How to return all prime factors of a number?

How to return all prime factors of a number？

如何判断一个数字n是否是质数？

1）如果该数n小于等于1,则False;不是质数,因为1不是质数,但是1and all other numbers are coprime;

2）然后i从2开始,依次判断到sqrt(n)  【A number that is the square root of a number】,If these numbers are not availablen整除,In other words there is a remainder,则说明n是质数;

i = 2

while i*i <= n:

        if n%i == 0: return False

        i += 1

return True

To test whether a number is prime,Why do we have to test if the number is only divisible by the square root of the number？

精彩 The analysis is excellent.

如果n不是质数,那么n可以分解成a*b,a和bAt least one should be less thansqrt（n）的 ,If both are greater than sqrt（n）Then multiply the two together,一定大于n.所以,从2到sqrt（n）一定有一个n的因数,If no such factor is found,说明n是质数;