使用素数分解的数的因子之和

给定数字N。任务是找到给定数字N的所有因子的总和。

例子：

Input : N = 12
Output : 28
All factors of 12 are: 1,2,3,4,6,12

Input : 60
Output : 168

方法：假设N = 1100，其思想是首先找到给定数N的素因式分解。
因此，素数分解为1100 = 2 ² * 5 ² * 11。

因此，计算所有因素之和的公式可以表示为：

(2⁰ + 2¹ + 2²) * (5⁰ + 5¹ + 5²) * (11⁰ + 11¹)
(upto the power of factor in factorization i.e. power of 2 and 5 is 2 and 11 is 1.)
= (1 + 2 + 2²) * (1 + 5 + 5²) * (1 + 11)
= 7 * 31 * 12
= 2604

So, sum of all factors of 1100 = 2604

下面是上述方法的实现：

C++

// C++ Program to find sum of all 
// factors of a given number
  
#include 
  
using namespace std;
  
// Using SieveOfEratosthenes to find smallest prime
// factor of all the numbers.
// For example, if N is 10,
// s[2] = s[4] = s[6] = s[10] = 2
// s[3] = s[9] = 3
// s[5] = 5
// s[7] = 7
void sieveOfEratosthenes(int N, int s[])
{
    // Create a boolean array "prime[0..n]" and
    // initialize all entries in it as false.
    vector prime(N + 1, false);
  
    // Initializing smallest factor equal to 2
    // for all the even numbers
    for (int i = 2; i <= N; i += 2)
        s[i] = 2;
  
    // For odd numbers less then equal to n
    for (int i = 3; i <= N; i += 2) {
        if (prime[i] == false) {
  
            // s(i) for a prime is the number itself
            s[i] = i;
  
            // For all multiples of current prime number
            for (int j = i; j * i <= N; j += 2) {
                if (prime[i * j] == false) {
                    prime[i * j] = true;
  
                    // i is the smallest prime factor for
                    // number "i*j".
                    s[i * j] = i;
                }
            }
        }
    }
}
  
// Function to find sum of all prime factors
int findSum(int N)
{
    // Declaring array to store smallest prime
    // factor of i at i-th index
    int s[N + 1];
  
    int ans = 1;
  
    // Filling values in s[] using sieve
    sieveOfEratosthenes(N, s);
      
    int currFactor = s[N]; // Current prime factor of N
    int power = 1; // Power of current prime factor
  
    while (N > 1) {
        N /= s[N];
  
        // N is now N/s[N]. If new N als has smallest
        // prime factor as currFactor, increment power
        if (currFactor == s[N]) {
            power++;
            continue;
        }
  
        int sum = 0;
          
        for(int i=0; i<=power; i++)
            sum += pow(currFactor,i);
          
        ans *= sum;
          
          
        // Update current prime factor as s[N] and
        // initializing power of factor as 1.
        currFactor = s[N];
        power = 1;
    }
  
    return ans;
}
  
// Driver code
int main()
{
    int n = 12;
  
    cout << "Sum of the factors is : ";
  
    cout << findSum(n);
      
    return 0;
}

Java

//Java Program to find sum of all 
//factors of a given number
public class GFG {
  
    //Using SieveOfEratosthenes to find smallest prime
    //factor of all the numbers.
    //For example, if N is 10,
    //s[2] = s[4] = s[6] = s[10] = 2
    //s[3] = s[9] = 3
    //s[5] = 5
    //s[7] = 7
    static void sieveOfEratosthenes(int N, int s[])
    {
     // Create a boolean array "prime[0..n]" and
     // initialize all entries in it as false.
     boolean[] prime = new boolean[N + 1];
       
     for(int i = 0; i < N+1; i++)
         prime[i] = false;
  
     // Initializing smallest factor equal to 2
     // for all the even numbers
     for (int i = 2; i <= N; i += 2)
         s[i] = 2;
  
     // For odd numbers less then equal to n
     for (int i = 3; i <= N; i += 2) {
         if (prime[i] == false) {
  
             // s(i) for a prime is the number itself
             s[i] = i;
  
             // For all multiples of current prime number
             for (int j = i; j * i <= N; j += 2) {
                 if (prime[i * j] == false) {
                     prime[i * j] = true;
  
                     // i is the smallest prime factor for
                     // number "i*j".
                     s[i * j] = i;
                 }
             }
         }
     }
    }
  
    //Function to find sum of all prime factors
    static int findSum(int N)
    {
     // Declaring array to store smallest prime
     // factor of i at i-th index
     int[] s = new int[N + 1];
  
     int ans = 1;
  
     // Filling values in s[] using sieve
     sieveOfEratosthenes(N, s);
        
     int currFactor = s[N]; // Current prime factor of N
     int power = 1; // Power of current prime factor
  
     while (N > 1) {
         N /= s[N];
  
         // N is now N/s[N]. If new N als has smallest
         // prime factor as currFactor, increment power
         if (currFactor == s[N]) {
             power++;
             continue;
         }
  
         int sum = 0;
            
         for(int i=0; i<=power; i++)
             sum += Math.pow(currFactor,i);
            
         ans *= sum;
            
            
         // Update current prime factor as s[N] and
         // initializing power of factor as 1.
         currFactor = s[N];
         power = 1;
     }
  
     return ans;
    }
  
    //Driver code
    public static void main(String[] args) {
          
        int n = 12;
  
         System.out.print("Sum of the factors is : ");
  
         System.out.print(findSum(n));
    }
}

Python 3

# Python 3 Program to find 
# sum of all factors of a
# given number 
  
# Using SieveOfEratosthenes 
# to find smallest prime 
# factor of all the numbers. 
# For example, if N is 10, 
# s[2] = s[4] = s[6] = s[10] = 2 
# s[3] = s[9] = 3 
# s[5] = 5 
# s[7] = 7 
def sieveOfEratosthenes(N, s) :
  
    # Create a boolean list "prime[0..n]" 
    # and initialize all entries in it 
    # as false.
    prime = [False] * (N + 1)
  
    # Initializing smallest 
    # factor equal to 2 for 
    # all the even numbers 
    for i in range(2, N + 1, 2) :
        s[i] = 2
  
    # For odd numbers less
    # then equal to n 
    for i in range(3, N + 1, 2) :
  
        if prime[i] == False :
  
            # s[i] for a prime is
            # the number itself 
            s[i] = i
  
            # For all multiples of 
            # current prime number 
            for j in range(i, (N + 1) // i, 2) :
                  
                if prime[i * j] == False :
                    prime[i * j] = True
                      
                    # i is the smallest
                    # prime factor for 
                    # number "i*j". 
                    s[i * j] = i
  
                #J += 2
  
# Function to find sum
# of all prime factors
def findSum(N) :
  
    # Declaring list to store 
    # smallest prime factor of
    # i at i-th index 
    s = [0] * (N + 1)
    ans = 1
  
    # Filling values in s[] using 
    # sieve function calling
    sieveOfEratosthenes(N, s)
  
    # Current prime factor of N
    currFactor = s[N]
  
    # Power of current prime factor
    power = 1
  
    while N > 1 :
        N //= s[N]
  
        # N is now N//s[N]. If new N 
        # also has smallest prime 
        # factor as currFactor, 
        # increment power
        if currFactor == s[N] :
            power += 1
            continue
  
        sum = 0
  
        for i in range(power + 1) :
            sum += pow(currFactor, i)
  
        ans *= sum
  
        # Update current prime factor 
        # as s[N] and initializing 
        # power of factor as 1. 
        currFactor = s[N]
        power = 1
          
    return ans 
  
# Driver Code
if __name__ == "__main__" :
  
    n = 12
    print("Sum of the factors is :", end = " ")
    print(findSum(n))
      
# This code is contributed by ANKITRAI1

C#

// C# Program to find sum of all 
// factors of a given number
using System;
  
class GFG 
{
  
// Using SieveOfEratosthenes to find smallest 
// prime factor of all the numbers.
// For example, if N is 10,
// s[2] = s[4] = s[6] = s[10] = 2
// s[3] = s[9] = 3
// s[5] = 5
// s[7] = 7
static void sieveOfEratosthenes(int N, int []s)
{
      
// Create a boolean array "prime[0..n]" and
// initialize all entries in it as false.
bool[] prime = new bool[N + 1];
  
for(int i = 0; i < N + 1; i++)
    prime[i] = false;
  
// Initializing smallest factor equal 
// to 2 for all the even numbers
for (int i = 2; i <= N; i += 2)
    s[i] = 2;
  
// For odd numbers less then equal to n
for (int i = 3; i <= N; i += 2)
{
    if (prime[i] == false) 
    {
  
        // s(i) for a prime is the
        // number itself
        s[i] = i;
  
        // For all multiples of current
        // prime number
        for (int j = i; j * i <= N; j += 2)
        {
            if (prime[i * j] == false) 
            {
                prime[i * j] = true;
  
                // i is the smallest prime factor 
                // for number "i*j".
                s[i * j] = i;
            }
        }
    }
}
}
  
// Function to find sum of all 
// prime factors
static int findSum(int N)
{
// Declaring array to store smallest 
// prime factor of i at i-th index
int[] s = new int[N + 1];
  
int ans = 1;
  
// Filling values in s[] using sieve
sieveOfEratosthenes(N, s);
  
int currFactor = s[N]; // Current prime factor of N
int power = 1; // Power of current prime factor
  
while (N > 1) 
{
    N /= s[N];
  
    // N is now N/s[N]. If new N als has smallest
    // prime factor as currFactor, increment power
    if (currFactor == s[N]) 
    {
        power++;
        continue;
    }
  
    int sum = 0;
      
    for(int i = 0; i <= power; i++)
        sum += (int)Math.Pow(currFactor, i);
      
    ans *= sum;
      
    // Update current prime factor as s[N] 
    // and initializing power of factor as 1.
    currFactor = s[N];
    power = 1;
}
  
return ans;
}
  
// Driver code
public static void Main() 
{
    int n = 12;
  
    Console.Write("Sum of the factors is : ");
  
    Console.WriteLine(findSum(n));
}
}
  
// This code is contributed by Shashank

PHP

 1) {
        $N /= $s[$N];
  
        // N is now N/s[N]. If new N als has smallest
        // prime factor as currFactor, increment power
        if ($currFactor == $s[$N]) {
            $power++;
            continue;
        }
        $sum = 0;
          
        for($i=0; $i<=$power; $i++)
            $sum += (int)pow($currFactor,$i);
          
        $ans *= $sum;
          
          
        // Update current prime factor as s[N] and
        // initializing power of factor as 1.
        $currFactor = $s[$N];
        $power = 1;
    }
  
    return $ans;
}
  
// Driver code
  
    $n = 12;
  
    echo "Sum of the factors is : ";
  
    echo findSum($n);
      
// This code is contributed by mits
?>

输出：

Sum of the factors is : 28