一个数的唯一素因数乘积的Python程序

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。

# Python program to find sum of given
# series.
  
def productPrimeFactors(n):
    product = 1
      
    for i in range(2, n+1):
        if (n % i == 0):
            isPrime = 1
              
            for j in range(2, int(i/2 + 1)):
                if (i % j == 0):
                    isPrime = 0
                    break
                  
            # condition if \'i\' is Prime number
            # as well as factor of num
            if (isPrime):
                product = product * i
                  
    return product  
      
      
      
# main()
n = 44
print (productPrimeFactors(n))
  
# Contributed by _omg

输出：

方法2（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

# Python program to find product of 
# unique prime factors of a number
  
import math
  
def productPrimeFactors(n):
    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 i in range (3, int(math.sqrt(n)), 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     
      
# main()
n = 44
print (int(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！