Thread: I asked the prof and still not direct answer

For example:
PowerInt (1) = 1!/1(2) =1
PowerInt (2) = 2!/2(2) = (1*2)/(2*2) = 2/4 =0.5
PowerInt (3) = 3!/3(2) = (1*2*3)/(3*3) = 6/9 =0.667
PowerInt (4) = 4!/4(2) = (1*2*3*4)/(4*4) = 24/16 =1.5

Poorly worded, I agree. However, from looking at what the latter portion of each line looks like, you should be able to come up with something that will churn out the desired results:

Code:

float p(int i)
{
    if( i < 10 && i > 1 )
    {
        int x = i*i, /* store the first number, squared */
             y; /* a counter */
        float z /* the sum */

        /*
        I will leave the inner workings to you.
        Use your counter to loop i times.
        Sum z with that number.
        */

        return z / (float)x;
    }
    return 1.0;
}

I believe that should give you what you need.

Quzah.

Originally posted by Speedy5
He wants you to find fractorials for numbers between 0 and 10 divided by that number times two.

By the number squared (from the examples given).

Would be better written as compute (n!)/(n^2) for 1 <= n <= 10.
My advice... Break up the operations.

Code:

#include <iostream>
#include <stdlib.h>

using namespace std;

double fac(double n);
double square(double x);
double PowerInt(double n);

int main()
{
  for(int i = 0; i <= 10; ++i)
    cout << i << ": " << PowerInt(i) << endl;
  system("pause");
  return 0;
}

double fac(double n)
{
    if(n < 2) return 1;
    return n * fac(n - 1);
}

double square(double x)
{
    return x * x;
}

double PowerInt(double n)
{
    if(n == 0) return 1;
    if( (n < 0)||(n > 10) ) return -1;
    return fac(n)/square(n);
}

/*Output

0: 1
1: 1
2: 0.5
3: 0.666667
4: 1.5
5: 4.8
6: 20
7: 102.857
8: 630
9: 4480
10: 36288
Press any key to continue . . .
*/

I was bored. Yeah yeah, I know recursion takes up lots of memory and yada, yada, yada. I use the recursive solution to finding factorials because I can remember it off the top of my head. For your class you probably ought to go with an iterative factorial function instead.

Yes...don't listen to the prof, listen to us...we make the rules

Originally posted by romeoz
I was reading over all these codes given on here and checking in my textbook and I dont understanf what y; does and float z;, I'm really lost, could you maybe explain that a little more into detail. I have never even seen this until now and I dont know where to start let alone what that code means that you wrote to help me out. sorry about that.

Here you go, an explanation:

Code:

float p(int i) // take an int, return a float
{
    if( i < 10 && i > 1 ) //check boundaries
    {
        // declare 2 integers, x and y
        int x = i*i,
             y;
        // declare a floating point number
        float z;

        // make a loop, use y as the loop
        // counter to move through it.
        // something like:
        for( y = ... ; ... ; ... )
            // do stuff here

        // you have to use a float to store
        // the sum, since you're returning
        // a decimal value
        return z / (float)x;
    }
    return 1.0;
}

I left the loop itself out intentionally, since you provided on effort of your own, only what you needed done. The loop was left out so you could learn how to do it on your own.

Not that that matter, since not matter what your question, some one will do the work for you. It always works that way.

Quzah.

Just a note, "float" means it can contain a decimal point (I was always confused by that...)

A couple things that I thought look odd in quzah's code:

Code:

 if( i < 10 && i > 1 ) //check boundaries

to make it more clear:

Code:

if (  (i<10)  &&  (i>1)  )

and the

Code:

y;

is actually being declared as an int,

Code:

int x = i*i,  y;

Code:

#include <iostream.h>

double long PowerInt(int Num);

int main(void)
{
  for(int i = 0; i < 5; i++) cout << PowerInt(i) << "\n";

  getchar();

  return 0;
}

double long PowerInt(int Num)
{
  unsigned Sq = Num * Num, FactResult = 1;

  if(Num < 0 || Num > 10) return -1; else if(!Sq) Sq++;

  for(int i = 1; i <= Num; i++) FactResult *= i;

  return (double long) FactResult / (double long) Sq;
}

I would just like to say that it's interesting to see how a few different people all looked at the exact same problem yet coded very different looking programs that still yield the exact same results.

joshdick - Just saw this line of code in your response
after I did mine.

if( (n < 0)||(n > 10) ) return -1;

Pretty similiar, good stuff.