# Find the value of N when F(N) = f(a)+f(b) where a+b is the minimum possible and a*b = N

Given an integer N, the task is to find the value of F(N) if:

1. F(1) = 0
2. F(2) = 2
3. F(N) = 0, if N is odd prime.
4. F(N) = F(a) + F(b), where a and b are factors of N and (a + b) is minimum among all factors. Also a * b = N

Examples:

Input: N = 5
Output: 0
Since 5 is an odd prime.

Input: N = 4
Output: 4
4 can be written as 2 * 2, hence f(2) + f(2) = 4

Input: N = 20
Output: 4
20 can be written as f(4) + f(5), and f(4) can be written as f(2) + f(2), which is 4.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The following steps can be followed to solve the problem:

• If N is 1 or 2, the answer is 0 or 2 respectively.
• On breaking the recurrence f(n) = f(a) + f(b), we get it is the number of times a number is divisible by 2.
• The answer for f(n) is 2 * (number of times a number if divisible by 2)

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the value of F(N) ` `int` `getValueOfF(``int` `n) ` `{ ` ` `  `    ``// Base cases ` `    ``if` `(n == 1) ` `        ``return` `0; ` `    ``if` `(n == 2) ` `        ``return` `1; ` ` `  `    ``int` `cnt = 0; ` ` `  `    ``// Count the number of times a number ` `    ``// if divisible by 2 ` `    ``while` `(n % 2 == 0) { ` `        ``cnt += 1; ` `        ``n /= 2; ` `    ``} ` ` `  `    ``// Return the summation ` `    ``return` `2 * cnt; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 20; ` `    ``cout << getValueOfF(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to return the value of F(N) ` `static` `int` `getValueOfF(``int` `n) ` `{ ` ` `  `    ``// Base cases ` `    ``if` `(n == ``1``) ` `        ``return` `0``; ` `    ``if` `(n == ``2``) ` `        ``return` `1``; ` ` `  `    ``int` `cnt = ``0``; ` ` `  `    ``// Count the number of times a number ` `    ``// if divisible by 2 ` `    ``while` `(n % ``2` `== ``0``) ` `    ``{ ` `        ``cnt += ``1``; ` `        ``n /= ``2``; ` `    ``} ` ` `  `    ``// Return the summation ` `    ``return` `2` `* cnt; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `n = ``20``; ` `    ``System.out.println (getValueOfF(n)); ` `} ` `} ` ` `  `// This code is contributed by ajit. `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the value of F(N) ` `def` `getValueOfF(n): ` ` `  `    ``# Base cases ` `    ``if` `(n ``=``=` `1``): ` `        ``return` `0` `    ``if` `(n ``=``=` `2``): ` `        ``return` `1` ` `  `    ``cnt ``=` `0` ` `  `    ``# Count the number of times a number ` `    ``# if divisible by 2 ` `    ``while` `(n ``%` `2` `=``=` `0``): ` `        ``cnt ``+``=` `1` `        ``n ``/``=` `2` ` `  `    ``# Return the summation ` `    ``return` `2` `*` `cnt ` ` `  `# Driver code ` `n ``=` `20` `print``(getValueOfF(n)) ` ` `  `# This code is contributed by mohit kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `         `  `// Function to return the value of F(N) ` `static` `int` `getValueOfF(``int` `n) ` `{ ` ` `  `    ``// Base cases ` `    ``if` `(n == 1) ` `        ``return` `0; ` `    ``if` `(n == 2) ` `        ``return` `1; ` ` `  `    ``int` `cnt = 0; ` ` `  `    ``// Count the number of times a number ` `    ``// if divisible by 2 ` `    ``while` `(n % 2 == 0) ` `    ``{ ` `        ``cnt += 1; ` `        ``n /= 2; ` `    ``} ` ` `  `    ``// Return the summation ` `    ``return` `2 * cnt; ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main () ` `{ ` `    ``int` `n = 20; ` `    ``Console.WriteLine(getValueOfF(n)); ` `} ` `} ` ` `  `// This code is contributed by akt_mit.  `

## PHP

 ` `

Output:

```4
```

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Improved By : mohit kumar 29, jit_t

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