Given a number n, we need to calculate the sum of divisors of factorial of the number.
Examples:
Input : 4 Output : 60 Factorial of 4 is 24. Divisors of 24 are 1 2 3 4 6 8 12 24, sum of these is 60. Input : 6 Output : 2418
A Simple Solution is to first compute the factorial of the given number, then count the number divisors of the factorial. This solution is not efficient and may cause overflow due to factorial computation.
Below is the implementation of the above approach:
C++
// C++ program to find sum of proper divisor of // factorial of a number #include <bits/stdc++.h>using namespace std;// function to calculate factorial int fact(int n){ if (n == 0) return 1; return n * fact(n - 1);}// function to calculate sum of divisor int div(int x){ int ans = 0; for (int i = 1; i<= x; i++) if (x % i == 0) ans += i; return ans;}// Returns sum of divisors of n! int sumFactDiv(int n){ return div(fact(n));}// Driver Codeint main(){ int n = 4; cout << sumFactDiv(n);}// This code is contributed // by Akanksha Rai |
C
// C program to find sum of proper divisor of // factorial of a number #include <stdio.h>// function to calculate factorial int fact(int n) { if (n == 0) return 1; return n * fact(n - 1);}// function to calculate sum of divisor int div(int x) { int ans = 0; for (int i = 1; i<= x; i++) if (x % i == 0) ans += i; return ans;}// Returns sum of divisors of n! int sumFactDiv(int n) { return div(fact(n));}// driver program int main() { int n = 4; printf("%d",sumFactDiv(n));} |
Java
// Java program to find sum of proper divisor of// factorial of a numberimport java.io.*;import java.util.*;public class Division{ // function to calculate factorial static int fact(int n) { if (n == 0) return 1; return n*fact(n-1); } // function to calculate sum of divisor static int div(int x) { int ans = 0; for (int i = 1; i<= x; i++) if (x%i == 0) ans += i; return ans; } // Returns sum of divisors of n! static int sumFactDiv(int n) { return div(fact(n)); } // driver program public static void main(String args[]) { int n = 4; System.out.println(sumFactDiv(n)); }} |
Python3
# Python 3 program to find sum of proper # divisor of factorial of a number # function to calculate factorial def fact(n): if (n == 0): return 1 return n * fact(n - 1)# function to calculate sum # of divisor def div(x): ans = 0; for i in range(1, x + 1): if (x % i == 0): ans += i return ans# Returns sum of divisors of n! def sumFactDiv(n): return div(fact(n))# Driver Coden = 4print(sumFactDiv(n))# This code is contributed# by Rajput-Ji |
C#
// C# program to find sum of proper // divisor of factorial of a numberusing System;class Division { // function to calculate factorial static int fac(int n) { if (n == 0) return 1; return n * fac(n - 1); } // function to calculate // sum of divisor static int div(int x) { int ans = 0; for (int i = 1; i <= x; i++) if (x % i == 0) ans += i; return ans; } // Returns sum of divisors of n! static int sumFactDiv(int n) { return div(fac(n)); } // Driver Code public static void Main() { int n = 4; Console.Write(sumFactDiv(n)); }} // This code is contributed by Nitin Mittal. |
PHP
<?php// PHP program to find sum of proper // divisor of factorial of a number // function to calculate factorial function fact($n){ if ($n == 0) return 1; return $n * fact($n - 1);}// function to calculate sum of divisor function div($x){ $ans = 0; for ($i = 1; $i<= $x; $i++) if ($x % $i == 0) $ans += $i; return $ans;}// Returns sum of divisors of n! function sumFactDiv($n){ return div(fact($n));}// Driver Code$n = 4;echo sumFactDiv($n);// This code is contributed // by Akanksha Rai?> |
Javascript
<script>// JavaScript program to find// sum of proper divisor of// factorial of a number// function to calculate factorialfunction fact(n){ if (n == 0) return 1; return n * fact(n - 1);}// function to calculate sum of divisorfunction div(x){ let ans = 0; for (let i = 1; i<= x; i++) if (x % i == 0) ans += i; return ans;}// Returns sum of divisors of n!function sumFactDiv(n){ return div(fact(n));}// Driver Code let n = 4; document.write(sumFactDiv(n));// This code is contributed by Surbhi Tyagi.</script> |
Output :
60
Time Complexity: O(n!)
Auxiliary Space: O(1)
An efficient solution is based on Legendre’s formula. Below are the steps.
- Find all prime numbers less than or equal to n (input number). We can use Sieve Algorithm for this. Let n be 6. All prime numbers less than 6 are {2, 3, 5}.
- For each prime number, p find the largest power of it that divides n!. We use Legendre’s formula for this purpose.
- The largest power of 2 that divides 6!, exp1 = 4.
- The largest power of 3 that divides 6!, exp2 = 2.
- The largest power of 5 that divides 6!, exp3 = 1.
- The result is based on the Divisor Function.
C++
// C++ program to find sum of divisors in n!#include<bits/stdc++.h>#include<math.h>using namespace std;// allPrimes[] stores all prime numbers less// than or equal to n.vector<int> allPrimes;// Fills above vector allPrimes[] for a given nvoid sieve(int n){ // 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. vector<bool> prime(n+1, true); // Loop to update prime[] 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; } } // Store primes in the vector allPrimes for (int p = 2; p <= n; p++) if (prime[p]) allPrimes.push_back(p);}// Function to find all result of factorial numberint factorialDivisors(int n){ sieve(n); // create sieve // Initialize result int result = 1; // find exponents of all primes which divides n // and less than n for (int i = 0; i < allPrimes.size(); i++) { // Current divisor int p = allPrimes[i]; // Find the highest power (stored in exp)' // of allPrimes[i] that divides n using // Legendre's formula. int exp = 0; while (p <= n) { exp = exp + (n/p); p = p*allPrimes[i]; } // Using the divisor function to calculate // the sum result = result*(pow(allPrimes[i], exp+1)-1)/ (allPrimes[i]-1); } // return total divisors return result;}// Driver program to run the casesint main(){ cout << factorialDivisors(4); return 0;} |
Java
// Java program to find sum of divisors in n! import java.util.*;class GFG{// allPrimes[] stores all prime numbers less // than or equal to n. static ArrayList<Integer> allPrimes=new ArrayList<Integer>(); // Fills above vector allPrimes[] for a given n static void sieve(int n) { // 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. boolean[] prime=new boolean[n+1]; // Loop to update prime[] for (int p = 2; p*p <= n; p++) { // If prime[p] is not changed, then it // is a prime if (prime[p] == false) { // Update all multiples of p for (int i = p*2; i <= n; i += p) prime[i] = true; } } // Store primes in the vector allPrimes for (int p = 2; p <= n; p++) if (prime[p]==false) allPrimes.add(p); } // Function to find all result of factorial number static int factorialDivisors(int n) { sieve(n); // create sieve // Initialize result int result = 1; // find exponents of all primes which divides n // and less than n for (int i = 0; i < allPrimes.size(); i++) { // Current divisor int p = allPrimes.get(i); // Find the highest power (stored in exp)' // of allPrimes[i] that divides n using // Legendre's formula. int exp = 0; while (p <= n) { exp = exp + (n/p); p = p*allPrimes.get(i); } // Using the divisor function to calculate // the sum result = result*((int)Math.pow(allPrimes.get(i), exp+1)-1)/ (allPrimes.get(i)-1); } // return total divisors return result; } // Driver program to run the cases public static void main(String[] args) { System.out.println(factorialDivisors(4)); } }// This code is contributed by mits |
Python3
# Python3 program to find sum of divisors in n! # allPrimes[] stores all prime numbers # less than or equal to n. allPrimes = []; # Fills above vector allPrimes[]# for a given n def sieve(n): # 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. prime = [True] * (n + 1); # Loop to update prime[] 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 for i in range(p * 2, n + 1, p): prime[i] = False; p += 1; # Store primes in the vector allPrimes for p in range(2, n + 1): if (prime[p]): allPrimes.append(p); # Function to find all result of factorial number def factorialDivisors(n): sieve(n); # create sieve # Initialize result result = 1; # find exponents of all primes which # divides n and less than n for i in range(len(allPrimes)): # Current divisor p = allPrimes[i]; # Find the highest power (stored in exp)' # of allPrimes[i] that divides n using # Legendre's formula. exp = 0; while (p <= n): exp = exp + int(n / p); p = p * allPrimes[i]; # Using the divisor function to # calculate the sum result = int(result * (pow(allPrimes[i], exp + 1) - 1) / (allPrimes[i] - 1)); # return total divisors return result; # Driver Codeprint(factorialDivisors(4));# This code is contributed by mits |
C#
// C# program to find sum of divisors in n! using System;using System.Collections;class GFG{// allPrimes[] stores all prime numbers less // than or equal to n. static ArrayList allPrimes=new ArrayList(); // Fills above vector allPrimes[] for a given n static void sieve(int n) { // 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. bool[] prime=new bool[n+1]; // Loop to update prime[] for (int p = 2; p*p <= n; p++) { // If prime[p] is not changed, then it // is a prime if (prime[p] == false) { // Update all multiples of p for (int i = p*2; i <= n; i += p) prime[i] = true; } } // Store primes in the vector allPrimes for (int p = 2; p <= n; p++) if (prime[p]==false) allPrimes.Add(p); } // Function to find all result of factorial number static int factorialDivisors(int n) { sieve(n); // create sieve // Initialize result int result = 1; // find exponents of all primes which divides n // and less than n for (int i = 0; i < allPrimes.Count; i++) { // Current divisor int p = (int)allPrimes[i]; // Find the highest power (stored in exp)' // of allPrimes[i] that divides n using // Legendre's formula. int exp = 0; while (p <= n) { exp = exp + (n/p); p = p*(int)allPrimes[i]; } // Using the divisor function to calculate // the sum result = result*((int)Math.Pow((int)allPrimes[i], exp+1)-1)/ ((int)allPrimes[i]-1); } // return total divisors return result; } // Driver program to run the cases static void Main() { Console.WriteLine(factorialDivisors(4)); } }// This code is contributed by mits |
PHP
<?php// PHP program to find sum of divisors in n! // allPrimes[] stores all prime numbers less // than or equal to n. $allPrimes=array(); // Fills above vector allPrimes[] for a given n function sieve($n) { global $allPrimes; // 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. $prime=array_fill(0,$n+1,true); // Loop to update prime[] 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; } } // Store primes in the vector allPrimes for ($p = 2; $p <= $n; $p++) if ($prime[$p]) array_push($allPrimes,$p); } // Function to find all result of factorial number function factorialDivisors($n) { global $allPrimes; sieve($n); // create sieve // Initialize result $result = 1; // find exponents of all primes which divides n // and less than n for ($i = 0; $i < count($allPrimes); $i++) { // Current divisor $p = $allPrimes[$i]; // Find the highest power (stored in exp)' // of allPrimes[i] that divides n using // Legendre's formula. $exp = 0; while ($p <= $n) { $exp = $exp + (int)($n/$p); $p = $p*$allPrimes[$i]; } // Using the divisor function to calculate // the sum $result = $result*(pow($allPrimes[$i], $exp+1)-1)/ ($allPrimes[$i]-1); } // return total divisors return $result; } // Driver program to run the cases print(factorialDivisors(4));// This code is contributed by mits?> |
Javascript
<script>// Javascript program to find sum of divisors in n!// allPrimes[] stores all prime numbers less// than or equal to n.let allPrimes=[];// Fills above vector allPrimes[] for a given nfunction sieve(n){ // 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. let prime = new Array(n + 1); for(let i = 0; i < n + 1; i++) prime[i] = true; // Loop to update prime[] for(let 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(let i = p * 2; i <= n; i += p) prime[i] = false; } } // Store primes in the vector allPrimes for(let p = 2; p <= n; p++) if (prime[p]) allPrimes.push(p);}// Function to find all result of// factorial numberfunction factorialDivisors( n){ // Create sieve sieve(n); // Initialize result let result = 1; // Find exponents of all primes which // divides n and less than n for(let i = 0; i < allPrimes.length; i++) { // Current divisor let p = allPrimes[i]; // Find the highest power (stored in exp)' // of allPrimes[i] that divides n using // Legendre's formula. let exp = 0; while (p <= n) { exp = exp + Math.floor(n/p); p = p*allPrimes[i]; } // Using the divisor function to calculate // the sum result = Math.floor(result * (Math.pow( allPrimes[i], exp + 1) - 1) / (allPrimes[i] - 1)); } // Return total divisors return result;}// Driver codedocument.write(factorialDivisors(4));// This code is contributed by rag2127</script> |
Output:
60
Time Complexity: O(n*log(log(n)))
Auxiliary Space: O(n)
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