Question

我需要编写一个代码，找到一系列数字中的所有素数，然后列出它们，以便说明哪些是素数，哪些不是素数，如果它们不是素数，则表示它们可以被分割为什么数字。看起来应该是这样的：

>>> Prime(1,10)
1 is not a prime number 
2 is a prime number
3 is a prime number
4 is divisible by 2
5 is a prime number
6 is divisible by 2, 3
7 is a prime number
8 is divisible by 2, 4
9 is divisible by 3

到目前为止，我有这个只能确定哪些数字是素数并将它们打印在列表中。我不知道如何做非素数并打印它可被整除的数字。我也得到1是素数。

def primeNum(num1, num2):
   for num in range(num1,num2):
    prime = True
    for i in range(2,num):
        if (num%i==0):
            prime = False
    if prime:
       print (num,'is a prime number')

Answer 1

使用筛子可以解决问题：

示例：

from __future__ import print_function def primes(): """Prime Number Generator Generator an infinite sequence of primes http://stackoverflow.com/questions/567222/simple-prime-generator-in-python """ # Maps composites to primes witnessing their compositeness. # This is memory efficient, as the sieve is not "run forward" # indefinitely, but only as long as required by the current # number being tested. # D = {} # The running integer that's checked for primeness q = 2 while True: if q not in D: # q is a new prime. # Yield it and mark its first multiple that isn't # already marked in previous iterations # yield q D[q * q] = [q] else: # q is composite. D[q] is the list of primes that # divide it. Since we've reached q, we no longer # need it in the map, but we'll mark the next # multiples of its witnesses to prepare for larger # numbers # for p in D[q]: D.setdefault(p + q, []).append(p) del D[q] q += 1 def factors(n): yield 1 i = 2 limit = n**0.5 while i <= limit: if n % i == 0: yield i n = n / i limit = n**0.5 else: i += 1 if n > 1: yield n def primerange(start, stop): pg = primes() p = next(pg) for i in xrange(start, stop): while p < i: p = next(pg) if p == i: print("{0} is prime".format(i)) else: print("{0} is not prime and has factors: {1}".format(i, ", ".join(map(str, set(factors(i))))))

<强>输出：

>>> primerange(1, 10) 1 is not prime and has factors: 1 2 is prime 3 is prime 4 is not prime and has factors: 1, 2 5 is prime 6 is not prime and has factors: 1, 2, 3 7 is prime 8 is not prime and has factors: 1, 2 9 is not prime and has factors: 1, 3

Answer 2

您可以将每个数字的除数存储在列表中，然后使用", ".join(yourList)

打印它

即：

def primeNum(num1, num2):
   for num in range(num1,num2):
       divisors = []
       for i in range(2,num):
           if (num%i == 0):
               divisors.append(str(i))
       if divisors:
           print ('%d is divisible by ' %num + ', '.join(divisors))
       else:
           print ('%d is a prime number' %num)

编辑：哑语法错误

Answer 3

只需在将prime设置为false的位置添加打印和中断。

更优雅的解决方案是创建单独的函数isPrime或在内部for循环中使用break和else。无论哪种方式都会使素数变得不必要。

你只能逐个划分它自己，所以它至少是一个素数，至少按照这个定义。

在python中找到素数

3 个答案: