直到n的数字的最小素因数

给定数字n ，请打印从1到n的所有数字的最小素数。整数n的最小素数因子是将数字除以最小的素数。所有偶数的最小素数是2。素数是它自己的最小素数(以及它自己的最大素数)。
注意：我们需要为1打印1。
例子 ：

Input : 6
Output : Least Prime factor of 1: 1
         Least Prime factor of 2: 2
         Least Prime factor of 3: 3
         Least Prime factor of 4: 2
         Least Prime factor of 5: 5
         Least Prime factor of 6: 2

我们可以使用Eratosthenes筛子的变体来解决上述问题。

1. 创建一个从2到n的连续整数列表：(2，3，4，…，n)。
2. 最初，让我等于2，最小的质数。
3. 通过以i的增量从2i计数到n来枚举i的倍数，并将它们标记为具有最小素数的i(如果尚未标记)。还要将i标记为i的最小质数(i本身是质数)。
4. 在未标记的列表中找到大于i的第一个数字。如果没有这样的数字，请停止。否则，让我现在等于这个新数字(这是下一个质数)，然后从步骤3开始重复。

下面是该算法的实现，其中minimum_prime []保存与各个索引相对应的最小素数因子的值。

C++

// C++ program to print the least prime factors
// of numbers less than or equal to
// n using modified Sieve of Eratosthenes
#include
using namespace std;
 
void leastPrimeFactor(int n)
{
    // Create a vector to store least primes.
    // Initialize all entries as 0.
    vector least_prime(n+1, 0);
 
    // We need to print 1 for 1.
    least_prime[1] = 1;
 
    for (int i = 2; i <= n; i++)
    {
        // least_prime[i] == 0
        // means it i is prime
        if (least_prime[i] == 0)
        {
            // marking the prime number
            // as its own lpf
            least_prime[i] = i;
 
            // mark it as a divisor for all its
            // multiples if not already marked
            for (int j = i*i; j <= n; j += i)
                if (least_prime[j] == 0)
                   least_prime[j] = i;
        }
    }
 
    // print least prime factor of
    // of numbers till n
    for (int i = 1; i <= n; i++)
        cout << "Least Prime factor of "
             << i << ": " << least_prime[i] << "\n";
}
 
// Driver program to test above function
int main()
{
    int n = 10;
    leastPrimeFactor(n);
    return 0;
}

Java

// Java program to print the least prime factors
// of numbers less than or equal to
// n using modified Sieve of Eratosthenes
 
import java.io.*;
import java.util.*;
 
class GFG
{
    public static void leastPrimeFactor(int n)
    {
         
        // Create a vector to store least primes.
        // Initialize all entries as 0.
        int[] least_prime = new int[n+1];
 
        // We need to print 1 for 1.
        least_prime[1] = 1;
 
        for (int i = 2; i <= n; i++)
        {
             
            // least_prime[i] == 0
            // means it i is prime
            if (least_prime[i] == 0)
            {
                 
                // marking the prime number
                // as its own lpf
                least_prime[i] = i;
 
                // mark it as a divisor for all its
                // multiples if not already marked
                for (int j = i*i; j <= n; j += i)
                    if (least_prime[j] == 0)
                        least_prime[j] = i;
            }
        }
 
        // print least prime factor of
        // of numbers till n
        for (int i = 1; i <= n; i++)
            System.out.println("Least Prime factor of " +
                               + i + ": " + least_prime[i]);
    }
    public static void main (String[] args)
    {
        int n = 10;
        leastPrimeFactor(n);
    }
}
 
// Code Contributed by Mohit Gupta_OMG <(0_o)>

Python 3

# Python 3 program to print the
# least prime factors of numbers
# less than or equal to n using
# modified Sieve of Eratosthenes
 
def leastPrimeFactor(n) :
     
    # Create a vector to store least primes.
    # Initialize all entries as 0.
    least_prime = [0] * (n + 1)
 
    # We need to print 1 for 1.
    least_prime[1] = 1
 
    for i in range(2, n + 1) :
         
        # least_prime[i] == 0
        # means it i is prime
        if (least_prime[i] == 0) :
             
            # marking the prime number
            # as its own lpf
            least_prime[i] = i
 
            # mark it as a divisor for all its
            # multiples if not already marked
            for j in range(i * i, n + 1, i) :
                if (least_prime[j] == 0) :
                    least_prime[j] = i
         
         
    # print least prime factor
    # of numbers till n
    for i in range(1, n + 1) :
        print("Least Prime factor of "
              ,i , ": " , least_prime[i] )
         
 
# Driver program
 
n = 10
leastPrimeFactor(n)
 
 
# This code is contributed
# by Nikita Tiwari.

C#

// C# program to print the least prime factors
// of numbers less than or equal to
// n using modified Sieve of Eratosthenes
using System;
 
class GFG
{
    public static void leastPrimeFactor(int n)
    {
        // Create a vector to store least primes.
        // Initialize all entries as 0.
        int []least_prime = new int[n+1];
 
        // We need to print 1 for 1.
        least_prime[1] = 1;
 
        for (int i = 2; i <= n; i++)
        {
            // least_prime[i] == 0
            // means it i is prime
            if (least_prime[i] == 0)
            {
                // marking the prime number
                // as its own lpf
                least_prime[i] = i;
 
                // mark it as a divisor for all its
                // multiples if not already marked
                for (int j = i*i; j <= n; j += i)
                    if (least_prime[j] == 0)
                        least_prime[j] = i;
            }
        }
 
        // print least prime factor of
        // of numbers till n
        for (int i = 1; i <= n; i++)
            Console.WriteLine("Least Prime factor of " +
                               i + ": " + least_prime[i]);
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 10;
         
        // Function calling
        leastPrimeFactor(n);
    }
}
 
// This code is contributed by Nitin Mittal

Javascript

// javascript program to print the least prime factors
// of numbers less than or equal to
// n using modified Sieve of Eratosthenes
function leastPrimeFactor( n)
{
 
    // Create a vector to store least primes.
    // Initialize all entries as 0.
    let least_prime = Array(n+1).fill(0);
 
    // We need to print 1 for 1.
    least_prime[1] = 1;
 
    for (let i = 2; i <= n; i++)
    {
 
        // least_prime[i] == 0
        // means it i is prime
        if (least_prime[i] == 0)
        {
         
            // marking the prime number
            // as its own lpf
            least_prime[i] = i;
 
            // mark it as a divisor for all its
            // multiples if not already marked
            for (let j = i*i; j <= n; j += i)
                if (least_prime[j] == 0)
                   least_prime[j] = i;
        }
    }
 
    // print least prime factor of
    // of numbers till n
    for (let i = 1; i <= n; i++)
       document.write( "Least Prime factor of "
             + i + ": " + least_prime[i] + "<br/>");
}
 
// Driver program to test above function
    let n = 10;
    leastPrimeFactor(n);
     
// This code is contributed by Rajput-Ji
 

输出

Least Prime factor of 1: 1
Least Prime factor of 2: 2
Least Prime factor of 3: 3
Least Prime factor of 4: 2
Least Prime factor of 5: 5
Least Prime factor of 6: 2
Least Prime factor of 7: 7
Least Prime factor of 8: 2
Least Prime factor of 9: 3
Least Prime factor of 10: 2