C++ Perfect Number Program

lpwilliams21 · #1 08-Oct-2006, 02:33

I need help!! I have a program to write the directions are" Write a function perfect that determines whether parameter number is a perfect number. Write a second function printFactor that prints the factors of a perfect number. Use the functions perfect and printFactor in a program that determines and prints all the perfect numbers between 1 and 1000 and the factors of each perfect number." I really need help!! If anyone can help me I hope it can happen by Monday morning!!!

Sokar · #2 08-Oct-2006, 09:52

Quote:

Originally Posted by lpwilliams21

I really need help!! If anyone can help me

I would say just pick one thing at a time to accomplish. The first thing you will need is a basic skeleton.

CPP / C++ / C Code:

#include <iostream>
#include <ostream>

using namespace std;

int main()
{

	return 0;
}

Now add one thing at a time. Getting what you have added working 100%. Then move onto the next thing.

So the first thing is what will you need next. Maybe input of the number to be tested.

CPP / C++ / C Code:

#include <iostream>
#include <ostream>

using namespace std;

int main()
{
	int var1 =0;
	cout << "Enter number : ";
	cin >> var1;

	return 0;
}

and so on ...

Then post back with specific question with what is giving you trouble.

lpwilliams21 · #3 08-Oct-2006, 12:02

The part was that is giving me that trouble is figuring out how to print the factors of the numbes and figuring out how to print out the perfect numbers from 1-1000! If you can help I would be really happy! this program is giving me the blues

ubergeek · #4 08-Oct-2006, 13:47

First of all, for the benefit of anyone who doesn't know what a perfect number is (including myself, until I looked this up):

Quote:

Originally Posted by Wikipedia

In mathematics, a perfect number is defined as an integer which is the sum of its proper positive divisors, that is, the sum of the positive divisors not including the number.

So, you need to

Code:

loop through i=1 to 1000
    find factors of i (including 1, but not including i)
    add up said factors
    if the sum == i, print out i and the set of factors
end loop

I gather you need some help with finding the factors of a number. The key is using the modulus operator to determine whether a certain number is divisible by another one. The modulus operator does division, and returns the remainder. For example, 5 % 4 = 1, but 8 % 2 = 0. If the remainder is 0, the numbers divide evenly.

Code:

loop through i=1 to number/2
    if number % i == 0
    add i and number/i to the list of factors
end loop

davekw7x · #5 08-Oct-2006, 16:01

Quote:

Originally Posted by ubergeek

The modulus operator does division, and returns the remainder. For example, 5 % 4 = 1, but 8 % 2 = 0. If the remainder is 0, the numbers divide evenly.

I was with you up to here, and I like the explanation. However, the pseudo code is wrong.

Quote:

Originally Posted by ubergeek

Code:

loop through i=1 to number/2
    if number % i == 0
    add i and number/i to the list of factors
end loop

It simply won't work for finding perfect numbers, since

1. If you start with i = 1 then the very first step incorrectly adds n to the list of factors, and you are dead from the start.

2. If you note that 1 is always a factor and, therefore you don't have to test for it, then you could add 1 to the list of factors before you do the loop, and start the loop with i = 2. This actually works for n = 6, the first perfect number. But sometimes that loop adds the same factor more than once.

To illustrate, I'll put the '1' on a line by itself, then as I go through the loop for i = 2 up to and not including number/2, I put each pair of factors from your code on a separate line.

For n = 28, a program based on your code would give something like the following:

Code:

The factors of 28:
  1
  2  14
  4   7
  7   4
sum = 39

In fact, 28 is a perfect number since the sum of its (unique) factors is 1+2+4+7+14 = 28

So, pseudo code for adding the unique factors of n could be something like:

Code:

sum = 0
loop through i=1 up to but not including n
    if n % i == 0
        sum += i
    end if
end loop

Now if you wanted to print the factors (after determining that n is, indeed a perfect number), then you could go through the same kind of loop, but you would simply print the value of i whenever you found that (n % i) == 0.

There may very well be more elegant ways (mathematically and programatically) to find perfect numbers, but I think it's most important to get the answers right at first. Then if insight gained by examining the method and the results of the simple-minded approach allows one to consider more efficient methods, I say, "Go for it!"

Regards,

Dave

yohosuff · #6 06-Oct-2008, 21:49

Hi,

I just recently came across a similar question, and here is the function I wrote to determine if a number is perfect or not:

CPP / C++ / C Code:

bool perfect(int number)
{
    int sum = 0;

    for(int i = 1; i <= number/2; ++i)
    {
        if(number%i == 0)
        {
            sum+=i;
        }

    }


    if(sum==number)
        return 1;
    else
        return 0;

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