Question

我似乎得到了所有素数，但一些复合数字也出现了。为什么呢？

p=[2,]
while len(p)<35:
        next = p[-1]+1
        for i in range(len(p)):
            if next%p[i]==0:
                next+=1
                i=0
        p.append(next)
print p

这是我得到的输出：

[2,3,5,7,11,13,16,17,19,23,27,29,31,35,37,41,43,47,50,53,59,61,65， 67,71,73,77,79,83,87,89,95,97,101,103]

Answer 1

因为当您在for循环中更新next时，您没有从主要列表的开头进行扫描，以检查新的next是否可以被整除。您可以尝试以下方式。

p=[2,]
number=3
while len(p)<35:
    if all(number%i!=0 for i in p):
        p.append(number)
    number += 1
print p

Answer 2

如果在每次迭代中打印i的值，您将看到问题。 i的值将继续增加。

Test 3 divided by 2 (p[0])
Test 4 divided by 2 (p[0])
Test 5 divided by 3 (p[1])
Test 6 divided by 2 (p[0])
Test 7 divided by 3 (p[1])
Test 7 divided by 5 (p[2])
Test 8 divided by 2 (p[0])
Test 9 divided by 3 (p[1])
Test 10 divided by 5 (p[2])
Test 11 divided by 7 (p[3])
Test 12 divided by 2 (p[0])
Test 13 divided by 3 (p[1])
Test 13 divided by 5 (p[2])
Test 13 divided by 7 (p[3])
Test 13 divided by 11 (p[4])
Test 14 divided by 2 (p[0])
Test 15 divided by 3 (p[1])
Test 16 divided by 5 (p[2])
Test 16 divided by 7 (p[3])
Test 16 divided by 11 (p[4])
Test 16 divided by 13 (p[5])
Test 17 divided by 2 (p[0])

每次代码评估行for i in range(len(p))时，它都不会考虑i的当前值，因为range(len(p))实际上是一个列表（或Python 3中的生成器），所以它只迭代列表，而不是像你期望的那样实际做i = i+1。

例如，考虑len(p)=3时的情况。然后range(len(p))将等同于[0,1,2]。 for循环首先取值0。然后，下一次迭代将具有值1，第三次将具有值2，无论您对循环体内i的值如何操作。

一种解决方案不是使用for i in range(len(p))，而是自己进行迭代：

next = p[-1]+1 # actually you can do next = next+1
i = 0
while i < len(p):
    if next%p[i]==0:
        next += 1
        i=0
    else:
        i+=1
p.append(next)

素数：为什么这段代码给了我一些复合数字？

2 个答案: