Day 19 Interfaces Hackerrank Solution in C++. The AdvancedArithmetic interface and the method declaration for the abstract divisorSum(n) method are provided for you in the editor below. Complete the implementation of the Calculator class, which implements the

**interface. The implementation for the divisorSum(n) method must return the sum of all divisors of n.***AdvancedArithmetic*This is a very simple problem(Interfaces) we have to just calculate the divisor sum and print the sum of the divisor. we can do this by running the loop(starting with 1 and ending with N) and putting the condition if number N is divided by I(where I = 1 to N) then the sum of the divisor is added each time when if N%I==0. Let's take an example and try to understand the problem step by step.

Example: Let's assume Number N is 24 and we have to calculate the Divisor sum so we divide the number by I=1 to I=24 with the help of "For Loop" Take a look at step by step.

**Step 1:**If(N % I ==0) then Sum or total is Sum = Sum + I Or Total = Total + I.

**Step 2:**Print the total outside the "For Loop". Here Divisor of 24 = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60.

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## Day 19 Interfaces Hackerrank Solution in C++

```
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
class AdvancedArithmetic
{
public:
virtual int divisorSum(int n) = 0;
};
class Calculator: public AdvancedArithmetic
{
public: int divisorSum(int n)
{
int sum = n;
for (int i = 1; i < ceil(n / 2) + 1; i++)
n % i ? 1 : sum += i;
return sum;
}
};
int main()
{
int n;
cin >> n;
AdvancedArithmetic *myCalculator = new Calculator();
int sum = myCalculator->divisorSum(n);
cout << "I implemented: AdvancedArithmetic\n" << sum;
return 0;
}
```

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