Pattern 286

#include <stdio.h>
int main()
{
  int i,j;
  int n=5;

  for(i=1; i<=n; i++)
  {
    for(j=n; j>=1; j--)
    {
      if(i>=j)
      {
        if(j%2==0)
          printf("* ");
        else
          printf("# ");
      }
      else
      {
        printf(" ");
      }

    }
    printf("\n");
  }
  return 0;
}

#include <iostream.h>
int main()
{
	
	int n=5;
	
	for(int i =1;i<=n;i++)
    {
     for(int j =n;j>=1;j--)
     {
      if(i>=j)	
      {
	   if(j%2==0)
	   cout<<"* ";
	   else
	   cout<<"# ";
	  }
	  else
	  {
	   cout<<" ";
	  }
      
      }
      cout<<endl;           
    }
	return 0;
}

class PatternProg
{
	public static void main(String args[])
	{

		int n = 5;

		for (int i = 1;i <= n;i++)
		{
		 for (int j = n;j >= 1;j--)
		 {
		  if (i >= j)
		  {
		   if (j % 2 == 0)
		   {
		   System.out.print("* ");
		   }
		   else
		   {
		   System.out.print("# ");
		   }
		  }
		  else
		  {
		   System.out.print(" ");
		  }

		 }
		  System.out.println();
		}
		
	}
}

using System;

class PatternProg
{
  public static void Main()
  {

    int n = 5;

    for (int i = 1; i <= n; i++)
    {
      for (int j = n; j >= 1; j--)
      {
        if (i >= j)
        {
          if (j % 2 == 0)
          {
            Console.Write("* ");
          }
          else
          {
            Console.Write("# ");
          }
        }
        else
        {
          Console.Write(" ");
        }

      }
      Console.WriteLine();
    }

    Console.ReadKey(true);
  }
}

n = 5

for x in range(1, n + 1):
  for y in range(n, 0, -1):
    if x >= y:
       if y % 2 == 0:
          print("* ", end="")
       else:
          print("# ", end="")
    else:
       print(" ", end="")
  print()