Count Divisors of n in O(n^1/3)
Given a number n, count all distinct divisors of it.
Examples:
Input : 18 Output : 6 Divisors of 18 are 1, 2, 3, 6, 9 and 18. Input : 100 Output : 9 Divisors of 100 are 1, 2, 4, 5, 10, 20, 25, 50 and 100
A Naive Solution would be to iterate all the numbers from 1 to sqrt(n), checking if that number divides n and incrementing number of divisors. This approach takes O(sqrt(n)) time.
C++
// C implementation of Naive method to count all // divisors #include <bits/stdc++.h> using namespace std; // function to count the divisors int countDivisors(int n) { int cnt = 0; for (int i = 1; i <= sqrt(n); i++) { if (n % i == 0) { // If divisors are equal, // count only one if (n / i == i) cnt++; else // Otherwise count both cnt = cnt + 2; } } return cnt; } /* Driver program to test above function */int main() { printf("Total distinct divisors of 100 are : %d", countDivisors(100)); return 0; } |
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Java
// JAVA implementation of Naive method // to count all divisors import java.io.*; import java.math.*; class GFG { // function to count the divisors static int countDivisors(int n) { int cnt = 0; for (int i = 1; i <= Math.sqrt(n); i++) { if (n % i == 0) { // If divisors are equal, // count only one if (n / i == i) cnt++; else // Otherwise count both cnt = cnt + 2; } } return cnt; } /* Driver program to test above function */ public static void main(String args[]) { System.out.println("Total distinct " + "divisors of 100 are : " + countDivisors(100)); } } /*This code is contributed by Nikita Tiwari.*/ |
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Python3
# Python3 implementation of Naive method # to count all divisors import math # function to count the divisors def countDivisors(n) : cnt = 0 for i in range(1, (int)(math.sqrt(n)) + 1) : if (n % i == 0) : # If divisors are equal, # count only one if (n / i == i) : cnt = cnt + 1 else : # Otherwise count both cnt = cnt + 2 return cnt # Driver program to test above function */ print("Total distinct divisors of 100 are : ", countDivisors(100)) # This code is contributed by Nikita Tiwari. |
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C#
// C# implementation of Naive method // to count all divisors using System; class GFG { // function to count the divisors static int countDivisors(int n) { int cnt = 0; for (int i = 1; i <= Math.Sqrt(n); i++) { if (n % i == 0) { // If divisors are equal, // count only one if (n / i == i) cnt++; // Otherwise count both else cnt = cnt + 2; } } return cnt; } // Driver program public static void Main() { Console.WriteLine("Total distinct" + " divisors of 100 are : " + countDivisors(100)); } } // This code is contributed by anuj_67. |
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PHP
<?php // PHP implementation of Naive // method to count all divisors // function to count the divisors function countDivisors($n) { $cnt = 0; for ($i = 1; $i <= sqrt($n); $i++) { if ($n % $i == 0) { // If divisors are equal, // count only one if ($n / $i == $i) $cnt++; // Otherwise count both else $cnt = $cnt + 2; } } return $cnt; } // Driver Code echo "Total distinct divisors of 100 are : ", countDivisors(100); // This code is contributed by Ajit ?> |
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Output :
Total distinct divisors of 100 are : 9
Optimized Solution (O(n^1/3))
- Split number n in two numbers x and y such that n=x*y where x contains only prime factors in range 2 <= x <= n(1/3) and y deals with higher prime factors greater than n(1/3).
- Count total factors of x using the naive trial division method. Let this count be F(x).
- Count total factors of y using the following three cases. Let this count be F(y).
- If y is a prime number then factors will be 1 and y itself. That implies, F(y) = 2.
- If y is square of a prime number, then factors will be 1, sqrt(y) and y itself. That implies, F(y) = 3.
- If y is the product of two distinct prime numbers, then factors will be 1, both prime numbers and number y itself. That implies, F(y) = 4.
- Since F(x*y) is a multiplicative function and gcd(x, y) = 1, that implies, F(x*y) = F(x)*F(y) which gives the count of total distinct divisors of n.
Note that, we have only these three cases for calculating factors of y since there can be at max two prime factors of y. If it would have had more than two prime factors, one of them would surely have been less than equal to n(1/3), and hence it would be included in x and not in y.
C++
// C++ program to count distinct divisors // of a given number n #include <bits/stdc++.h> using namespace std; void SieveOfEratosthenes(int n, bool prime[], bool primesquare[], int a[]) { // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // Not a prime, else true. for (int i = 2; i <= n; i++) prime[i] = true; // Create a boolean array "primesquare[0..n*n+1]" // and initialize all entries it as false. A value // in squareprime[i] will finally be true if i is // square of prime, else false. for (int i = 0; i <= (n * n + 1); i++) primesquare[i] = false; // 1 is not a prime number prime[1] = false; for (int p = 2; p * p <= n; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == true) { // Update all multiples of p for (int i = p * 2; i <= n; i += p) prime[i] = false; } } int j = 0; for (int p = 2; p <= n; p++) { if (prime[p]) { // Storing primes in an array a[j] = p; // Update value in primesquare[p*p], // if p is prime. primesquare[p * p] = true; j++; } } } // Function to count divisors int countDivisors(int n) { // If number is 1, then it will have only 1 // as a factor. So, total factors will be 1. if (n == 1) return 1; bool prime[n + 1], primesquare[n * n + 1]; int a[n]; // for storing primes upto n // Calling SieveOfEratosthenes to store prime // factors of n and to store square of prime // factors of n SieveOfEratosthenes(n, prime, primesquare, a); // ans will contain total number of distinct // divisors int ans = 1; // Loop for counting factors of n for (int i = 0;; i++) { // a[i] is not less than cube root n if (a[i] * a[i] * a[i] > n) break; // Calculating power of a[i] in n. int cnt = 1; // cnt is power of prime a[i] in n. while (n % a[i] == 0) // if a[i] is a factor of n { n = n / a[i]; cnt = cnt + 1; // incrementing power } // Calculating the number of divisors // If n = a^p * b^q then total divisors of n // are (p+1)*(q+1) ans = ans * cnt; } // if a[i] is greater than cube root of n // First case if (prime[n]) ans = ans * 2; // Second case else if (primesquare[n]) ans = ans * 3; // Third case else if (n != 1) ans = ans * 4; return ans; // Total divisors } // Driver Program int main() { cout << "Total distinct divisors of 100 are : " << countDivisors(100) << endl; return 0; } |
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Java
// JAVA program to count distinct // divisors of a given number n import java.io.*; class GFG { static void SieveOfEratosthenes(int n, boolean prime[], boolean primesquare[], int a[]) { // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // Not a prime, else true. for (int i = 2; i <= n; i++) prime[i] = true; /* Create a boolean array "primesquare[0..n*n+1]" and initialize all entries it as false. A value in squareprime[i] will finally be true if i is square of prime, else false.*/ for (int i = 0; i < ((n * n) + 1); i++) primesquare[i] = false; // 1 is not a prime number prime[1] = false; for (int p = 2; p * p <= n; p++) { // If prime[p] is not changed, // then it is a prime if (prime[p] == true) { // Update all multiples of p for (int i = p * 2; i <= n; i += p) prime[i] = false; } } int j = 0; for (int p = 2; p <= n; p++) { if (prime[p]) { // Storing primes in an array a[j] = p; // Update value in // primesquare[p*p], // if p is prime. primesquare[p * p] = true; j++; } } } // Function to count divisors static int countDivisors(int n) { // If number is 1, then it will // have only 1 as a factor. So, // total factors will be 1. if (n == 1) return 1; boolean prime[] = new boolean[n + 1]; boolean primesquare[] = new boolean[(n * n) + 1]; // for storing primes upto n int a[] = new int[n]; // Calling SieveOfEratosthenes to // store prime factors of n and to // store square of prime factors of n SieveOfEratosthenes(n, prime, primesquare, a); // ans will contain total number // of distinct divisors int ans = 1; // Loop for counting factors of n for (int i = 0;; i++) { // a[i] is not less than cube root n if (a[i] * a[i] * a[i] > n) break; // Calculating power of a[i] in n. // cnt is power of prime a[i] in n. int cnt = 1; // if a[i] is a factor of n while (n % a[i] == 0) { n = n / a[i]; // incrementing power cnt = cnt + 1; } // Calculating the number of divisors // If n = a^p * b^q then total // divisors of n are (p+1)*(q+1) ans = ans * cnt; } // if a[i] is greater than cube root // of n // First case if (prime[n]) ans = ans * 2; // Second case else if (primesquare[n]) ans = ans * 3; // Third case else if (n != 1) ans = ans * 4; return ans; // Total divisors } // Driver Program public static void main(String args[]) { System.out.println("Total distinct divisors" + " of 100 are : " + countDivisors(100)); } } /*This code is contributed by Nikita Tiwari*/ |
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Python3
# Python3 program to count distinct # divisors of a given number n def SieveOfEratosthenes(n, prime,primesquare, a): # Create a boolean array "prime[0..n]" # and initialize all entries it as # true. A value in prime[i] will finally # be false if i is not a prime, else true. for i in range(2,n+1): prime[i] = True # Create a boolean array "primesquare[0..n*n+1]" # and initialize all entries it as false. # A value in squareprime[i] will finally be # true if i is square of prime, else false. for i in range((n * n + 1)+1): primesquare[i] = False # 1 is not a prime number prime[1] = False p = 2 while(p * p <= n): # If prime[p] is not changed, # then it is a prime if (prime[p] == True): # Update all multiples of p i = p * 2 while(i <= n): prime[i] = False i += p p+=1 j = 0 for p in range(2,n+1): if (prime[p]==True): # Storing primes in an array a[j] = p # Update value in primesquare[p*p], # if p is prime. primesquare[p * p] = True j+=1 # Function to count divisors def countDivisors(n): # If number is 1, then it will # have only 1 as a factor. So, # total factors will be 1. if (n == 1): return 1 prime = [False]*(n + 2) primesquare = [False]*(n * n + 2) # for storing primes upto n a = [0]*n # Calling SieveOfEratosthenes to # store prime factors of n and to # store square of prime factors of n SieveOfEratosthenes(n, prime, primesquare, a) # ans will contain total # number of distinct divisors ans = 1 # Loop for counting factors of n i=0 while(1): # a[i] is not less than cube root n if(a[i] * a[i] * a[i] > n): break # Calculating power of a[i] in n. cnt = 1 # cnt is power of # prime a[i] in n. while (n % a[i] == 0): # if a[i] is a factor of n n = n / a[i] cnt = cnt + 1 # incrementing power # Calculating number of divisors # If n = a^p * b^q then total # divisors of n are (p+1)*(q+1) ans = ans * cnt i+=1 # if a[i] is greater than # cube root of n n=int(n) # First case if (prime[n]==True): ans = ans * 2 # Second case elif (primesquare[n]==True): ans = ans * 3 # Third case elif (n != 1): ans = ans * 4 return ans # Total divisors # Driver Code if __name__=='__main__': print("Total distinct divisors of 100 are :",countDivisors(100)) # This code is contributed # by mits |
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C#
// C# program to count distinct // divisors of a given number n using System; class GFG { static void SieveOfEratosthenes(int n, bool[] prime, bool[] primesquare, int[] a) { // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // Not a prime, else true. for (int i = 2; i <= n; i++) prime[i] = true; /* Create a boolean array "primesquare[0..n*n+1]" and initialize all entries it as false. A value in squareprime[i] will finally be true if i is square of prime, else false.*/ for (int i = 0; i < ((n * n) + 1); i++) primesquare[i] = false; // 1 is not a prime number prime[1] = false; for (int p = 2; p * p <= n; p++) { // If prime[p] is not changed, // then it is a prime if (prime[p] == true) { // Update all multiples of p for (int i = p * 2; i <= n; i += p) prime[i] = false; } } int j = 0; for (int p = 2; p <= n; p++) { if (prime[p]) { // Storing primes in an array a[j] = p; // Update value in // primesquare[p*p], // if p is prime. primesquare[p * p] = true; j++; } } } // Function to count divisors static int countDivisors(int n) { // If number is 1, then it will // have only 1 as a factor. So, // total factors will be 1. if (n == 1) return 1; bool[] prime = new bool[n + 1]; bool[] primesquare = new bool[(n * n) + 1]; // for storing primes upto n int[] a = new int[n]; // Calling SieveOfEratosthenes to // store prime factors of n and to // store square of prime factors of n SieveOfEratosthenes(n, prime, primesquare, a); // ans will contain total number // of distinct divisors int ans = 1; // Loop for counting factors of n for (int i = 0;; i++) { // a[i] is not less than cube root n if (a[i] * a[i] * a[i] > n) break; // Calculating power of a[i] in n. // cnt is power of prime a[i] in n. int cnt = 1; // if a[i] is a factor of n while (n % a[i] == 0) { n = n / a[i]; // incrementing power cnt = cnt + 1; } // Calculating the number of divisors // If n = a^p * b^q then total // divisors of n are (p+1)*(q+1) ans = ans * cnt; } // if a[i] is greater than cube root // of n // First case if (prime[n]) ans = ans * 2; // Second case else if (primesquare[n]) ans = ans * 3; // Third case else if (n != 1) ans = ans * 4; return ans; // Total divisors } // Driver Program public static void Main() { Console.Write("Total distinct divisors" + " of 100 are : " + countDivisors(100)); } } // This code is contributed by parashar. |
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PHP
<?php // PHP program to count distinct // divisors of a given number n function SieveOfEratosthenes($n, &$prime, &$primesquare, &$a) { // Create a boolean array "prime[0..n]" // and initialize all entries it as // true. A value in prime[i] will finally // be false if i is not a prime, else true. for ($i = 2; $i <= $n; $i++) $prime[$i] = true; // Create a boolean array "primesquare[0..n*n+1]" // and initialize all entries it as false. // A value in squareprime[i] will finally be // true if i is square of prime, else false. for ($i = 0; $i <= ($n * $n + 1); $i++) $primesquare[$i] = false; // 1 is not a prime number $prime[1] = false; for ($p = 2; $p * $p <= $n; $p++) { // If prime[p] is not changed, // then it is a prime if ($prime[$p] == true) { // Update all multiples of p for ($i = $p * 2; $i <= $n; $i += $p) $prime[$i] = false; } } $j = 0; for ($p = 2; $p <= $n; $p++) { if ($prime[$p]) { // Storing primes in an array $a[$j] = $p; // Update value in primesquare[p*p], // if p is prime. $primesquare[$p * $p] = true; $j++; } } } // Function to count divisors function countDivisors($n) { // If number is 1, then it will // have only 1 as a factor. So, // total factors will be 1. if ($n == 1) return 1; $prime = array_fill(false, $n + 1, NULL); $primesquare = array_fill(false, $n * $n + 1, NULL); // for storing primes upto n $a = array_fill(0, $n, NULL); // Calling SieveOfEratosthenes to // store prime factors of n and to // store square of prime factors of n SieveOfEratosthenes($n, $prime, $primesquare, $a); // ans will contain total // number of distinct divisors $ans = 1; // Loop for counting factors of n for ($i = 0;; $i++) { // a[i] is not less than cube root n if ($a[$i] * $a[$i] * $a[$i] > $n) break; // Calculating power of a[i] in n. $cnt = 1; // cnt is power of // prime a[i] in n. while ($n % $a[$i] == 0) // if a[i] is a // factor of n { $n = $n / $a[$i]; $cnt = $cnt + 1; // incrementing power } // Calculating the number of divisors // If n = a^p * b^q then total // divisors of n are (p+1)*(q+1) $ans = $ans * $cnt; } // if a[i] is greater than // cube root of n // First case if ($prime[$n]) $ans = $ans * 2; // Second case else if ($primesquare[$n]) $ans = $ans * 3; // Third case else if ($n != 1) $ans = $ans * 4; return $ans; // Total divisors } // Driver Code echo "Total distinct divisors of 100 are : ". countDivisors(100). "\n"; // This code is contributed // by ChitraNayal ?> |
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Output :
Total distinct divisors of 100 are : 9
Time Complexity: O(n1/3)
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