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.

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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.

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Python3

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# 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

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C#

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// 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. 

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PHP

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<?php
//PHP implementation of the approach
// Function to return the value of F(N)
  
function 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
    $n = 20;
    echo getValueOfF($n);
  
// This code is contributed by Tushil..
?>

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Output:

4


My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

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Improved By : mohit kumar 29, jit_t