⚡ Match Highlights: dsa/superperfect number

Given an integer n. Check whether the number n is superperfect number or not. A superperfect number is a positive integer which satisfies ?²(n) = ?(?(n)) = 2n, where ? is divisor summatory function.

Input: n = 16
Output: yes
Explanation:
16 is a superperfect number as ?(16) = 1 + 2 + 4 + 8 + 16 = 31,
and ?(31) = 1 + 31 = 32, 
thus ?(?(16)) = 32 = 2 × 16.

Input: n = 8
Output: no 
Explanation:
?(8) = 1 + 2 + 4 + 8 = 15
and ?(15) = 1 + 3 + 5 + 15 = 24
thus ( ?(?(8)) = 24 ) ? (2 * 8 = 26)

We strongly recommend that you click here and practice it, before moving on to the solution.

The idea is simply straightforward. We just iterate from 1 to sqrt(n) and find sum of all divisors of n, lets we call this sum as n1. Now we again need to iterate from 1 to sqrt(n1) and find sum of all divisors. After that we just need to check whether the resulted sum is equal to 2*n or not.

C++

// C++ program to check whether number is
// superperfect or not
#include<bits/stdc++.h>
using namespace std;

// Function to calculate sum of all divisors
int divSum(int num)
{
    // Final result of summation of divisors
    int result = 0;

    // find all divisors which divides 'num'
    for (int i=1; i*i <= num; ++i)
    {
        // if 'i' is divisor of 'num'
        if (num%i == 0)
        {
            // if both divisors are same then add
            // it only once else add both
            if (i == (num/i))
                result += i;
            else
                result += (i + num/i);
        }
    }

    return result;
}

// Returns true if n is Super Perfect else false.
bool isSuperPerfect(int n)
{
    // Find the sum of all divisors of number n
    int n1 = divSum(n);

    // Again find the sum of all divisors of n1
    // and check if sum is equal to n1
    return (2*n == divSum(n1));
}

//Driver code
int main()
{
    int n = 16;
    cout << (isSuperPerfect(n) ? "Yes\n" : "No\n");

    n = 6;
    cout << (isSuperPerfect(n) ? "Yes\n" : "No\n");
    return 0;
}

Java

// Java program to check whether number is
// superperfect or not

public class Divisors
{
    // Function to calculate sum of all divisors
    static int divSum(int num)
    {
        // Final result of summation of divisors
        int result = 0;
        
        // find all divisors which divides 'num'
        for (int i=1; i*i <= num; ++i)
        {
            // if 'i' is divisor of 'num'
            if (num%i == 0)
            {
                // if both divisors are same then add
                // it only once else add both
                if (i == (num/i))
                    result += i;
                else
                    result += (i + num/i);
            }
        }
        return result;
    }
    
    // Returns true if n is Super Perfect else false.
    static boolean isSuperPerfect(int n)
    {
        // Find the sum of all divisors of number n
        int n1 = divSum(n);
        // Again find the sum of all divisors of n1
        // and check if sum is equal to n1
        return (2*n == divSum(n1));
    }
    
    public static void main (String[] args)
    {
        int n = 16;
        System.out.printf((isSuperPerfect(n) ? "Yes\n" : "No\n"));
        
        n = 6;
        System.out.printf((isSuperPerfect(n) ? "Yes\n" : "No\n"));
    }
}

// This code is contributed by Saket Kumar

Python3

# Python program to check whether number
# is superperfect or not
import math

# Function to calculate sum of all divisors
def divSum(num):

    # Final result of summation of divisors
    result = 0

    # find all divisors which divides 'num'
    sq = int(math.sqrt(num))
    for i in range(1, sq+1):

        # if 'i' is divisor of 'num'
        if num %i == 0:

            # if both divisors are same then add
            # it only once else add both
            if i == (num//i):
                result += i
            else:
                result += (i + num//i)



    return result

# Returns true if n is superperfect else false
def isSuperPerfect(n):

    # Find the sum of all divisors of number n
    n1 = divSum(n)

    # Again find the sum of all divisors of n1
    return divSum(n1) == 2*n


#Driver code
n = 16
print ('Yes' if isSuperPerfect(n) else 'No')

n = 6
print ('Yes' if isSuperPerfect(n) else 'No')

// C# program to check whether number is
// superperfect or not
using System;

class Divisors
{
    // Function to calculate sum of all divisors
    static int divSum(int num)
    {
        // Final result of summation of divisors
        int result = 0;
        
        // find all divisors which divides 'num'
        for (int i = 1; i * i <= num; ++i)
        {
            // if 'i' is divisor of 'num'
            if (num % i == 0)
            {
                // if both divisors are same then add
                // it only once else add both
                if (i == (num / i))
                    result += i;
                else
                    result += (i + num / i);
            }
        }
        return result;
    }
    
    // Returns true if n is Super Perfect else false.
    static bool isSuperPerfect(int n)
    {
        // Find the sum of all divisors of number n
        int n1 = divSum(n);
        // Again find the sum of all divisors of n1
        // and check if sum is equal to n1
        return (2 * n == divSum(n1));
    }
    
    public static void Main ()
    {
        int n = 16;
        Console.WriteLine((isSuperPerfect(n) ? "Yes" : "No"));
        
        n = 6;
        Console.WriteLine((isSuperPerfect(n) ? "Yes" : "No"));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to check whether
// number is superperfect or not

// Function to calculate
// sum of all divisors
function divSum($num)
{
    
    // Final result of 
    // summation of divisors
    $result = 0;

    // find all divisors
    // which divides 'num'
    for ($i = 1; $i * $i <= $num; ++$i)
    {
        
        // if 'i' is divisor
        // of 'num'
        if ($num % $i == 0)
        {
            
            // if both divisors 
            // are same then add
            // it only once else 
            // add both
            if ($i == ($num / $i))
                $result += $i;
            else
                $result += ($i + $num/$i);
        }
    }

    return $result;
}

// Returns true if n is 
// Super Perfect else false.
function isSuperPerfect($n)
{
    
    // Find the sum of all 
    // divisors of number n
    $n1 = divSum($n);

    // Again find the sum 
    // of all divisors of n1
    // and check if sum is 
    // equal to n1
    return (2 * $n == divSum($n1));
}

    // Driver code
    $n = 16;
    $hh = (isSuperPerfect($n) ? "Yes\n" : "No\n");
        echo($hh);

    $n = 6;
    $hh=(isSuperPerfect($n) ? "Yes\n" : "No\n");
        echo($hh);

// This code is contributed by AJit
?>

JavaScript

<script>

// JavaScript program for the above approach
 
    // Function to calculate sum of all divisors 
    function divSum(num) 
    { 
    
        // Final result of summation of divisors 
        let result = 0; 
          
        // find all divisors which divides 'num' 
        for (let i = 1; i * i <= num; ++i) 
        { 
        
            // if 'i' is divisor of 'num' 
            if (num % i == 0) 
            { 
            
                // if both divisors are same then add 
                // it only once else add both 
                if (i == (num/i)) 
                    result += i; 
                else
                    result += (i + num/i); 
            } 
        } 
        return result; 
    } 
      
    // Returns true if n is Super Perfect else false. 
    function isSuperPerfect(n) 
    { 
    
        // Find the sum of all divisors of number n 
        let n1 = divSum(n); 
        
        // Again find the sum of all divisors of n1 
        // and check if sum is equal to n1 
        return (2*n == divSum(n1)); 
    } 

// Driver Code    
      let n = 16; 
      document.write((isSuperPerfect(n) ? "Yes\n" : "No\n") + "<br />"); 
          
       n = 6; 
      document.write((isSuperPerfect(n) ? "Yes\n" : "No\n") + "<br />");

// This code is contributed by splevel62.
</script>

Output:
Yes
No

Time complexity: O(sqrt(n + n1)) where n1 is sum of divisors of n.
Auxiliary space: O(1)
Facts about Supernumbers:

If n is an even superperfect number, then n must be a power of 2 i.e., 2^k such that 2^k+1 - 1 is a Mersenne prime.
It is not known whether there are any odd superperfect numbers. An odd superperfect number n would have to be a square number such that either n or ?(n) is divisible by at least three distinct primes. There are no odd superperfect numbers below 7×10²⁴

Reference:
https://en.wikipedia.org/wiki/Superperfect_number

Superperfect Number

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