质数定义：质数（prime number）又称素数。质数定义为在大于1的自然数中，除了1和它本身以外不再有其他因数。

• 示例解决方案1

有很多方法可以解决这个问题，下面是一些例子:这是一个不同的功能分解来解决问题。

def get_number(prompt):    '''Returns integer value for input. Prompt is displayed text'''    return int(input(prompt))

def is_prime(number):    '''Returns True for prime numbers, False otherwise'''    # Edge Cases    if number == 1:        prime = False    elif number == 2:        prime = True    # All other primes    else:        prime = True        for check_number in range(2, (number / 2) + 1):            if number % check_number == 0:                prime = False                break    return prime

def print_prime(number):    prime = is_prime(number)    if prime:        descriptor = ""    else:        descriptor = "not "    print(number, " is ", descriptor, "prime.", sep="", end="\n\n")

# never ending loop

while 1 == 1:    print_prime(get_number("Enter a number to check. Ctl-C to exit."))

• 示例解决方案2

import sysnumber = input("Please enter a number" + "\n" + ">>>")number = int(number)if number > 0:    for x in range (2, number): #this range excludes number and 1, both of which number is divisible by        if number % x != 0: #If number isn't evenly divisible by x, start over with the next one            continue        elif number % x == 0: #If number is evenly divisible by x, it can't be prime            sys.exit("The number is not prime.")    sys.exit("The number is prime.") #number wasn't evenly divisible by any x, so it's primeelif number == 0:    sys.exit("The number is not prime.") #According to the Google, 0 is not primeelse:#if number is less than 0, the number is not prime (according to the Google).    sys.exit("The number is not prime.")

• 示例解决方案3

我觉得这是最简洁的一种

def prime(n):    factor=[i for i in range(2,n) if n%i ==0]    if len(factor)==0:        print(str(n)+"is prime!")    elif n==2:print(str(n)+" " +"is prime!")    else:print("not prime!")

以上就是解决找质数的三种方法，碰到一个问题要多想如何更加简洁高效的实现功能，而不是实现了就行了，对自己要更高要求，这样才会进步的更快。