Question

def primeF (n):
    x = 2
    factors: List[int] = []
    while (x < n):
        y = 2
        flag = 0
        while (y < x):
            if (x % y == 0):
                flag = 1
            y = y + 1
        if (n % x == 0 and flag == 0):
            factors.append(x)
    x = x + 1
    return factors
primeF(6)

运行此代码时始终会出现错误，但我不确定为什么。错误消息是：

回溯（最近通话最近）：
文件“ C：/Users/User/untitled/EulerQs.py”，第41行，在 print（primeF（6））文件“ C：/Users/User/untitled/EulerQs.py”，第38行，位于primeF中 factor.append（x）MemoryError

我不知道这意味着什么。如果有任何区别，我正在使用PyCharm。

Answer 1

如评论中所述，崩溃的原因是外部while循环永远不会结束，因为x的增量在该循环之外。只需更改缩进量即可将x = x + 1放入循环内。

另一有用的事情是自动写出一系列数字的结果。然后，每次n本身为质数时，该函数都会给出一个空列表。因此，在外部while循环的停止条件下，还应该允许x == n。

更改此代码，代码如下：

def primeF (n):
    x = 2
    factors = []
    while (x <= n):
        y = 2
        flag = 0
        while (y < x):
            if (x % y == 0):
                flag = 1
            y = y + 1
        if (n % x == 0 and flag == 0):
            factors.append(x)
        x = x + 1
    return factors

for i in range(11):
    print(i, "has as prime factors:", primeF(i))

输出：

0 has as prime factors: []
1 has as prime factors: []
2 has as prime factors: [2]
3 has as prime factors: [3]
4 has as prime factors: [2]
5 has as prime factors: [5]
6 has as prime factors: [2, 3]
7 has as prime factors: [7]
8 has as prime factors: [2]
9 has as prime factors: [3]
10 has as prime factors: [2, 5]

一些进一步改进程序的想法：

在if-test中，括号并非总是必需的；习惯上将它们排除在外，以使内容更具可读性。
在大多数编程语言中，标志称为布尔值，其值为True和False。另外，给他们起有意义的名字也有助于理解程序。
现在，您要测试每个小于x的{{1}}是否为质数，然后再测试其是否为除数。如果您首先测试除数，那么您不需要检查素数就可以赢得一些时间。另外，一旦发现n不是素数，就可以跳出循环。

以下是建议更改的代码：

我编写的python3主因子函数无法按预期工作

1 个答案: