[英]Python - count the number of prime divisors without using range
I have to write a function that counts the total number of prime divisors of a given positive integer n.我必须编写一个函数来计算给定正整数 n 的素数除数的总数。 Since I have just started learning python a week ago, I am not allowed to use for loop with range.由于我一周前才开始学习python,因此不允许使用带有范围的for循环。 Below is what I have so far:以下是我到目前为止所拥有的:
def count_prime_divisors(n):
return num_of_divisors(n, 1)
def num_of_divisors(n, i):
if i > n:
return 0
if n % i == 0 and num_of_divisors(i, 1) == 2:
return 1 + num_of_divisors(n, i+1)
else:
return num_of_divisors(n, i+1)
So I know that the exceeding maximum recursion depth error occurs at this line if n % i == 0 and num_of_divisors(i, 1) == 2
but I don't know how to check if the divisor is prime so that the function can work appropriately.所以我知道if n % i == 0 and num_of_divisors(i, 1) == 2
会在这一行发生超过最大递归深度错误if n % i == 0 and num_of_divisors(i, 1) == 2
但我不知道如何检查除数是否为素数,以便函数可以适当地工作。 Maybe I should write another helper function?也许我应该编写另一个辅助函数? Can someone help me with this?有人可以帮我弄这个吗? Any help is much appreciated :( Thank you!非常感谢任何帮助:(谢谢!
You can try this.你可以试试这个。
def prime_fac(n,count=0,prime_num=2):
if n==1:
print('Number of prime factors',count)
return
elif n%prime_num==0:
print(prime_num)
return prime_fac(n//prime_num,count+1,prime_num)
else:
return prime_fac(n,count,prime_num+1)
prime_fac(100)
prime_fac(12)
prime_fac(999)
output输出
2
2
5
5
Number of prime factors 4
2
2
3
Number of prime factors 3
3
3
3
37
Number of prime factors 4
Use this if want to store the count of prime factors and unique prime factors .如果要存储质因数和唯一质因数的计数,请使用此选项。
def prime_fac(n,count=0,prime_num=2,prime_set=set()):
if n==1:
print('Number of prime factors',count)
return count,prime_set
elif n%prime_num==0:
print(prime_num)
prime_set=prime_set|{prime_num} #set union operation
return prime_fac(n//prime_num,count+1,prime_num,prime_set)
else:
return prime_fac(n,count,prime_num+1,prime_set)
a=prime_fac(100)
print(a)
output输出
2
2
5
5
Number of prime factors 4
(4, {2, 5}) #a looks like this where 4 is prime factors count and {2,5} is unique prime factors.
if you are using Linux, this should work (no loops, no ranges, no nothing =):如果您使用的是 Linux,这应该可以工作(没有循环,没有范围,没有什么 =):
>>> import os
>>>
>>> def factor(num) :
... output = os.popen( 'factor %d' % num ).read()
... return map( int, output.split(':')[-1].split())
...
>>> print factor(128)
[2, 2, 2, 2, 2, 2, 2]
>>>
I hope you know how to take a length of the list using len()
我希望你知道如何使用len()
获取列表的长度
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