Finding Perfect Number with C++

[Freyster98]

I am supposed to write a program to test a user-provided number to see if it is perfect. Also, I have to use a subroutine to do the test, and call to that in the main program. Finally, I must use a boolean to control the final output. I understand that it would be SO MUCH easier if I could just write a simple program to do this without the boolean and subroutine, but an assignment is an assignment. Here is what I have so far...it runs, but no matter what number I enter when running the program, it comes back as perfect. Can anyone see what I am doing wrong? Any help would be appreciated.

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

void Perfect(int number,int&ans,bool&valid)
{
	int sum=0;
	valid=false;
	for(int i=1;i<=number/2;i++)
	{
		if(number%i==0)
			sum+=i;
	}
	if(sum==number)
		valid=true;
		ans=sum;
}

void Perfect(int,int&,bool&);
char ch;

void main()
{
	int num,answer; 
	bool found;
	cout<<"Enter the number you want to test to see if it is perfect: ";
	cin>>num;
	Perfect(num,answer,found);
	if(found=true)
	{
		cout<<num<<" is a perfect number.";
	}
	else
	{
		cout<<num<<" is NOT a perfect number.";
	}
	cin>>ch;
}

 

[mgb_phys]

if(found=true)

doesn't do what you think it does!

 

[Freyster98]

if(found=true)

doesn't do what you think it does!

I can't tell you how many times I've gone over this and completely missed that. Thank you for taking the time to help, I really appreciate it.

 

[mgb_phys]

I can't tell you how many times I've gone over this and completely missed that.

Been there - done that - bought the Tshirt!

A good tip is to always put the constant first. eg.
if ( 2=count) or if ( true=found) then the compiler will warn you.

Even better is not to compare with true/false eg.
if (found) or if (!found) is much better.

 

[CRGreathouse]

Alternate C++ code:

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

bool perfect(int n)
{
	int sum=0;
	int lim = (int)sqrt(n);
	for(int i=1; i <= lim; i++)
	{
		if(n%i == 0)
			sum += i + n/i;
	}
	return sum == n;
}

void main()
{
	int num;
	cout << "Enter the number you want to test to see if it is perfect: ";
	cin >> num;
	if(perfect(num))
		cout<<num<<" is a perfect number.";
	else
		cout<<num<<" is NOT a perfect number.";
}

Alternate function perfect (replace in the above):

bool isPrime(int n) {
	if ((n&1) == 0)
		return n == 2;
	if (n%3 == 0)
		return n == 3;
	if (n < 25)
		return n > 1;
	int lim = (int)sqrt(n);
	for (int i = 5; i <= lim; i += 4)
	{
		if (n%i == 0)
			return false;
		i += 2;
		if (n%i == 0)
			return false;
	}
	return true;
}

bool perfect(int n)
{
	int s = 0;
	while ((n&1) == 0)
	{
		s++;
		n >>= 1;
	}

	if (1 << s != (n+1) >> 1)
		return false;
	return isPrime(n);
}

Alternate function perfect (replace in the above; assumes 32-bit signed ints):

bool perfect(int n)
{
	if (n > 496)
		return n == 8128 || n == 33550336;
	return n == 6 || n == 28 || n == 496;
}

 

[WaleedAhmed]

"Guys here is a short one if you appreciate"

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


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

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

if(sum==number)
return true;
else
return false;


}


int main(){

int num;
bool q;
count<<"Enter the number you want to test to see if it is perfect: ";
cin>>num;
q=perfect(num);
if(q)
count<<"perfect number.\n";
else
count<<"NOT.\n";
return 0;
}