Question

Please Write the task in c++

Task A perfect number is an integer that is equal to the sum of its divisors (where 1 is considered a divisor). For example, 6 is perfect because its divisors are 1, 2, and 3, and 1 + 2 + 3 is 6. Similarly, 28 is perfect because it equals 1 + 2 + 4 + 7 + 14. A quite good number is an integer whose badness—the size of the difference between the sum of its divisors and the number itself—is not greater than a specified value. For example, if the maximum badness is set at 3, there are 12 quite good numbers less than 100: 2, 3, 4, 6, 8, 10, 16, 18, 20, 28, 32, and 64. Your task is to write a C++ program that determines numbers of a designated maximum badness that are less than a specified value. The limiting value and maximum badness are provided as command line arguments when the program is executed.

Background A simple strategy is to try all possible numbers from 2 up to the specified limit. For each candidate number, calculate the sum of the candidate's divisors. If the difference between the sum and the candidate is not greater than the maximum badness, the number is considered good enough.

Level 1: Small perfect numbers Begin by writing a program that prints perfect numbers (badness of 0) up to values less than 10,000. Separate the output of each number by a single space. For example, quitegood 100 should print 6 28.

Answer 1

//C++ program

#include<iostream>
#include<stdlib.h>
#include<math.h>
using namespace std;
  
bool isQuiteGood( int num , int badness)
{
int total = 0;
for ( int i=1; i< num; i++)
{
if (num%i==0) total+=i;
}
if (abs(total -num) <= badness) return true;

return false;
}

// Driver program
int main(int argc, char *argv[])
{   
   if(argc<=2) {
cout<<"insufficient command line arguments\n";
exit(1);
}
   int limit , badness ;
limit = atoi(argv[1]);
badness = atoi(argv[2]);
  
for (int n =2; n<limit; n++) {
if (isQuiteGood(n , badness)) cout<<n<<" ";
   }

return 0;
}