在python中查找前N个素数

Question

我是编程世界的新手.我只是在python中编写这段代码来生成N个素数.用户应输入N的值,即打印出的素数总数.我写了这段代码,但它没有抛出所需的输出.相反,它打印素数直到第N个数字.例如:用户输入N = 7的值.所需输出:2,3,5,7,11,13,19实际输出:2,3,5,7

好心提醒.

i=1
x = int(input("Enter the number:"))
for k in range (1, (x+1), 1):
    c=0
    for j in range (1, (i+1), 1):
        a = i%j
        if (a==0):
            c = c+1

    if (c==2):
          print (i)
    else:
          k = k-1

    i=i+1

Answer 1

使用正则表达式:)

#!/usr/bin/python

import re, sys


def isPrime(n):
    # see http://www.noulakaz.net/weblog/2007/03/18/a-regular-expression-to-check-for-prime-numbers/
    return re.match(r'^1?$|^(11+?)\1+$', '1' * n) == None


N = int(sys.argv[1]) # number of primes wanted (from command-line)
M = 100              # upper-bound of search space
l = list()           # result list

while len(l) < N:
    l += filter(isPrime, range(M - 100, M)) # append prime element of [M - 100, M] to l
    M += 100                                # increment upper-bound

print l[:N] # print result list limited to N elements

Answer 2

大卫·艾普斯坦(David Eppstein)实施的超级快速筛选- 我的PC上前1000个素数需要0.146秒:

def gen_primes():
    """ Generate an infinite sequence of prime numbers.
    """
    # 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

primes = gen_primes()


x = set()
y = 0
a = gen_primes()
while y < 10000:
  x |= set([a.next()])
  y+=1

print "x contains {:,d} primes".format(len(x))
print "largest is {:,d}".format(sorted(x)[-1])

Answer 3

作为参考,各种所述解决方案之间存在非常显着的速度差异.这是一些比较代码.Lennart指出的解决方案被称为"历史性",Ants提出的解决方案被称为"天真",而RC的解决方案被称为"regexp".

from sys import argv
from time import time

def prime(i, primes):
    for prime in primes:
        if not (i == prime or i % prime):
            return False
    primes.add(i)
    return i

def historic(n):
    primes = set([2])
    i, p = 2, 0
    while True:
        if prime(i, primes):
            p += 1
            if p == n:
                return primes
        i += 1

def naive(n):
    from itertools import count, islice
    primes = (n for n in count(2) if all(n % d for d in range(2, n)))
    return islice(primes, 0, n)

def isPrime(n):
    import re
    # see http://tinyurl.com/3dbhjv
    return re.match(r'^1?$|^(11+?)\1+$', '1' * n) == None

def regexp(n):
    import sys
    N = int(sys.argv[1]) # number of primes wanted (from command-line)
    M = 100              # upper-bound of search space
    l = list()           # result list

    while len(l) < N:
        l += filter(isPrime, range(M - 100, M)) # append prime element of [M - 100, M] to l
        M += 100                                # increment upper-bound

    return l[:N] # print result list limited to N elements

def dotime(func, n):
    print func.__name__
    start = time()
    print sorted(list(func(n)))
    print 'Time in seconds: ' + str(time() - start)

if __name__ == "__main__":
    for func in naive, historic, regexp:
        dotime(func, int(argv[1]))

我机器上n = 100的输出是:

naive
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541]
Time in seconds: 0.0219371318817
historic
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541]
Time in seconds: 0.00515413284302
regexp
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541]
Time in seconds: 0.0733318328857

如你所见,这是一个非常大的差异.这里再次为1000(删除了主要输出):

naive
Time in seconds: 1.49018788338
historic
Time in seconds: 0.148319005966
regexp
Time in seconds: 29.2350409031

Answer 4

这条线k = k-1不符合你的想法.它没有效果.改变k不会影响循环.在每次迭代时,k都会分配给范围的下一个元素,因此您k在循环内所做的任何更改都将被覆盖.

Answer 5

你想要的是这样的：

x = int(input("Enter the number:"))
count = 0
num = 2
while count < x:
     if isnumprime(x):
        print(x)
        count += 1
     num += 1

我将由您来实现isnumprime()；）提示：您只需要使用所有先前找到的素数来测试除法。