Question

I need some help revising this. It keeps only displaying 0s as the temp. Thank you.

// A program to determine whether the input number is a perfect number
// A perfect number is defined by the sum of all its positive divisors excluding itself
// 28: 1+2+3+7+14 = 28. 

int perfect, limit, divisor;

cout << "Please enter a positive integer in order to define whether it is a perfect integer or not: " ;
 cin >> perfect;
 cout << endl;

 int temp = 0;
 int prevtemp = 0;
  limit = 1;
  divisor = 1;
 while (limit < perfect)
 {

  if ((perfect % divisor) == 0)
   {
   divisor = prevtemp;
   temp = prevtemp + temp;
   }

  limit++;
  divisor++;
 }

 if (perfect == temp)
  cout << "Your number is a perfect number!" << endl;
 else
  cout << "Your number is not a perfect number" << endl;

 return 0;

Answer 1

I’m not sure, but I’d guess that in the code:

if ((perfect % divisor) == 0)
    divisor = prevtemp;

you intended this to be prevtemp=divisor instead. That fixes an obvious problem, but still leaves quite a bit that doesn’t look like it’s doing that you probably intended. For example, I can’t quite figure out what limit is intended to accomplish — you initialize it and increment it, but as far as I can see, you never use its value (well, I guess you use it, but its value is always the same as divisor‘s so I’m not sure why you think you need both, or how limit makes any sense as its name).

Edit: It would make sense to have a limit. In particular, factors always come in pairs: one that’s less than or equal to the square root of the number, and one that matches the first that’s always greater than or equal to the square root of the number. As such, you don’t need to scan all the way up to the number itself looking for factors — you can set the square root of the number as the limit, and scan only up to that point. For each factor you find up to that point, the matching factor will be perfect/divisor. Since you’ve already gotten one working example, I guess I might as well just hope this isn’t homework, and post an example as well:

bool is_perfect(int number) { 
    int limit = sqrt((double)number);
    int sum = 1;

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