Question

我要求创建一个名为primes(x)的方法，该方法返回一个列表，其中包含从2开始一直到x的所有素数。

这是我的代码：

def primes(x):
    i = 2
    while(i < x):
        j = 2
        while(j <= (i/j)):
            if not(i%j): 
                break
            j = j + 1
            if (j > i/j): 
                print(i, " is prime")
        i = i + 1

预计将打印：

>>> primes(50)
(2, ' is prime')
(3, ' is prime')
(5, ' is prime')
(7, ' is prime')
(11, ' is prime')
(13, ' is prime')
(17, ' is prime')
(19, ' is prime')
(23, ' is prime')
(29, ' is prime')
(31, ' is prime')
(37, ' is prime')
(41, ' is prime')
(43, ' is prime')
(47, ' is prime')

然而，它没有打印2和3，它只打印：

(5, ' is prime')
(7, ' is prime')
(11, ' is prime')
(13, ' is prime')
(17, ' is prime')
(19, ' is prime')
(23, ' is prime')
(29, ' is prime')
(31, ' is prime')
(37, ' is prime')
(41, ' is prime')
(43, ' is prime')
(47, ' is prime')

Answer 1

这是这一行：

while(j <= (i/j)):

当i为2或3时，j = 2的起始值导致条件第一次评估为假。

Answer 2

在while(j <= (i/j))中，当i为2或3时，while循环不会激活。

Answer 3

了解

i = 2
while(i < x):
    j = 2
    while(j <= (i/j)):

由于j = 2大于i/j，因此i仅递增而不打印。 i = 3也一样。

Answer 4

如果问题是要打印2和3，那么这样就很容易了;）

def primes(x):
    i = 2
    print(2, " is prime")
    print(3, " is prime")

    while(i < x):
        j = 2

        while(j <= (i/j)):
            if not(i%j): 
                break
            j = j + 1
            if (j > i/j): 
                print(i, " is prime")
        i = i + 1

我不知道为什么它不打印2和3

4 个答案: