Question

找到第10,001个素数。

我试图在不使用复制和粘贴我不理解的代码的情况下解决Euler项目问题。我编写了代码，以查找数字是否为质数，并试图对其进行修改以运行前10,001个质数。我以为这会一直持续到我休息之前，但它并没有按预期工作。请记住，我尝试了其他几种方式对此进行编码，而这是我认为可行的方法。我想我对以前的生活有些混乱。

import math
import itertools
counter = 0
counters = 0
for i in itertools.count():
    for n in range (2, math.ceil(i/2)+1):
        if i%n == 0:
            counter+=1
    if counter == 0 or i == 2:
        counters+=1
    if counters == 10001:
        break
    else:
        pass
print(i)

Answer 1

您非常接近正确的解决方案。请记住下次再告诉我们您究竟期望什么以及获得了什么。不要写类似的东西

但是它没有按预期工作

为了使它运行并使其更快，您必须进行一些更改！

import math
import itertools
counter = 0
counters = 0
nth_prime = 0
for i in itertools.count():
    # ignore numbers smaller than 2 because they are not prime
    if i < 2:
        continue
    # The most important change: if you don't reset your counter to zero, all but the first prime number will be counted as not prime because your counter is always greater than zero
    counter = 0
    for n in range (2, math.ceil(i/2)+1):
        if i%n == 0:
            counter = 1
            # if you find out your number is not prime, break the loop to save calculation time
            break;

    # you don't need to check the two here because 2%2 = 0, thus counter = 1    
    if counter == 0:
        counters+=1
    else:
        pass
    if counters == 10001:
        nth_prime = i
        break
print(nth_prime)

该代码花了我36.1秒的时间才能找到104743作为第10.001个素数。

寻找主要的项目欧拉

1 个答案: