Question

该函数应返回m和n之间的素数列表。如果m和/或n是素数，它们也应该包含在列表中。

这就是我所做的：
Code

这是应该是的输出：
请忽略前两个输出，因为我已经完成了它们。 Output file

Answer 1

import sys
sys.setrecursionlimit (30000)

"""The checkprime function uses the prime factor theorum to determine if a number is prime by checking it against a list of
prime numbers and ensuring it is not divisible by said numbers, therefore ensuring it is a prime, it requires a list of all prime numbers
smaller than the number being checked itself"""

def checkprime(primelist,value,recursions):
    #Decrease the index that is being looked at
    recursions -= 1
    #If the prime is not on 
    if recursions == -1:
        return(True)
    if value%primelist[recursions] == 0:
        return(False)
    return(checkprime(primelist,value,recursions))

'''The compileprimes function returns a list of all the prime numbers below the maximum value, using a list of primes,the function starts with a list of the first few prime numbers as a 
default value and works up from there, its start value is naturally set to the following prime to avoid unneccesary checks'''
def compileprimes(Up_To,primelist=[2,3,5,7],start=11):
    #Base case: If the value to compare is greater than the maximum value return the list of primes
    if start > Up_To:
        return(primelist)
    
    #Use the checkprime function to check if the 'start' value is a prime
    if checkprime(primelist,start,len(primelist)):
        #If the 'start' value is prime, append it to the list of primes
        primelist.append(start)
    #Increase by two to skip all even numbers
    start += 2
    
    #Recursive step
    return(compileprimes(Up_To,primelist,start))

compileprimes 函数将递归地创建一个质数列表，直到和低于您输入的数字。例如。

'''
>>> compileprimes(23)
[2,3,5,7,11,13,17,23]

'''

您可能会遇到的一个问题是，再次调用该函数时，可能必须将 primelist 数组分配给默认值，但是该方法被认为比之前提出的解决方案高效得多，因为它没有除以任何非质数（它们本身是质数的乘积），它的工作速度也应该比前面的例子快得多。删除 n 值下的任何数字很容易，所以我将把它留给你

Answer 2

def primes(m, n):
    list_of_primes = []        # Keep track of primes
    for num in range(m, n + 1):    # make inclusive of upper bound
        is_prime = True            # Assume number is prime as starting condition
        for i in range(2, num):    # Check remainder from number 2 up to num
            if num % i ==  0:
                is_prime = False   
                break
        if is_prime and num > 1:          #EDIT HERE
            list_of_primes.append(num)   #add to the list of primes if True
    return list_of_primes

请注意，有多种方法可以改进此算法，例如不检查偶数等，但这符合您问题中定义的条件。

返回素数

2 个答案: