Python练习题答案: 检查素数【难度：1级】--景越Python编程实例训练营,1000道上机题等你来挑战

检查素数【难度:1级】:

答案1:

def is_prime(n):
  return n > 1 and all(n % i for i in xrange(2, n))​

答案2:

def is_prime(n):
  'Return True if n is a prime number otherwise return False'
  return  n > 1 and all(n % i for i in xrange(2, n))​

答案3:

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

答案4:

def is_prime(n):
    'Return True if n is a prime number otherwise return False'
    # 0 and 1 are not primes
    if n < 2: 
        return False
    
    # 2 is only even prime so reject all others
    if n == 2: 
        return True
        
    if not n &amp; 1: 
        return False
    
    # test other possiblities
    for i in range(3, n):
        if (n % i) == 0 and n != i: 
            return False
  
  
    return True​

答案5:

def is_prime(n):
  'Return True if n is a prime number otherwise return False'
  return n > 1 and not any(n % x == 0 for x in xrange(2, int(n**0.5)+1))​

答案6:

def is_prime(n):
    'Return True if n is a prime number otherwise return False'
  
    if n <=1: return False
    return all(n % i for i in xrange(2, n))​

答案7:

def is_prime(n):
  if n <= 1:
      return False
  return all(n % i for i in xrange(2, n))​

答案8:

def is_prime(n):
    """Return True if n is a prime number otherwise return False"""
    return n > 1 and all(n % i for i in xrange(int(n ** 0.5), 1, -1))​

答案9:

def is_prime(num):
    'Return True if n is a prime number otherwise return False'
    return num > 1 and not any(num % n == 0 for n in range(2, num))​

答案10:

def is_prime(num):
  return num > 1 and not any(num % n == 0 for n in range(2,num))​

答案11:

from math import sqrt

def is_prime(n):
    if n < 2:
        return False
    elif n == 2:
        return True
    elif n % 2 == 0:
        return False
    else:
        for i in range(3, int(sqrt(n)) + 1, 2):
            if n % i == 0:
                return False
        else:
            return True​

答案12:

def is_prime(n):
    if n < 2:
        return False

    for i in range(3, int(n**0.5)+1, 2):
        if n%i == 0:
            return False
    return True​

答案13:

def is_prime(n):
  if n == 2:
      return True
  if n % 2 == 0 or n == 1:
      return False
  return all(n % i != 0 for i in range(3, int(n**0.5) + 1, 2))​