打印给定数目的所有质因数的高效程序
原文:https://www . geesforgeks . org/print-all-prime-factors-of-a-number/
给定一个数字 n，写一个高效的函数打印 n 的所有质因数，比如输入数字是 12，那么输出应该是“2 2 3”。如果输入的数字是 315，那么输出应该是“3 3 5 7”。
以下是寻找所有主要因素的步骤。 1) 当 n 可被 2 整除时，打印 2 并除以 2。 2) 第 1 步之后，n 一定是奇数。现在开始一个从 i = 3 到 n 的平方根的循环，当我除 n 时，打印 I，用 I 除 n，在我未能除 n 后，将 I 增加 2 并继续。 3) 如果 n 是素数且大于 2，那么通过以上两步 n 不会变成 1。所以如果大于 2 就打印 n。
C++
// C++ program to print all prime factors
#include <bits/stdc++.h>
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 <stdio.h>
# include <math.h>
// 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 语言(一种计算机语言，尤用于创建网站)
// 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 when
// 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);
}
}
计算机编程语言
# 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 and 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP Efficient program to print all
// prime factors of a given number
// function to print all prime
// factors of a given number n
function primeFactors($n)
{
// Print the number of
// 2s that divide n
while($n % 2 == 0)
{
echo 2," ";
$n = $n / 2;
}
// n must be odd at this
// point. So we can skip
// one element (Note i = i +2)
for ($i = 3; $i <= sqrt($n);
$i = $i + 2)
{
// While i divides n,
// print i and divide n
while ($n % $i == 0)
{
echo $i," ";
$n = $n / $i;
}
}
// This condition is to
// handle the case when n
// is a prime number greater
// than 2
if ($n > 2)
echo $n," ";
}
// Driver Code
$n = 315;
primeFactors($n);
// This code is contributed by aj_36
?>
java 描述语言
<script>
// JavaScript program to print all prime factors
// A function to print all prime
// factors of a given number n
function primeFactors(n)
{
// Print the number of 2s that divide n
while (n % 2 == 0)
{
document.write(2 + " ");
n = Math.floor(n / 2);
}
// n must be odd at this point.
// So we can skip one element
// (Note i = i +2)
for(let i = 3;
i <= Math.floor(Math.sqrt(n));
i = i + 2)
{
// While i divides n, print i
// and divide n
while (n % i == 0)
{
document.write(i + " ");
n = Math.floor(n / i);
}
}
// This condition is to handle the
// case when n is a prime number
// greater than 2
if (n > 2)
document.write(n + " ");
}
// Driver code
let n = 315;
primeFactors(n);
// This code is contributed by Surbhi Tyagi.
</script>
输出:
3 3 5 7
这是如何工作的？ 步骤 1 和 2 处理复合数，步骤 3 处理素数。为了证明完整的算法是有效的，我们需要证明步骤 1 和 2 实际上处理了复合数。很明显，步骤 1 处理偶数。在步骤 1 之后，所有剩余的质因数必须是奇数(两个质因数的差必须至少为 2)，这解释了为什么 I 增加 2。
现在主要部分是，循环运行到 n 的平方根，而不是 n。为了证明这种优化是有效的，让我们考虑复合数的以下性质。
每个复合数至少有一个质因数小于或等于本身的平方根。 这个性质可以用反例来证明。设 a 和 b 是 n 的两个因子，使得 ab = n，如果两者都大于√n，那么 a.b > √n， √n，这与表达式“a * b = n”相矛盾。
在上述算法的步骤 2 中，我们运行一个循环，并在循环 中执行以下操作:a)找到最小质因数 I(必须小于√n)， b)通过重复用 I 除 n 来从 n 中移除所有出现的 I，即 c)对除 n 和 i = i + 2 重复步骤 a 和 b。重复步骤 a 和 b，直到 n 变成 1 或质数。
相关文章: 使用筛子 O(log n)进行多查询的素分解 感谢 Vishwas Garg 提出上述算法。如果发现有不正确的地方，请写评论，或者想分享更多关于以上讨论话题的信息
版权属于：月萌API www.moonapi.com，转载请注明出处
