高效程序可打印给定数字的所有素数

给定数字n，编写一个有效函数以打印n的所有素数。例如，如果输入数字为12，则输出应为“ 2 2 3”。如果输入数字为315，则输出应为“ 3 3 5 7”。

以下是查找所有主要因素的步骤。
1)当n被2整除时，打印2并将n除以2。
2)步骤1之后，n必须为奇数。现在从i = 3到n的平方根开始循环。当我除以n时，打印i并将n除以i。在我未能将n除以后，将i加2并继续。
3)如果n是质数且大于2，那么经过以上两个步骤，n不会变为1。因此，如果大于2，则打印n。

C++

// C++ program to print all prime factors 
#include 
using namespace std;
  
// A function to print all prime 
// factors of a given number n 
void primeFactors(int n) 
{ 
    // Print the number of 2s that divide n 
    while (n % 2 == 0) 
    { 
        cout << 2 << " "; 
        n = n/2; 
    } 
  
    // n must be odd at this point. So we can skip 
    // one element (Note i = i +2) 
    for (int i = 3; i <= sqrt(n); i = i + 2) 
    { 
        // While i divides n, print i and divide n 
        while (n % i == 0) 
        { 
            cout << i << " "; 
            n = n/i; 
        } 
    } 
  
    // This condition is to handle the case when n 
    // is a prime number greater than 2 
    if (n > 2) 
        cout << n << " "; 
} 
  
/* Driver code */
int main() 
{ 
    int n = 315; 
    primeFactors(n); 
    return 0; 
} 
  
// This is code is contributed by rathbhupendra

C

// Program to print all prime factors
# include 
# include 
  
// A function to print all prime factors of a given number n
void primeFactors(int n)
{
    // Print the number of 2s that divide n
    while (n%2 == 0)
    {
        printf("%d ", 2);
        n = n/2;
    }
  
    // n must be odd at this point.  So we can skip 
    // one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i+2)
    {
        // While i divides n, print i and divide n
        while (n%i == 0)
        {
            printf("%d ", i);
            n = n/i;
        }
    }
  
    // This condition is to handle the case when n 
    // is a prime number greater than 2
    if (n > 2)
        printf ("%d ", n);
}
  
/* Driver program to test above function */
int main()
{
    int n = 315;
    primeFactors(n);
    return 0;
}

Java

// Program to print all prime factors
import java.io.*;
import java.lang.Math;
  
class GFG
{
    // A function to print all prime factors
    // of a given number n
    public static void primeFactors(int n)
    {
        // Print the number of 2s that divide n
        while (n%2==0)
        {
            System.out.print(2 + " ");
            n /= 2;
        }
  
        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            while (n%i == 0)
            {
                System.out.print(i + " ");
                n /= i;
            }
        }
  
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n > 2)
            System.out.print(n);
    }
  
    public static void main (String[] args)
    {
        int n = 315;
        primeFactors(n);
    }
}

Python

# Python program to print prime factors
  
import math
  
# A function to print all prime factors of 
# a given number n
def primeFactors(n):
      
    # Print the number of two's that divide n
    while n % 2 == 0:
        print 2,
        n = n / 2
          
    # n must be odd at this point
    # so a skip of 2 ( i = i + 2) can be used
    for i in range(3,int(math.sqrt(n))+1,2):
          
        # while i divides n , print i ad divide n
        while n % i== 0:
            print i,
            n = n / i
              
    # Condition if n is a prime
    # number greater than 2
    if n > 2:
        print n
          
# Driver Program to test above function
  
n = 315
primeFactors(n)
  
# This code is contributed by Harshit Agrawal

C#

// C# Program to print all prime factors
using System;
  
namespace prime
{
public class GFG
{     
                  
    // A function to print all prime 
    // factors of a given number n
    public static void primeFactors(int n)
    {
        // Print the number of 2s that divide n
        while (n % 2 == 0)
        {
            Console.Write(2 + " ");
            n /= 2;
        }
  
        // n must be odd at this point. So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.Sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            while (n % i == 0)
            {
                Console.Write(i + " ");
                n /= i;
            }
        }
  
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n > 2)
            Console.Write(n);
    }
      
    // Driver Code
    public static void Main()
    {
        int n = 315;
        primeFactors(n);
    }
  
} 
}
  
// This code is contributed by Sam007

PHP

 2)
        echo $n," ";
}
  
    // Driver Code
    $n = 315;
    primeFactors($n);
  
// This code is contributed by aj_36
?>

输出：

3 3 5 7

这是如何运作的?
步骤1和2负责合成数字，而步骤3负责质数。为了证明完整的算法有效，我们需要证明步骤1和2实际上处理了复合数。显然，第1步要处理偶数。在第1步之后，所有剩余质数必须为奇数(两个质数之差必须至少为2)，这解释了为什么我要加2。
现在最主要的是，循环一直运行到n的平方根，直到n为止。为了证明这种优化有效，让我们考虑以下合成数字的属性。
每个复合数至少具有一个小于或等于其平方根的素数。
可以使用计数器语句证明此属性。令a和b为n的两个因数，使得a * b = n。如果两者均大于＆Sqrt; n，则ab>＆Sqrt; n，*＆Sqrt; n，这与表达式“ a * b = n”相矛盾。

在上述算法的第2步中，我们运行一个循环并在循环中进行后续操作
a)找到最小素数因子i(必须小于＆Sqrt; n，)
b)通过将n重复除以i，从n中删除所有出现的i。
c)重复步骤a和b，除以n，i = i +2。重复步骤a和b，直到n变为1或质数。

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