Perfect Numbers

Corrupt Shadow · #1 27-Mar-2008, 17:27

Don't you love classes that are required for your major that have nothing to do with what you'll be doing with your degree? Yay!

Okay, I have to write a complete program to write to the file perfect.out all perfect numbers between 2 and a specified named integer constant LIMIT (which we give a value of 1000).

This is what I have so far... but I have no clue how to use these functions in my main to get what I need to get. PLEASE HELP!!! I'm going crazy!

CPP / C++ / C Code:

#include <iostream>
#include <cstdlib>

using namespace std;

int sumOfPositiveDivisors (int number)
{
 int sum = 0, denom = number, divisor = (number / denom);
    for (;;) {
        if ((number % denom) == 0) {
            sum += divisor;
        }
        denom--;
        if (denom == 1)
            return sum;   
}

bool isPerfect (int number)
{
     return number == sumOfPositiveDivisors (number);
}

int main()
{
    return 0;
{

C++_Bandit · #2 27-Mar-2008, 19:38

By perfect number, you mean the mathematical definition: a number whose divisors summed together equal that number. For example 6 whose divisors are 1, 2, and 3.
1+2+3 = 6
or 28
1+2+4+7+14 = 28
Right?

Corrupt Shadow · #3 27-Mar-2008, 20:02

Yes, that is correct.

mamntc02 · #4 28-Mar-2008, 01:59

An example of using your functions could be:

CPP / C++ / C Code:

int main(){
	int number = 20;
	cout << "Number  " << number << (isPerfect(number) ? " is perfect" : " is not perfect")<< endl;

	return 0;
}

However, your code won't compile, because in your sumOfPositiveDivisors you missed some closing braces.
Anyway, there's an easier way to whether a number is perfect. For example:

CPP / C++ / C Code:

bool isPerfect(int num)
{
    int sum = 0 ;
    int i;
 
    for( i = 1; i<num; i++ )
        if( num%i == 0 )
        	sum += i;
    if( sum == num )
        return true;
    else
        return false;
 
}

Well, it's not a optimized code, but works

Your code seems to be too much complicated, thought it has an only error, besides braces: Your divisor variable doesn't change with denom. It must stay in for loop.

Hope it helps!

Corrupt Shadow · #5 29-Mar-2008, 08:13

I'd love to use that for isPerfect, but our professor specifically told us that isPerfect must call sumOfPositiveDivisors. Can I use that code in place of sumOfPositiveDivisors to the point where it looks like this?

CPP / C++ / C Code:

#include <iostream>
#include <cstdlib>

using namespace std;

int sumOfPositiveDivisors (int number)
{
 int sum = 0 ;
    int i;
 
    for( i = 1; i<number; i++ )
        if( number%i == 0 )
        	sum += i;
    if( sum == number )
        return true;
    else
        return false;
}

bool isPerfect (int number);
{
     return number == sumOfPositiveDivisors (number);
}

int main();
{
    int number = 20;
	cout << "Number  " << number << (isPerfect(number) ? " is perfect" : " is not perfect")<< endl;

	return 0;
}

Corrupt Shadow · #6 30-Mar-2008, 14:17

I updated my code a bit... I hate not knowing what the heck I'm doin. For an engineer-to-be, this is very irritating... HAHA!

CPP / C++ / C Code:

#include <iostream>
#include <cstdlib>
#include <fstream.h>
using namespace std;

int sumOfPositiveDivisors (int number)
{
 int sum = 0 ;
    int i;
 
    for( i = 1; i<number; i++ )
        if( number%i == 0 )
        	sum += i;
    if( sum == number )
        return true;
    else
        return false;
}

bool isPerfect (int number);
{
     return number == sumOfPositiveDivisors (number);
}

int main();
{
    const int LIMIT = 10000
    int number;
	return isPerfect == 2 << LIMIT;
	ofstream outFile("perfect.cc",ios::app);
    outFile << "New line\n";
    outFile.close();
    return 0;
}

mamntc02 · #7 31-Mar-2008, 01:28

Hi,

I'm sure your program doesn't compile. Your must learn to read compiler error messages. For example, I've compiled your first code (without file-management) and the compiler prints:

Quote:

prova2.cpp:21: error: expected unqualified-id before '{' token
prova2.cpp:26: error: expected unqualified-id before '{' token

I admit is not a good message, but when you take a look at lines 21 and 26 you see:

CPP / C++ / C Code:

bool isPerfect (int number);
{

and

CPP / C++ / C Code:

int main();
{

You have an extra ; because it's a method definition.
Anyway, it won't compile, because you are comparing an integer (the number) with a bool value (true or false) in isPerfect. It makes no sense.

As I said in my previous message, your first code was good, though has a couple of errors: the braces you missed and not update divisor in each iteration. So, using your first code, this should work:

CPP / C++ / C Code:

int sumOfPositiveDivisors (int number)
{
	int sum = 0, denom = number, divisor = (number / denom);
	for (;;) {
		if ((number % denom) == 0) {
			divisor = (number / denom);
			sum += divisor;
		}
		denom--;
		if (denom == 1)
			return sum;   
	}
}

Regards.

Corrupt Shadow · #8 31-Mar-2008, 05:37

CPP / C++ / C Code:

#include <iostream>
#include <cstdlib>
using namespace std;

int sumOfPositiveDivisors (int number)
{
	int sum = 0, denom = number, divisor = (number / denom);
	for (;;) {
		if ((number % denom) == 0) {
			divisor = (number / denom);
			sum += divisor;
		}
		denom--;
		if (denom == 1)
			return sum;   
	}
}

bool isPerfect (int number)
{
     return number == sumOfPositiveDivisors (number);
}

int main()
{
    const int LIMIT = 10000;
	return isPerfect == 2 << LIMIT;
	ofstream outFile("perfect.cc");
    outFile.close();
    return 0;
}

I still get errors in my MAIN:

Quote:

33 [Warning] left shift count >= width of type
34 variable `std::ofstream outFile' has initializer but incomplete type

What I need to do is print all perfect numbers (using isPerfect) from 2 to LIMIT (which we declare as 10,000) to the file "perfect.cc"

mamntc02 · #9 31-Mar-2008, 06:35

Of course, it won't compile!! What are you attempting to do with this?

CPP / C++ / C Code:

return isPerfect == 2 << LIMIT;

To print all perfect numbers up to a given number, you need a loop. For example:

CPP / C++ / C Code:

int main()
{
	const int LIMIT = 10000;

	ofstream outFile("perfect.cc", ios::app);
	
	for (int i = 2; i < LIMIT ; i++){
		if (isPerfect(i))
			outFile << i << endl;
	}

	outFile.close();
	return 0;
}

This is not the best loop possible, because perfect numbers are not odd, but I think it works

Regards

Corrupt Shadow · #10 31-Mar-2008, 06:43

Geez... I feel dumb now. ;)

I was thinking you could tell it to print all isPerfects from 2 to LIMIT in a return.

Duuurrrr...

The only problem Dev C++'s compiler is telling me is this:
35 variable `std::ofstream outFile' has initializer but incomplete type

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