Hacker Rank solution For Day 19: Interfaces

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Check This:- Hacker rank solution for Strings, Classes, STL, Inheritance in C++.

Explanation:- This is a very simple problem(Interfaces) we have to just calculate divisor sum and print the sum of the divisor. we can do this by run the loop(started with 1 and end with N) and put the condition if number N is divided by I(where I = 1 to N) than 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.

Also Check:- Geeksforgeeks solution for School, Basic, Easy, Medium, Hard in C++.

Tip:- Copy the colored code or full code(According to Requirement ) and paste it into hacker rank editor.All solution provided here are in C++ (CPP) if any reader wants these solutions in C, and Java comments below or Email me with your query like " day n solution in C / C++ / Java. Check the end of the post solutions with the full explanation
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#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
class AdvancedArithmetic
        virtual int divisorSum(int n)=0;

struct Calculator : public AdvancedArithmetic
    int divisorSum(int n)
    int total=0,i;
        for(i=1; i<=n; i++)
            if(n%i == 0)
        return total;

int main()
    int n;
    cin >> n;
    AdvancedArithmetic *myCalculator = new Calculator(); 
    int sum = myCalculator->divisorSum(n);
    cout << "I implemented: AdvancedArithmetic\n" << sum;
    return 0;


Hacker Rank solution For Day 19: Interfaces

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