7. Write a function with one integer parameter that will return the sum of the divisors of its argument.
C++ CODE
[Note: Explanation of the program code is provided within the comments.]
Program code:
//including header files
#include<bits/stdc++.h>
using namespace std;
// Function to calculate sum of all the divisors
int divSum(int num)
{
//declaring required variables
int result = 0;
// find all divisors which divides 'num'
//for loop
for (int i=2; i<=sqrt(num); i++)
{
// if 'i' is divisor of 'num'
if (num%i==0)
{
// if both divisors are same then adding it only once
if (i==(num/i))
result += i;
//otherwise
else
//adding here
result += (i + num/i);
}
}
// Adding 1 to the result as 1 is also a divisor
return (result + 1);
}
// main method
int main()
{
//declaring required variable
int num = 12;
//calling function and displaying results
cout << "Sum of all divisors of "<<num<<" is : " <<divSum(num);
return 0;
} //end of main method
Step by step
Solved in 2 steps with 1 images
