在python中找到前N个素数 [英] To find first N prime numbers in python

问题描述

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，即要打印的素数总数.我已经编写了这段代码，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.

I am new to the programming world. I was just writing this code in python to generate N prime numbers. User should input the value for N which is the total number of prime numbers to print out. I have written this code but it doesn't throw the desired output. Instead it prints the prime numbers till the Nth number.

例如:用户输入 N = 7 的值.

For example: User enters the value of N = 7.

期望输出:2、3、5、7、11、13、19

Desired output: 2, 3, 5, 7, 11, 13, 19

实际输出:2、3、5、7

Actual output: 2, 3, 5, 7

请多多指教.

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

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

    i = i + 1

推荐答案

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

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