打印任意数的素数分解的函数/ Python

Question

我正在寻找帮助编写一个函数，它接受一个正整数n作为输入，并将它的素因式分解打印到屏幕上。输出应将这些因素集合到单个字符串中，以便像prime_factorization(60)这样的调用的结果将字符串“60 = 2 x 2 x 3 x 5”打印到屏幕上。以下是我到目前为止所掌握的内容。更新:我取得了进展，并弄清楚了如何找到素因式分解。但是，我仍然需要帮助打印正确的方式如上所述。

""""
Input is a positive integer n
Output is its prime factorization, computed as follows: 

"""
import math
def prime_factorization(n):

    while (n % 2) == 0:
        print(2)
    
        # Turn n into odd number 
        n = n / 2

    for i in range (3, int(math.sqrt(n)) + 1, 2):
    
        while (n % i) == 0:
            print(i)
            n = n / I

    if (n > 2):
        print(n)
    

prime_factorization(60)

请注意，我正在尝试打印它，因此如果输入为60，则输出为“60 =2x2x3x5”

Answer 1

使用列表存储所有因子，然后将它们一起以所需的格式打印为字符串。

import math
def prime_factorization(n):
    factors = []  # to store factors
    while (n % 2) == 0:
        factors.append(2)
    
        # Turn n into odd number 
        n = n / 2

    for i in range (3, int(math.sqrt(n)) + 1, 2):
    
        while (n % i) == 0:
            factors.append(i)
            n = n / I

    if (n > 2):
        factors.append(n)

    print(" x ".join(str(i) for i in factors))  # to get the required string
    

prime_factorization(60)

Answer 2

这是一种使用f-string来完成此操作的方法。此外，你需要做整数除法(用//)，以避免在你的答案中得到浮点数。

""""
Input is a positive integer n
Output is its prime factorization, computed as follows:
"""
import math

def prime_factorization(n):
    n_copy = n
    prime_list = []
    while (n % 2) == 0:
        prime_list.append(2)
        # Turn n into odd number
        n = n // 2

    for i in range(3, int(math.sqrt(n)) + 1, 2):
        while (n % i) == 0:
            prime_list.append(i)
            n = n // i

    if (n > 2):
        prime_list.append(n)

    print(f'{n_copy} =', end = ' ')
    for factor in prime_list[:-1]:
        print (f'{factor} x', end=' ' )
    print(prime_list[-1])


prime_factorization(60)
#output: 60 = 2 x 2 x 3 x 5

Answer 3

您应该始终将计算与表示分开。您可以将该函数构建为一个生成器，该生成器通过增加除数(2，然后是赔率)来除数。当你找到一个合适的，输出它并继续除法的结果。这只会产生质数因子。

然后使用该函数获取要打印的数据，而不是尝试混合打印和格式化。

def primeFactors(N):
    p,i = 2,1               # prime divisor and increment
    while p*p<=N:           # no need to go beyond √N 
        while N%p == 0:     # if is integer divisor
            yield p         # output prime divisor
            N //= p         # remove it from the number
        p,i = p+i,2         # advance to next potential divisor 2, 3, 5, ... 
    if N>1: yield N         # remaining value is a prime if not 1

输出：

N=60
print(N,end=" = ")
print(*primeFactors(N),sep=" x ")

60 = 2 x 2 x 3 x 5