Java程序查找数字的唯一素数的乘积

给定一个数 n，我们需要找到它所有唯一素因子的乘积。质因数：它基本上是质数本身的一个数的因数。

例子：

Input: num = 10
Output: Product is 10
Explanation:
Here, the input number is 10 having only 2 prime factors and they are 5 and 2.
And hence their product is 10.

Input : num = 25
Output: Product is 5
Explanation:
Here, for the input to be 25  we have only one unique prime factor i.e 5.
And hence the required product is 5.

方法1（简单）
使用从 i = 2 到 n 的循环并检查 i 是否是 n 的因数，然后检查 i 是否是素数本身，如果是，则将产品存储在产品变量中并继续此过程直到 i = n。

// Java program to find product of
// unique prime factors of a number.
  
public class GFG {
    public static long productPrimeFactors(int n)
    {
        long product = 1;
  
        for (int i = 2; i <= n; i++) {
            // Checking if 'i' is factor of num
            if (n % i == 0) {
  
                // Checking if 'i' is a Prime number
                boolean isPrime = true;
                for (int j = 2; j <= i / 2; j++) {
                    if (i % j == 0) {
                        isPrime = false;
                        break;
                    }
                }
  
                // condition if 'i' is Prime number
                // as well as factor of num
                if (isPrime) {
                    product = product * i;
                }
            }
        }
        return product;
    }
  
    public static void main(String[] args)
    {
        int n = 44;
        System.out.print(productPrimeFactors(n));
    }
}
  
// Contributed by _omg

输出：

方法二（高效）
这个想法是基于高效程序来打印给定数字的所有素因子

// Java program to find product of
// unique prime factors of a number.
import java.util.*;
import java.lang.*;
  
public class GFG {
    public static long productPrimeFactors(int n)
    {
        long product = 1;
        // Handle prime factor 2 explicitly so that
        // can optimally handle other prime factors.
        if (n % 2 == 0) {
            product *= 2;
            while (n % 2 == 0)
                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 <= Math.sqrt(n); i = i + 2) {
            // While i divides n, print i and
            // divide n
            if (n % i == 0) {
                product = product * i;
                while (n % i == 0)
                    n = n / i;
            }
        }
  
        // This condition is to handle the case when n
        // is a prime number greater than 2
        if (n > 2)
            product = product * n;
  
        return product;
    }
  
    public static void main(String[] args)
    {
        int n = 44;
        System.out.print(productPrimeFactors(n));
    }
}
  
// Contributed by _omg

输出：

更多详细信息，请参阅关于一个数的唯一素因数乘积的完整文章！