//Exercises - that is all
//Like the Jeffrey's Tube in StarTrek: GNDN (Goes Nowhere, Does Nothing)
#include <iostream.h>

double bresenham(int n, double a, double b, double d)
// a small variation of this algorithm is used for drawing straight lines
// on graphic displays and plotters
{
	for( int i = n;  i > 0;  i--)
	{	
		if   (d <= 0)
			 d += a;
		else
			d -= b;
	}
	return d;
}

int sum_squares(int n)
//The pattern of this algorithm is more important than the calculation

{	int i, sum;
	sum = 0;
	for( i = 1; i <= n; i++ )
	{
		sum += i*i;
	}
	return sum;
}

int pow(int m, int n)// Given n and m returns m to the power n
//This is a useful function to have in one's personal library
//However there is faster algorithm than this one.
{	int i, p;
	p = 1;
	for( i = 1; i <= n; i++ )
		p *= m;

	return p;

}

int factorial( int n)
/* 	Given an integer n  calculate the factorial of n  using local integer p.
		1. if n is less than or equal to 0 then 
			1.1. output an error message
			1.2. set p to the default value 0
		2. Otherwise
			2.1. Set the product p to 1 
			2.2. For each integer i from 2 to n
				2.2.1.  multiply p by i
		3. return the value of p
*/

{
	int p; 
	if( n <= 0 )
	{
		cerr << "Error in factorial( "<<n<<"). Argument must be >0\n";
			// notice that 'cerr' is the right place for errors
			// even tho' it makes little difference in theory
		p = 0;
	}// here n was <= 0 and p is 0
	else // n > 0
	{
		p = 1; // here p  is 1
		for(int i = 2; i<=n; i++)
		{	 // here p is 1*2*3*...*(i-1)
			p *= i;
			//here p is 1*2*3*...*(i-1)*i
		}// end for
		//here i was n and and so p is 1*2*3*...*(n-1)*n
	} //here n was >0 and  p is 1*2*3*...*(n-1)*n
	return p;

}
//--------------test harness----------------
int main()
{
	cout << sum_squares(3) <<endl;
	cout << pow(2, 3) <<endl;
	cout << factorial(3) <<endl;
	cout << factorial(-3) <<endl;
	return 0;
}