若干阶乘除数之和

给定一个数字n，我们需要计算该数字的阶乘除数之和。

例子：

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

一个简单的解决方案是首先计算给定数的阶乘，然后计算阶乘的数除数。该解决方案效率不高，并且可能会因阶乘计算而导致溢出。

下面是上述方法的实现：

C++

// C++ program to find sum of proper divisor of 
// factorial of a number 
#include 
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 Code
int 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 
// 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 number
import 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 Code
n = 4
print(sumFactDiv(n))
  
# This code is contributed
# by Rajput-Ji

C#

// C# program to find sum of proper 
// divisor of factorial of a number
using 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.

输出 ：

60

一个有效的解决方案基于Legendre的公式。以下是步骤。

1. 查找所有小于或等于n(输入数字)的质数。我们可以为此使用筛网算法。令n为6。所有小于6的质数均为{2，3，5}。
2. 对于每个素数p，找出最大的幂除以n!。为此，我们在勒让德的公式公式中使用以下公式。

   除以p的最大乘方的值是每个项的下限值n / p + n /(p2)+ n /(p3)+……
   令这些值为exp1，exp2，exp3，.。使用上面的公式，我们得到n = 6的下面的值。

   • 2的最大幂除以6！，exp1 = 4。
   • 3的最大幂除以6！，exp2 = 2。
   • 5的最大幂除以6 !, exp3 = 1。
3. 结果基于除数函数。

   C++

   // C++ program to find sum of divisors in n!
   #include
   #include
   using namespace std;
     
   // allPrimes[] stores all prime numbers less
   // than or equal to n.
   vector allPrimes;
     
   // Fills above vector allPrimes[] for a given n
   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.
       vector 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 number
   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[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 cases
   int 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 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. 
       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 Code
   print(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
   

   输出：

   60