Almost Perfect Number

Last Updated : 25 Feb, 2024

Given a number n, check it is the Almost Perfect number or not. Almost perfect number is a natural number whose sum of all divisors including 1 and the number itself is equal to 2n - 1.
Example :

Input: n = 16
Output: Yes
Explanation: sum of divisors = 1 + 2 + 4 + 8 + 16 = 31 = 2n - 1

Input: n = 9
Output: No
Explanation: sum of divisors = 1 + 3 + 9 ? 2n - 1

C++

// CPP program to check if a number // is almost perfect. #include <bits/stdc++.h> using namespace std;  bool isAlmostperfect(int n) {     int divisors = 0;      for (int i = 1; i <= n; i++) {          // store sum of divisors of n         if (n % i == 0)             divisors += i;     }      // sum of divisors = 2*n - 1     if (divisors == 2 * n - 1)         return true;      return false; }  int main() {     int n = 16;     if (isAlmostperfect(n))         cout << "Yes";     else         cout << "No"; }

Java

// Java program to check if a  // number is almost perfect.  class GFG {      // Function to check number is  // almost perfect or not static boolean isAlmostperfect(int n) {     int divisors = 0;      for (int i = 1; i <= n; i++)     {          // store sum of divisors of n         if (n % i == 0)             divisors += i;     }      // sum of divisors = 2*n - 1     if (divisors == 2 * n - 1)         return true;      return false; }  // Driver Code public static void main(String[] args) {     int n = 16;     if (isAlmostperfect(n))         System.out.println("Yes");     else         System.out.println("No"); } }  // This code is contributed by // Smitha Dinesh Semwal.

Python3

# Python program to check if a number # is almost perfect. def isAlmostperfect(n):      divisors = 0     for i in range(1, n+1):          # store sum of divisors of n         if (n % i == 0):             divisors = divisors + i      # sum of divisors = 2*n - 1     if (divisors == 2 * n - 1):         return True     else:         return False  # Driver code n = 16 if (isAlmostperfect(n)):     print ("Yes") else:     print ("No")  # This code is contributed by # Manish Shaw (manishshaw1)

// C# program to check if a  // number is almost perfect. using System;  class GFG {          // Function to check number is      // almost perfect or not     static bool isAlmostperfect(int n)     {         int divisors = 0;              for (int i = 1; i <= n; i++)         {                  // store sum of divisors of n             if (n % i == 0)                 divisors += i;         }              // sum of divisors = 2 * n - 1         if (divisors == 2 * n - 1)             return true;              return false;     }      // Driver Code     static public void Main ()     {         int n = 16;                  if (isAlmostperfect(n))             Console.WriteLine("Yes");         else             Console.WriteLine("No");     } }  // This code is contributed by Ajit.

JavaScript

<script>  // Javascript program to check if a number // is almost perfect.  function isAlmostperfect( n) {     let divisors = 0;      for (let i = 1; i <= n; i++) {          // store sum of divisors of n         if (n % i == 0)             divisors += i;     }      // sum of divisors = 2*n - 1     if (divisors == 2 * n - 1)         return true;      return false; }      // Driver Code     let n = 16;     if (isAlmostperfect(n))         document.write("Yes");     else         document.write("No");         </script>

PHP

<?php // PHP program to check if a  // number is almost perfect.  // function to check  // almost perfect function isAlmostperfect($n) {     $divisors = 0;      for ($i = 1; $i <= $n; $i++)     {          // store sum of          // divisors of n         if ($n % $i == 0)             $divisors += $i;     }      // sum of divisors = 2*n - 1     if ($divisors == 2 * $n - 1)         return true;      return false; }  // Driver code $n = 16; if (isAlmostperfect($n))     echo("Yes"); else     echo("No");  // This code is contributed by Ajit. ?>

Output

Yes

The almost perfect numbers are found to be of the form 2^k(k = 0, 1, 2, 3, 4, ..). However it is not proved.
Time Complexity: O(n)
Auxiliary Space: O(1)

Almost Perfect Number

jaideeppyne1997

Improve

Article Tags :

Practice Tags :

Almost Perfect Number

Similar Reads