为什么每个数都是素数？

Question

这是我的代码：

from time import sleep

n = 5
k = 2
found_factors = 0

while True:
   kinn = n/k

    if (n == k) and found_factors == 0:
        print("Found prime:", n)
        n += 1
        k = 2
        found_factors = 0
        sleep(0.1)
        continue
    elif (n == k) and found_factors > 0:
        n += 1
        k = 2
        found_factors = 0
        continue


    if isinstance(kinn, int) == True:
        found_factors += 1

    k += 1

该程序旨在查找从 5 开始的素数。但由于某种原因，它输出的每个数字都是素数！

为什么会这样？

Answer 1

from time import sleep

n = 5.0 # change one of n/k to float so division will not truncate
k = 2
found_factors = 0

while True:
   kinn = n/k # note: in python 2 dividing ints truncates the remainder

    if (n == k) and found_factors == 0:
        print("Found prime:", n)
        n += 1
        k = 2
        found_factors = 0
        sleep(0.1)
        continue
    elif (n == k) and found_factors > 0:
        n += 1
        k = 2
        found_factors = 0
        continue

    # kinn used to truncate to an int, which is why this statement was always hit.
    # By changing the type of n to float, we test for remainder of n/k instead
    if n % k == 0:
        found_factors += 1

    k += 1

Answer 2

n/k 总是 returns 一个浮点数，即使 k 除 n，所以例如 6/2 是 3.0 这是一个float，并且不是 int 的实例，但是，还有另一种检查数字是否为整数的方法，使用 n 和 n 之间的模运算符 % k求除法余数：

if n % k == 0:
        found_factors += 1

通过在你的程序中替换它，它输出：

Found prime: 5
Found prime: 7
Found prime: 11
Found prime: 13
Found prime: 17
Found prime: 19
Found prime: 23
Found prime: 29
Found prime: 31
Found prime: 37
Found prime: 41
Found prime: 43
Found prime: 47
Found prime: 53
Found prime: 59
Found prime: 61
Found prime: 67
# ........ more

为什么每个数都是素数？

Why is every number prime?

python

primes