Friendly Pair in C++

November 25, 2022

friendly pair or not in c++

C Program to check Friendly Pair or not

We will write a code to Check Friendly pair in C++

Two numbers num1 & num2 are friendly pairs if the following holds true -

(Sum of divisors of num1)/num1= (Sum of divisors of num2)/num2

Algorithm:-

For two numbers num1 and num2

  1. Create two variables, sum1 and sum2
  2. Find the sum of divisors for num1 and num2 and assign the respective sums to sum1 and sum2 variables
  3. Find the ratio of sum1/num1 and sum2/num2
  4. If both ratios are same then these are friendly pair numbers
Friendly Pair in C++

Strong number

Perfect number

Harshad number

Automorphic number

Abundant number

Method 1

C++ Code:-

Run 
#include <iostream>
using namespace std;

int getDivisorsSum(int num){
    
    int sum = 0;
    
    for(int i = 1; i < num; i++){
        if(num % i == 0)
            sum = sum + i;
    }
    return sum;
}

int main ()
{
    int num1 = 30, num2 = 140;
    
    int sum1 = getDivisorsSum(num1);
    int sum2 = getDivisorsSum(num2);
    
    if(sum1/num1 == sum2/num2)
        cout << num1 << " & " << num2 << " are friendly pairs";
    else
        cout << num1 << " & " << num2 << " are not friendly pairs";

    
}
// Time complexity: O(N)
// Space complexity: O(1)

Output

30 & 140 are friendly pairs

Method 2

C++ Code:-

Run 
#include <iostream>
#include <math.h>
using namespace std;

int getDivisorsSum(int n){
    
    int sum = 0;
    
    for(int i = 1; i < sqrt(n); i++) 
    { 
        if (n % i == 0) 
        { 
            // For n : (1, n) will always be pair of divisor 
            // acc to def., we must ignore adding n itself as divisor 
            // when calculating for abundant number 
            if(i == 1) 
                sum = sum + i; 

            // Example For 100 (10,10) will be one of the pair 
            // But, we should add value 10 to the sum just once 
            else if(i == n/i) 
                sum = sum + i; 

            // add both pairs as divisors 
            // For any divisor i, n/i will also be a divisor 
            else 
                sum = sum + i + n/i; 
        } 
    } 
    return sum; 
} 

int main ()
{
    int num1 = 30, num2 = 140;
    
    int sum1 = getDivisorsSum(num1);
    int sum2 = getDivisorsSum(num2);
    
    if(sum1/num1 == sum2/num2)
        cout << num1 << " & " << num2 << " are friendly pairs";
    else
        cout << num1 << " & " << num2 << " are not friendly pairs";

    
}
// Time complexity: O(√N)
// Space complexity: O(1)

Output

30 & 140 are friendly pairs

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