Perfect Number Program in C++

June 29, 2023

Perfect Number in C++

In this post, we will learn different ways of coding Perfect Number in C++.

A perfect number is a number in which the sum of the proper positive divisors of the number is equal to the number itself.

For Example: 28
Divisors : 1 + 2 + 4 + 7 + 14 = 28

Methods

Method 1: Iterating between [1, num] to find factors using for loop
Method 2: Iterating between [1, num] to find factors using while loop
Method 3: Iterating between [1, num/2] to find factors using
Method 4: Using Function and Iterating between [1, num/2] to find factors using
Method 5: Using Recursion
Method 6: Iterating between [1, √num] to find factors

Method 1

For an input num

Initialize sum = 0
Using a for loop Iterate between [1, num-1] to find all divisors
Add divisors to sum
If sum == num, then it’s a perfect number

C++ Code:-

Run

#include <iostream>
using namespace std;

int main ()
{
    int n = 28, sum = 0;
    
    for(int i = 1; i < n; i++){
        if(n % i == 0)
            sum = sum + i;
    }
    
    if(sum == n)
        cout << n << " is a perfect number";
    else
        cout << n << " is not a perfect number";
    

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

Output

28 is a perfect number

Finding Prime Factors of a number

Perfect number

Method 2

For an input num

Initialize sum = 0
Using a while loop Iterate between [1, num-1] to find all divisors
Add divisors to sum
If sum == num, then its a perfect number

C++ Code:-

Run

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

#include <iostream>
using namespace std;

int main ()
{
    int num = 28, sum = 0;
    
    int i = 1;
    while(i < num)  
    {
        // check if i is a divisor
        if(num % i == 0)  
            sum = sum + i;
            
        i++;  
    }  
    
    if(sum == num)
        cout << num << " is a perfect number";
    else
        cout << num << " is not a perfect number";
    
    return 0;
}

Output

28 is a perfect number

Method 3

If we iterate b/w [1, num/2]
All the divisors of the number can be found in this range
Example : for 28: divisors {1, 2, 4, 7, 14}
All are before num/2 i.e. 14

C++ Code:-

Run

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

#include <iostream>
using namespace std;

int main ()
{
    int num = 28, sum = 0;
    
    int i = 1;
    
    // If we iterate b/w [1, num/2]
    // all the divisors of the number can be found in this range
    // example : for 28: divsors {1, 2, 4, 7, 14}
    // all are before num/2 i.e. 14    
    while(i <= num/2)  
    {
        // checking if i is a divisor
        if(num % i == 0)  
            sum = sum + i;  
        i++;  
    }  
    
    if(sum == num)
        cout << num << " is a perfect number";
    else
        cout << num << " is not a perfect number";
    
    return 0;
}

Output

28 is a perfect number

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Method 4

This method uses the same approach as above. However, we do it via a function and a for loop.

C++ Code:-

Run

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

#include <iostream>
using namespace std;

bool checkPerfectNum(int n)
{
    int sum = 0;
    
    for (int i = 1; i <= n/2; i++){
        if (n % i == 0)
            sum = sum + i;
        }

    if (sum == n)
        return true;

    return false;
}

int main ()
{
    int num = 28;
    
    if(checkPerfectNum(num))
        cout << num << " is a perfect number";
    else
        cout << num << " is not a perfect number";
    
    return 0;
}

Output

28 is a perfect number

Method 5

This method uses recursion in C++

C++ Code:-

Run

// Time complexity: O(N)
// Space complexity: O(1)
// Auxiliary Space complexity: O(N) due to function call stack

#include <iostream>
using namespace std;

static int sum = 0;
int getSumDivisors(long num, int i)
{
    // since, all factors can be found will num/2
    // we will only check in this range
    if(i <= num/2)
    {
        if(num % i ==0)
            sum = sum + i;

        i++;
        // recursively call isPerfect
        getSumDivisors(num, i);
    }

    //returns the sum
    // when i becomes > num/2
    return sum;
}

int main ()
{
    int num = 28;
    
    if(getSumDivisors(num, 1) == num)
        cout << num << " is a perfect number";
    else
        cout << num << " is not a perfect number";
    
    return 0;
}

Output

28 is a perfect number

Method 6

This method uses observations that all factors come in pairs.

All Factors come in pairs For n = a * b (For each a there exists a unique b)

Example 1 : 100
(1,100), (2, 50), (4, 25), (5, 20), (10, 100)

Example 2 : 28
(1, 28), (2, 14), (4, 7)
Note : We will need to ignore pair of 1. As it will be the number itself.

Shorten the Loop We can shorten the loop running between [1, num] to [1, √num]

Since we will find all pairs before √num (n = sqrt(n) * sqrt(n))

Example: For 28, all pairs can be found before √28 = 5.2

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 num itself as divisor
            // when calculating for perfect number
            if(i == 1)
                sum = sum + i;
            
            // Example For 100 (10,10)  will be one pair
            // But, we should add value 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 n = 28, sum = 0;
    
    if(n == getDivisorsSum(n))
        cout << n << " is a perfect number";
    else
        cout << n << " is not a perfect number";
    

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

Output

28 is a perfect number

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One comment on “Perfect Number Program in C++”

Mayank Ashtekar
#include
using namespace std;
int main(int argc, char const *argv[])
{
int n;
cout << "enter the number" <> n;
int ans = 0;
for (int i = 1; i < n; i++)
{
if (n % i == 0)
{
ans = ans + i;
}
}

if (n == ans)
{
cout << n << " is a perfect number " << endl;
}
else
cout << n << " is not a perfect number " << endl;

return 0;
}

0

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