我想检查一个数字是否是素数，但是下面的代码不起作用，当我输入 65 时它显示，这是一个素数

Question

def prime (num) :
    if num == 1 or num == 0 or num == 2:
        return 'This is not a prime number'
    for number in range(2,num):
        if num % number == 0 :
            return 'This is not a prime number'
        else:
            return 'This is a prime number'

This is the code and this is not working result for 65 is prime, why?

Answer 1

The problem is you are iterating from 2 till that number. So it goes into the loop, number is 2 , and 65 % 2 != 0 . Hence It returns 'This is a prime number' . Instead try this:

def prime(num):
    if num == 1 or num == 0:
        return 'This is not a prime number'
    for number in range(2, num):
        if num % number == 0 :
            return 'This is not a prime number'
    return 'This is a prime number'

What this does is it keeps looping until remainder is 0. If the remainder is never 0, it returns 'This is a prime number'.

PS 2 is a prime number.

Answer 2

there's an elegant algo to solve this problem https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

here's my implementation of its caching version

primes = [1, 2]
def prime(n):
    if n <= 1:
        return 1
    elif n < len(primes):
        return primes[n]
    else:
        p = prime(n-1)+1

        while True:
            for i in range(2, n-1):
                if p%prime(i) == 0:
                    p += 1
                    break
            else:
                primes.append(p)
                return p

for i in range(100):
print(prime(i))