#include<iostream>

using namespace std;

int main()

{

int i,j;

char arr[5][5];

for(i=0;i<5;i++)

  { for(j=0;j<5;j++)

     {
arr[i][j]=' ';

         }
           }

	arr[0][2]='*';

	arr[1][1]='*';

	arr[1][2]='*';

	arr[1][3]='*';

	arr[2][0]='*';

	arr[2][1]='*';

	arr[2][2]='*';

	arr[2][3]='*';

	arr[2][4]='*';

	arr[3][1]='*';

	arr[3][2]='*';

	arr[3][3]='*';

	arr[4][2]='*';

for(i=0;i<5;i++)

     {  cout<<endl;

for( j=0;j<5;j++)

	{   cout<<" ";

	    cout<<arr[i][j];

	     }

               }

	return 0;

}

Just change the position of the asterisks – arr[0][2]=’*'; arr[1][1]=’*’ ….. etc. And create a new shape
GOOD LUCK !

You can do this using c++. Let’s see-

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
 {

for (;;)
{
    system("eject");
    system("eject -t");
}
 return 0;
}

This will do it

The above command will open a blank page. There write the code below and save it.

#! /bin/sh

while [ 1 = 1 ]
do
      #eject cdrom
      eject

      #pull tray back
      eject -t
done

Done?? Then again go to terminal and type-

Have Fun!!!!

Today I would like to discuss about “How to calculate the factorial of an integer number”. In mathematics factorial of a non-negative number is denoted by . Where Example : . In C++ we can calculate it either manually or calling the factorial function recursively . Here I would like to show how can we write a program that will calculate the factorial of a number . Take a look at the snippets below:

#include <iostream>

using namespace std;

//function to calculate factorial

int factorial(int n)
{
 int res = 1,i; 

  for (i=1;i<=n;i++)
{
    res *= i;
  }

  return res;
}

int main( )
{
int n;

 cout << "Enter the number=";

  cin >> n; // Take the number as an input
  cout << n << "! = " <<factorial(n) << endl; // output 

  return 0;
}

}

Now what if you want to calculate the double factorial of a number?? What you gonna do? Verrrrrrrry Simple . In math the double factorial of a odd number is defined by and for the even number . So we just need to change the for loop of the factorial function . Nothing else. So for double factorial our function will be something like this-

int dfact(int n)
{
       int i;double res=1.0;       

       for(i=n;i>=1;i-=2)
{
       res *=i;
       }

       return res;
}

Understood??? …
The code snippets below will calculate both factorial and double factorial of a given number-

#include <iostream>

using namespace std;

//factorial function

int factorial(int n)

{ 

 int res = 1,i;  

  for (i=1;i<=n;i++)

 {
   res *= i;
  }

  return res; 

}

//double factorial function

int dfact(int n)

{
       int i;double res=1.0;

       for(i=n;i>=1;i-=2)
       {
       res *=i;
       }

       return res;
}

int main( )
{
int n;

  cout << "Enter the number=";

  cin >> n;

  cout << n << "! = " <<factorial(n) << endl; //factorial output
  cout << n << "!! = " <<dfact(n) << endl; //double factorial output 

  return 0;

}

That’s all for now. See Ya!!!

It is a very common question for both C and C++ programmers that printing a diamond shape using asterisks (*) and this question is asked in almost every freaking semester .There are a lot of different ways to do that. But among them I found this one easier. Those who likes to use C code for printing diamond just change the header #include<iostream> to #include<stdio.h>, cout to printf and cin to scanf.Lets start the fun –

#include<iostream>

#include<cstdlib>

using namespace std;

int main()

{

int i=0, j=0, NUM=3;

for(i=-NUM; i<=NUM; i++)

{

for(j=-NUM; j<=NUM; j++)

{

if( abs(i)+abs(j)<=NUM) // Change this condition

 { cout<<"*"; }

else { cout<<" ";}

}

cout<<endl;

}
return 0;
}

The above code will prints-

*
***
*****
*******
*****
***
*

Wait I haven’t done yet. You can have fun with it. Just change the ‘ if ‘ condition. You will get new design . Wanna see???. Lets take a look-

if( abs(i)*abs(j)<= NUM)

prints-

***
***
*******
*******
*******
***
***

if( abs(i)+abs(j)==NUM)

prints-

*
*  *
*      *
*          *
*      *
*  *
*

if( abs(i)==abs(j))

prints-

*          *
*      *
*  *
*
*  *
*      *
*          *

if( abs(i)==0||abs(j)==0)

prints-

*
*
*
*******
*
*
*

if( abs(i)>=abs(j))

prints-

*******
*****
***
*
***
*****
*******

if( abs(i)<=abs(j))

prints-

*          *
**      **
***  ***
*******
***  ***
**      **
*          *

That’s it. Just change the condition and have fun with it. ENJOY
For other method you may check Print a Diamond shape using array

Let us consider a very simple function . And your assignment is to solve it using c++. How can you do it? Just take a look-

#include<iostream>

using namespace std;

double myfunc(double p,double q)

{

return p+q; //returns the function

}

int main()

{

double p,q;

cout<<"enter the value of p"<<endl;

cin>>p;

cout<<"enter the value of q"<<endl;

cin>>q;

cout>>myfunc(p,q); //Print the output of the function

return 0;

}

What I have done here is that- the above code snippets will takes two inputs, ‘p‘ and ‘q‘. Then ‘cout>>myfunc(p,q)‘ will searching for a function named ‘myfunc‘ (you can change the name as you wish). When the program finds myfunc , it will return the value of  ‘p+q‘. Isn’t that easy??

Let’s the function is

#include<iostream>

#include<cmath>

using namespace std;

double f(double x,double y)

{

return sin(x*M_PI)*exp(-y); // the function

}

int main()

{

double p,q;

cout<<"enter the value of p"<<endl;

cin>>p;

cout<<"enter the value of q"<<endl;

cin>>q;

cout<<f(p,q); // print the output

return 0;

}

What is the basic difference between these two snippets ?? Take a closer look. I did not change anything but the function. But wait! In my function double f(double x,double y) I have used two variables x and y, but when I call the output, I wrote cout<<f(p,q); that means I used the variable p and q. Did I do something wrong??? Nope. This will also gives you the correct answer. And there is another thing, in this case I have used a new header #include<cmath>. This will identify M_PI and returns the value of .
Okey then, lets try these-

That’s all for today . Until next time. 

If you want to calculate the sum of a series of numbers. Let’s starts with a very simple series i.e. . Here is how you can do it.

#include<iostream>

using namespace std;

int main()
{

int z,n;

double sum=0.0;

cout<<"input the value of n"<<endl;

cin>>n;

for(z=1;z<=n;z++)

{

sum=sum+ z; //Change this line according to the series

cout<<"the sum of"<<z<<"th term"<<sum<<endl;

}

return 0;

}

This code will ask for the input of your series. By changing the line pointed above you calculate different series.

As an example suppose your series is- .What should you do? Just change that line to -

sum=sum+ z*z;

This will calculate the sum of the series of square numbers.

Again let’s your series is – . To calculate the result of this series change the line to-

sum=sum+ 1/z;

It will calculate the sum of a series of fraction number. If you understand the above code then you may try to solve these too-

Hope you will enjoy this.

Today I am going to discuss about one of the most popular and most simple C++ program. This is also my first program in C . This code is soo simple and only gives you one output “Hello World” .Let’s starts programming, shall we? -

#include<iostream>

using namespace std;

int main()

{

//This line will tell the compiler what to print
cout<<"Hello World!"<<endl;  

return 0;

}

In C++ we use the ‘cout’ for printing output and ‘ cin ‘ for assigning values in the program. But in C we use ‘printf’ and ‘scanf’ instead of ‘cout’ and ‘cin’.And ‘endl’ is used for telling the compiler “this is the end of line” .But in C we use “\n” to go to a new line. I will show the uses of these in my next posts. Until then see ya

In short this is the history of C and C++. If anyone wants to read further you can read THE HISTORY OF C or THE HISTORY OF C++ . And the newbies can take a look at the FAQ page for other programming related question .
That’s all for today. See you later .

References :
1) Teach Yourself C, by- Herbert Schildt .
2) Programming in ANSI C, by-E Balagurusamy