Python Prime数字检查器

Question

我一直试图编写一个程序,它将采用一个输入的数字,并检查它是否是素数.到目前为止,我所做的代码完全可以正常工作,如果这个数字实际上是素数.如果数字不是素数,那就很奇怪了.我想知道是否有人能告诉我代码的问题.

a=2
num=13
while num > a :
  if num%a==0 & a!=num:
    print('not prime')
    a=a+1
  else:
    print('prime')
    a=(num)+1

当24被输入时给出的结果是:不是素数不是素数而不是素数

我如何修复每个偶数的每个奇数而不是素数的报告素数的错误

Answer 1

You need to stop iterating once you know a number isn't prime. Add a break once you find prime to exit the while loop.

Making only minimal changes to your code to make it work:

a=2
num=13
while num > a :
  if num%a==0 & a!=num:
    print('not prime')
    break
  i += 1
else: # loop not exited via break
  print('prime')

Your algorithm is equivalent to:

for a in range(a, num):
    if a % num == 0:
        print('not prime')
        break
else: # loop not exited via break
    print('prime')

If you throw it into a function you can dispense with break and for-else:

def is_prime(n):
    for i in range(3, n):
        if n % i == 0:
            return False
    return True

Even if you are going to brute-force for prime like this you only need to iterate up to the square root of n. Also, you can skip testing the even numbers after two.

With these suggestions:

import math
def is_prime(n):
    if n % 2 == 0 and n > 2: 
        return False
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        if n % i == 0:
            return False
    return True

Note that this code does not properly handle 0, 1, and negative numbers.

我们通过使用all生成器表达式替换for循环来使这更简单.

import math
def is_prime(n):
    if n % 2 == 0 and n > 2: 
        return False
    return all(n % i for i in range(3, int(math.sqrt(n)) + 1, 2))

Answer 2

def isprime(n):
    '''check if integer n is a prime'''

    # make sure n is a positive integer
    n = abs(int(n))

    # 0 and 1 are not primes
    if n < 2:
        return False

    # 2 is the only even prime number
    if n == 2: 
        return True    

    # all other even numbers are not primes
    if not n & 1: 
        return False

    # range starts with 3 and only needs to go up 
    # the square root of n for all odd numbers
    for x in range(3, int(n**0.5) + 1, 2):
        if n % x == 0:
            return False

    return True

取自:

http://www.daniweb.com/software-development/python/code/216880/check-if-a-number-is-a-prime-number-python

Answer 3

def is_prime(n):
    return all(n%j for j in xrange(2, int(n**0.5)+1)) and n>1

Answer 4

您的代码的两个主要问题是:

1. 在指定一个非素数的数字之后,你继续检查其余的除数,即使你已经知道它不是素数,这可能导致它在打印"非素数"后打印"素数".提示:使用`break'语句.
2. 在检查了需要检查的所有除数之前,指定一个数字素数,因为您在循环内打印"素数" .因此,您可以多次获得"素数",一次针对每个除数均未达到被测试的数量.提示:else只有当循环退出而没有中断时,才使用带循环的子句来打印"prime".

一些非常显着的低效率:

1. 你应该跟踪你已经找到的最重要的数字,并且只能除以那些数字.当你已经除以2时为什么除以4？如果一个数字可以被4整除,它也可以被2整除,所以你已经捕获它并且没有必要除以4.
2. 您只需要测试最多被测试数字的平方根,因为任何大于该值的因子都需要乘以小于该值的数字,并且在您到达较大的数字时已经测试过.