stay

第一篇，放点干货。资源已经上传，下载积分设置为0。 openssh8.1p1的rpm包，供rhel6使用。已经在rhel6.8环境验证，上传后直接yum update *.rpm或yum install *.rpm使用。 自己做的spec文件，已经在自己目前所在公司的生产系统使用。 持续更新，如有任何问题，可直接微博下留言联系。 外链：https://pan.baidu.com/s/1q_WnUzsnGy3B-6llhzoHDw 提取码：78lt 来源： CSDN 作者： stay_hd 链接： https://blog.csdn.net/stay_hd/article/details/104345263

天津某百货大楼内部相继出现 5 例新冠肺炎确诊病例，从起初的 3 个病例来看，似乎找不到任何流行病学上的关联性。在这种背景之下，作为技术人员可以通过什么技术来找寻病例之间的联系呢？ 摘要 最初，nCoV 新冠病毒的扩散过程是由一个人（节点）向各其他人（节点）扩散的树状结构，但随着病毒的变异和人员交叉感染、“百家宴”、“联欢会”、“春运”等人员聚集，疫情扩散为网状结构。可以使用图数据库来存储相关人员、地理位置、感染时间等数据，本文将使用图数据库 Nebula Graph 作为工具，带大家一起探讨疫情的传播路径，并找到相关的疑似病例。 天津案例简述 下面用 Usr1、Usr2、Usr3、Usr4、Usr5 来代指这 5 例病例，看一下他们的行为轨迹： Usr1 信息: Usr1 于 1 月 24 日开始发热，在 1 月 22 日至 1 月 30 日期间在天津百货大厦 A 区工作，于 1 月 31 日确诊； Usr2 信息：Usr2 为 Usr1 丈夫，于 1 月 25 日开始出现腹泻症状，于 2 月 1 日确诊； Usr3 信息：Usr3 于 1 月 18 日接触过一个疑似病例，而后在天津百货大厦 B 区工作，于 1 月 24 日开始发热，于 2 月 1 日确诊； Usr4 信息：Usr4 于 1 月 12 日、13 日接触过疑似病例，而后在天津百货大厦 C 区工作，于 1 月 21

```java import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; /** * @program: suanfa * @description: * @author: 国宇航 * @create: 2019-12-31 11:07 **/ public class onezerozeroeight { public static void main(String[] args) throws IOException { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in)); //当前要到的楼层 int cur = 0; //当前停留的楼层 int stay = 0; int sum = 0; String[] split = bufferedReader.readLine().split(" "); for (int i = 0;i<Integer.parseInt(split[0]);i++){ cur = Integer.parseInt(split[i+1]); if(stay<=cur){ sum+=(cur-stay)*6+5

题目如下： You have a pointer at index 0 in an array of size arrLen . At each step, you can move 1 position to the left, 1 position to the right in the array or stay in the same place (The pointer should not be placed outside the array at any time). Given two integers steps and arrLen , return the number of ways such that your pointer still at index 0 after exactly steps steps. Since the answer may be too large, return it modulo 10^9 + 7 . Example 1: Input: steps = 3, arrLen = 2 Output: 4 Explanation: There are 4 differents ways to stay at index 0 after 3 steps. Right, Left, Stay Stay, Right, Left

https://codeforces.com/contest/1236/problem/D 题意：一个n*m格子矩阵，放一个人偶在左上角向右走，只能在每个格子最多右转一次，有k个障碍物。求是否能够一次走完矩阵的所有非障碍物格然后停留在任意位置。 题解：在每个格子最多右转一次，相当于每个格子只能走一次，否则就出不来了。容易想到障碍物必须也是占据一些蛇形的片段，并且留下的位置刚好可以让人偶走一个蛇形的绕到中心，但是怎么判断障碍物的形状呢？百思不得其解遂看题解。题解表示观察到人偶撞墙或者撞障碍物必转向（自己走过的路也是墙）。然后暴力模拟一遍，每次在数据结构里面搜索前进方向上最近的墙/障碍物，走到它面前。至多进行n+m次。 所以选择一个结构就是set，维护行和列分别两堆set，一个障碍物同时插入行列两个set里面。然后墙/自己走过形成的墙就设置一个最值把多走的部分截断就可以了。判断答案的方法就是暴力统计走过的格子数是否等于空格数。 注：实现的时候卡在了样例4，因为会让人偶走回头路，但是也不能够简单让人偶移动0格就退出，比如一条竖线的矩阵。修复的方法是在人偶的后方一个格子塞一个障碍物。然后卡在样例9，看了dalao的代码才知道错在哪里，首先while循环的停止条件错了，不应该是d>u||r>l，应该是d>=u&&r>=l，因为是闭区间。这个改进再加上后面堵一个障碍物就可以防止原路返回了

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: EDIT : Sample project available on github . I'm using Neo4J (Rest graph database, hosted in grapheneDb) and Spring Data in our backend project. <bean id="graphDatabaseService" class="org.springframework.data.neo4j.rest.SpringCypherRestGraphDatabase"> I have a simple one-to-many relationship between two entities: User and Stay . EDIT : I thought this wasn't relevant for the issue, but after seeing a similar problem in SDN4, I think I need to update the question (there is a basic @NodeEntity class, and both entities are extending this base

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I'm trying to build an xml document in a JUnit test. doc=docBuilder.newDocument(); Element root = doc.createElement("Settings"); doc.appendChild(root); Element label0 = doc.createElement("label_0"); root.appendChild(label0); String s=doc.getTextContent(); System.out.println(s); Yet the document stays empty (i.e. the println yields null .) I don'thave a clue why that is. The actual problem is that a subsequent XPath expression throws the error: Unable to evaluate expression using this context . 回答1: The return value of getTextContent on

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have a form that looks like this and I would like to stay on the same page, when Submit is clicked, but still have receiver.pl executed. How should that be done? 回答1: The easiest answer: jQuery. Do something like this: $(document).ready(function(){ var $form = $('form'); $form.submit(function(){ $.post($(this).attr('action'), $(this).serialize(), function(response){ // do something here on success },'json'); return false; }); }); If you want to add content dynamically and still need it to work, and also with more than one form, you can do

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have next working code in my SpringMVC controller: @RequestMapping(value = "/register", method = RequestMethod.GET) public void registerForm(Model model) { model.addAttribute("registerInfo", new UserRegistrationForm()); } @RequestMapping(value = "/reg", method = RequestMethod.POST) public String create( @Valid @ModelAttribute("registerInfo") UserRegistrationForm userRegistrationForm, BindingResult result) { if (result.hasErrors()) { return "register"; } userService.addUser(userRegistrationForm); return "redirect:/"; } In short create

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I noticed that while streaming audio from a remote server through 3G (mobile) connection and while the WIFI is disconnected or OFF, as soon as WIFI is activated and connected, connection through 3G is dropped. I want the app keep using 3G even if WIFI is connected too now. I want to do this to keep continuity. (User may opt-in/out to/from this behaviour). Is there a special flag, lock, etc.. For this purpose? 回答1: This isn't possible on devices before Android 5.0 (Lollipop). The OS only keeps one interface up at a time, and applications don