Pattern 240

Pattern 240 post thumbnail image

C

#include <stdio.h>

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

  int x = 1;

  for(i = 1; i <= n; i++)
  {
    for(j = 1; j <= n; j++)
    {
      if ((i + j) % 2 == 1)
      {
        printf("%2d ", x++);
      }
      else
      {
        printf("%2c ",'*');
      }
    }
    printf("\n");
  }
  return 0;
}

C++

#include <iostream.h>
#include <iomanip.h>

int main()
{
  int n = 5;
  int x = 1;

  for(int i = 1; i <= n; i++)
  {
    for(int j = 1; j <= n; j++)
    {
      if ((i + j) % 2 == 1)
      {
        cout<<x++<<" ";
      }
      else
      {
        cout<<"* ";
      }
    }
    cout<<endl;
  }
  return 0;
}

Java

class PatternProg
{
	public static void main(String args[])
	{
	  int n = 5;
	  int x = 1;

	  for (int i = 1; i <= n; i++)
	  {
		for (int j = 1; j <= n; j++)
		{
		  if ((i + j) % 2 == 1)
		  {
			System.out.printf("%2d ",x++);
		  }
		  else
		  {
			System.out.printf("%2c ",'*');
		  }
		}
		System.out.println();
	  }
	  
	}
}

C#

using System;

class PatternProg
{
  public static void Main()
  {
    int n = 5;
    int x = 1;

    for (int i = 1; i <= n; i++)
    {
      for (int j = 1; j <= n; j++)
      {
        if ((i + j) % 2 == 1)
        {
          Console.Write("{0,2:D} ", x++);
        }
        else
        {
          Console.Write("{0,2} ", '*');
        }
      }
      Console.WriteLine();
    }

    Console.ReadKey(true);
  }
}

Python

n = 5
d = 1

for x in range(1, n + 1):
  for y in range(1, n + 1):
    if (x + y) % 2 == 1:
        print(str(d)+" ", end="")
        d += 1
    else:
        print("*  ", end="")
  print()
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