Pattern 345

#include <stdio.h>

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

  for(i = 1; i <= n; i++)
  {
    for (j = 1; j <= n; j++)
    {
      if (i == 1 || i == n || j == 1 || j == n)
        printf("1 ");
      else
        printf("0 ");
    }
    printf("\n");
  }
  return 0;
}

#include <iostream.h>

int main()
{
  int n = 5;
  

  for(int i = 1; i <= n; i++)
  {
    for(int j = 1; j <= n; j++)
    {
      if (i == 1 || i == n || j == 1 || j == n)
        cout<<"1 ";
      else
        cout<<"0 ";
    }
    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 = 1; j <= n; j++)
		{
		  if (i == 1 || i == n || j == 1 || j == n)
		  {
			System.out.print("1 ");
		  }
		  else
		  {
			System.out.print("0 ");
		  }
		}
		System.out.println();
	  }
	  
	}
}

using System;

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


    for (int i = 1; i <= n; i++)
    {
      for (int j = 1; j <= n; j++)
      {
        if (i == 1 || i == n || j == 1 || j == n)
        {
          Console.Write("1 ");
        }
        else
        {
          Console.Write("0 ");
        }
      }
      Console.WriteLine();
    }

    Console.ReadKey(true);

  }
}

n = 5

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