Count of Fibonacci divisors of a given number

Given a number N, the task is to find the number of divisors of N which belongs to the fibonacci series.

Examples:

Input: N = 12
Output: 3
Explanation:
1, 2 and 3 are the 3 divisors of 12 which are in the Fibonacci series.
Hence, the answer is 3.

Input: N = 110
Output: 4
Explanation:
1, 2, 5 and 55 are 4 divisors of 110 which are in the Fibonacci series.

Efficient Approach:



  1. Create a hash table to store all the Fibonacci numbers till N, for checking in O(1) time.
  2. Find all divisors of N in O(∛N)
  3. For each divisor, check if it is a Fibonacci number as well. Count the number of such divisors and print them.

Below is the implementation of the above approach:

C++

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// C++ program to count number of divisors
// of N which are Fibonacci numbers
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to create hash table
// to check Fibonacci numbers
void createHash(
    set<int>& hash, int maxElement)
{
    int prev = 0, curr = 1;
    hash.insert(prev);
    hash.insert(curr);
  
    while (curr <= maxElement) {
        int temp = curr + prev;
        hash.insert(temp);
        prev = curr;
        curr = temp;
    }
}
  
// Function to count number of divisors
// of N which are fibonacci numbers
int countFibonacciDivisors(int n)
{
    set<int> hash;
    createHash(hash, n);
  
    int cnt = 0;
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
  
            // If divisors are equal,
            // check and count only one
            if ((n / i == i)
                && (hash.find(n / i)
                    != hash.end()))
                cnt++;
  
            // Otherwise check and count both
            else {
                if (hash.find(n / i)
                    != hash.end())
                    cnt++;
                if (hash.find(n / (n / i))
                    != hash.end())
                    cnt++;
            }
        }
    }
    return cnt;
}
  
// Driver code
int main()
{
    int n = 12;
  
    cout << countFibonacciDivisors(n);
  
    return 0;
}

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Java

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// Java program to count number of divisors
// of N which are Fibonacci numbers
import java.util.*;
  
class GFG{
   
// Function to create hash table
// to check Fibonacci numbers
static void createHash(
    HashSet<Integer> hash, int maxElement)
{
    int prev = 0, curr = 1;
    hash.add(prev);
    hash.add(curr);
   
    while (curr <= maxElement) {
        int temp = curr + prev;
        hash.add(temp);
        prev = curr;
        curr = temp;
    }
}
   
// Function to count number of divisors
// of N which are fibonacci numbers
static int countFibonacciDivisors(int n)
{
    HashSet<Integer> hash = new HashSet<Integer>();
    createHash(hash, n);
   
    int cnt = 0;
    for (int i = 1; i <= Math.sqrt(n); i++) {
        if (n % i == 0) {
   
            // If divisors are equal,
            // check and count only one
            if ((n / i == i)
                && (hash.contains(n / i)))
                cnt++;
   
            // Otherwise check and count both
            else {
                if (hash.contains(n / i))
                    cnt++;
                if (hash.contains(n / (n / i)))
                    cnt++;
            }
        }
    }
    return cnt;
}
   
// Driver code
public static void main(String[] args)
{
    int n = 12;
   
    System.out.print(countFibonacciDivisors(n)); 
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 program to count number of divisors
# of N which are Fibonacci numbers
from math import sqrt,ceil,floor
  
# Function to create hash table
# to check Fibonacci numbers
def createHash(maxElement):
    prev = 0
    curr = 1
    d = dict()
    d[prev] = 1
    d[curr] = 1
  
    while (curr <= maxElement):
        temp = curr + prev
        d[temp] = 1
        prev = curr
        curr = temp
    return d
  
# Function to count number of divisors
# of N which are fibonacci numbers
def countFibonacciDivisors(n):
    hash = createHash(n)
  
    cnt = 0
    for i in range(1, ceil(sqrt(n))):
        if (n % i == 0):
  
            # If divisors are equal,
            # check and count only one
            if ((n // i == i)
                and (n // i in hash)):
                cnt += 1
  
            # Otherwise check and count both
            else:
                if (n // i in hash):
                    cnt += 1
                if (n // (n // i) in hash):
                    cnt += 1
    return cnt
  
# Driver code
n = 12
  
print(countFibonacciDivisors(n))
  
# This code is contriuted by mohit kumar 29

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C#

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// C# program to count number of divisors
// of N which are Fibonacci numbers
using System;
using System.Collections.Generic;
  
class GFG{
    
// Function to create hash table
// to check Fibonacci numbers
static void createHash(
    HashSet<int> hash, int maxElement)
{
    int prev = 0, curr = 1;
    hash.Add(prev);
    hash.Add(curr);
    
    while (curr <= maxElement) {
        int temp = curr + prev;
        hash.Add(temp);
        prev = curr;
        curr = temp;
    }
}
    
// Function to count number of divisors
// of N which are fibonacci numbers
static int countFibonacciDivisors(int n)
{
    HashSet<int> hash = new HashSet<int>();
    createHash(hash, n);
    
    int cnt = 0;
    for (int i = 1; i <= Math.Sqrt(n); i++) {
        if (n % i == 0) {
    
            // If divisors are equal,
            // check and count only one
            if ((n / i == i)
                && (hash.Contains(n / i)))
                cnt++;
    
            // Otherwise check and count both
            else {
                if (hash.Contains(n / i))
                    cnt++;
                if (hash.Contains(n / (n / i)))
                    cnt++;
            }
        }
    }
    return cnt;
}
    
// Driver code
public static void Main(String[] args)
{
    int n = 12;
    
    Console.Write(countFibonacciDivisors(n)); 
}
}
   
// This code is contributed by PrinciRaj1992

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Output:

3

Time Complexity: O(∛N)

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