
  {
    for(int j = n; j >= i; j--)
    {
      cout<<x--<<" ";
    }
    cout<<endl;
  }
  return 0;
}

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

	  int x = (n * (n + 1)) / 2;

	  for (int i = n; i >= 1; i--)
	  {
		for (int j = n; j >= i; j--)
		{
		  System.out.printf("%2d ",x-- );
		}
		System.out.println();
	  }
	  
	}
}

using System;

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

    int x = (n * (n + 1)) / 2;

    for (int i = n; i >= 1; i--)
    {
      for (int j = n; j >= i; j--)
      {
        Console.Write("{0,2:D} ", x--);
      }
      Console.WriteLine();
    }

    Console.ReadKey(true);
  }
}

n = 5

d = (n * (n + 1)) // 2

for x in range(n, 0, -1):
 for y in range(n, x - 1, -1):
   print("{:2d} ".format(d), end="")
   d -= 1
 print()