Almost Perfect Number

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 ` `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#

 `// 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. `

PHP

 ` `

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.

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