当F(N)= f(a)+ f(b)时找到N的值，其中a + b是最小可能值，而a * b = N

给定整数N ，任务是在以下情况下找到F(N)的值：

1. F(1)= 0
2. F(2)= 2
3. 如果N是奇质数，则F(N)= 0。
4. F(N)= F(a)+ F(b)，其中a和b是N的因数，而(a + b)最小。也是a * b = N

例子：

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.

方法：可以按照以下步骤解决问题：

• 如果N为1或2，则答案分别为0或2。
• 在打破递归f(n)= f(a)+ f(b)时，我们得到它是一个被2整除的次数。
• f(n)的答案是2 *(如果被2整除，则为该数字的倍数)

下面是上述方法的实现：

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.

输出：

4