帮助ProjectEuler＃200？[重复]-HelpwithProjectEuler#200?[duplicate]

I'm a Project Euler administrator. Please do not post information that can spoil the problem for others, particularly code and answers, even half-functioning code. Please edit your question accordingly. EDIT: Thank you for doing so!

我是Project Euler管理员。请不要发布可能破坏他人问题的信息,特别是代码和答案,甚至是半功能的代码。请相应地修改您的问题。编辑:谢谢你这样做!

It's not unusual for solvers to use the web to search for information on solving a problem, and it would take away some fun if they stumbled upon such a spoiler. (Yes, I know there are sites with lots of solutions ready-made, but at least they're generally for the lowered numbered easy problems.)

解算器使用网络搜索解决问题的信息并不罕见,如果他们偶然发现这样的剧透,它会带来一些乐趣。 (是的,我知道有些网站有很多现成的解决方案,但至少它们通常用于降低编号的简单问题。)

We have forums for discussing difficulties with problems and getting hints, which are aggressively edited for spoilers.

我们有论坛讨论问题的困难和提示,这些都是为剧透主义者积极编辑的。

Aren't you supposed to think of a clever solution which doesn't take a day or even an hour to run? :D I think the problem is isProbablePrime, which doesn't guarantee that a number is a prime. It just says that a prime which was found could be a prime with a certain probability. You should use an algorithm which is certain that it has found a prime.

难道你不应该想到一个不需要一天甚至一小时才能运行的聪明解决方案吗? :D我认为问题是isProbablePrime,它不保证数字是素数。它只是说发现的素数可能是具有一定概率的素数。您应该使用一种确定已找到素数的算法。

The first answer is incorrect because isProbablyPrime isn't always correct (hence the Probably). It's slow, in part, because you are using BigInteger. All the values involved will fit in a long. Why not use a long?

第一个答案是不正确的,因为isProbablyPrime并不总是正确的(因此可能)。它的速度很慢,部分原因是因为你使用的是BigInteger。所涉及的所有价值都适合长期。为什么不用多久?

There may be some simple refactorings you could do that would save some time. It seems that the bulk of the time is spent in the nested for loop. The order of magnitude is n squared. You could reduce the complexity by not nesting loops. Another problem is that you are finding more potential results than are required. You only need to find 200. The reason you are needing to find more is due to the fact that you are not finding the potential results in their numeric order. The TreeSet does keep the results in order, but the algorithm would be faster if it were able to stop when the 200th result was found.

可能会有一些简单的重构,可以节省一些时间。似乎大部分时间都花在嵌套for循环中。数量级为n平方。您可以通过不嵌套循环来降低复杂性。另一个问题是您发现了比所需更多的潜在结果。您只需要找到200.您需要找到更多的原因是因为您没有在数字顺序中找到可能的结果。 TreeSet确实按顺序保存结果,但如果在找到第200个结果时它能够停止,则算法会更快。