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小于或等于 x 的素数个数

[英]Number of primes less than or equal to x

π(x) = Number of primes ≤ x π(x) = 素数个数 ≤ x

Below code gives number of primes less than or equal to N下面的代码给出了小于或等于 N 的素数

It works perfect for N<=100000,它适用于 N<=100000,

  • Input - Output Table输入 - 输出表

    | Input | Output | |-------------|---------------| | 10 | 4✔ | | 100 | 25✔ | | 1000 | 168✔ | | 10000 | 1229✔ | | 100000 | 9592✔ | | 1000000 | 78521✘ |

    However, π(1000000) = 78498然而,π(1000000) = 78498

     import time def pi(x): nums = set(range(3,x+1,2)) nums.add(2) #print(nums) prm_lst = set([]) while nums: p = nums.pop() prm_lst.add(p) nums.difference_update(set(range(p, x+1, p))) #print(prm_lst) return prm_lst if __name__ == "__main__": N = int(input()) start = time.time() print(len(pi(N))) end= time.time() print(end-start)

You can read from this thread the fastest way like below and with this function for n = 1000000 I find correctly 78498 prime numbers.您可以像下面这样以最快的方式从该线程中读取数据,并使用此函数在n = 1000000正确找到78498素数。 (I change one line in this function) (我在此功能中更改了一行)

From:从:

return ([2] + [i for i in range(3,n,2) if sieve[i]])

To:到:

return len([2] + [i for i in range(3,n,2) if sieve[i]])

Finally:最后:

def primes(n):
    sieve = [True] * n
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
    return len([2] + [i for i in range(3,n,2) if sieve[i]])

inp =  [10, 100, 1000, 10000, 100000, 1000000]
for i in inp:
    print(f'{i}:{primes(i)}')

Output:输出:

10:4
100:25
1000:168
10000:1229
100000:9592
1000000:78498

You code is only correct if nums.pop() returns a prime, and that in turn will only be correct if nums.pop() returns the smallest element of the set.只有当nums.pop()返回素数时,您的代码才是正确的,而只有当nums.pop()返回集合中的最小元素时,代码才是正确的。 As far as I know this is not guaranteed to be true.据我所知,这不能保证是真的。 There is a third-party module called sortedcontainers that provides a SortedSet class that can be used to make your code work with very little change.有一个名为sortedcontainers的第三方模块,它提供了一个 SortedSet 类,可用于使您的代码只需很少的更改即可工作。

import time

import sortedcontainers
from operator import neg


def pi(x):
    nums = sortedcontainers.SortedSet(range(3, x + 1, 2), neg)
    nums.add(2)
    # print(nums)
    prm_lst = set([])
    while nums:
        p = nums.pop()
        prm_lst.add(p)
        nums.difference_update(set(range(p, x + 1, p)))
    # print(prm_lst)
    return prm_lst


if __name__ == "__main__":
    N = int(input())
    start = time.time()
    print(len(pi(N)))
    end = time.time()
    print(end - start)

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