错误找到了，在对已经配置好侦听器的AlwaysON组添加跨网段的节点的时候，先要删除侦听器，然后再把跨网段的节点加入到AlwaysON组，再重新创建侦听器，此时，新的侦听器包含两个网段的IP
3条回答

采纳
点赞 评论 复制链接分享

采纳
你需要检查以下的几点配置是否存在问题。
 检查onpremise subnet and Azure virtual network subnet之间能否正常通信。
 在步骤1正常通信的前提下，检查可用性组侦听器中配置的subnet, ip address (推荐使用静态ip)是否正确。注意可用性组侦听器中需要添加两个不同subnet的ip address。
 检查步骤2中配置的ip address是否有效。
点赞 评论 复制链接分享 
采纳
先要给下site to site的网络配置谢谢
点赞 评论 复制链接分享
提交
相关推荐
 4年前回答 1 已采纳 Problem Description Bubu's bookshelf is in a mess! Help him! There are n books on his bookshelf. We define the mess value to be the number of segments of consecutive equalheight books. For example, if the book heights are 30, 30, 31, 31, 32, the mess value is 3, that of 30, 32, 32, 31 is also 3, but the mess value of 31, 32, 31, 32, 31 is 5  it's indeed in a mess! Bubu wants to reduce the mess value as much as possible, but he's a little bit tired, so he decided to take out at most k books, then put them back somewhere in the shelf. Could you help him? Input There will be at most 20 test cases. Each case begins with two positive integers n and k (1 <= k <= n <= 100), the total number of books, and the maximum number of books to take out. The next line contains n integers, the heights of each book, from left to right. Each height is an integer between 25 and 32, inclusive. The last test case is followed by n = k = 0, which should not be processed. Output For each test case, print the case number and the minimal final mess value. Print a blank line after the output of each test case. Sample Input 5 2 25 25 32 32 25 5 1 25 26 25 26 25 0 0 Sample Output Case 1: 2 Case 2: 3
 回答 2 已采纳 我的python版本2.7.12，系统win10，使用help命令有问题。 例如，help(int) ![图片说明](https://imgask.csdn.net/upload/201706/03/1496482109_999708.png) 进入help模式能够按照提示查找帮助 ![图片说明](https://imgask.csdn.net/upload/201706/03/1496482254_439844.png) 请问是什么原因
 启动tomcat，maven报错：[ERROR] [Help 1] http://cwiki.apache.org/confluence/display/MAVEN/PluginConfigurationException javajavaeemaventomcateclipse2年前回答 2 已采纳 ``` [INFO] Scanning for projects... [INFO] [INFO]  [INFO] Building taotaomanageweb 1.0.0SNAPSHOT [INFO]  [INFO] [INFO] >>> tomcat7mavenplugin:2.2:run (defaultcli) > processclasses @ taotaomanageweb >>> [WARNING] The artifact org.apache.commons:commonsio:jar:1.3.2 has been relocated to commonsio:commonsio:jar:1.3.2 [INFO] [INFO]  mavenresourcesplugin:2.7:resources (defaultresources) @ taotaomanageweb  [INFO] Using 'UTF8' encoding to copy filtered resources. [INFO] Copying 7 resources [INFO] [INFO]  mavencompilerplugin:3.2:compile (defaultcompile) @ taotaomanageweb  [INFO] Nothing to compile  all classes are up to date [INFO] [INFO] [Help 1] [ERROR] [ERROR] To see the full stack trace of the errors, rerun Maven with the e switch. [ERROR] Rerun Maven using the X switch to enable full debug logging. [ERROR] [ERROR] For more information about the errors and possible solutions, please read the following articles: [ERROR] [Help 1] http://cwiki.apache.org/confluence/display/MAVEN/PluginConfigurationException ``` 烦烦烦，几个这种Bug,刚刚弄完一个，现在又出现一个，repository换了一个新的，前面的几个这样的解决了，现在又冒出一个这个问题。求大佬帮忙！
 回答 4 已采纳 ![图片说明](https://imgask.csdn.net/upload/201811/12/1542021051_792846.png) mvn v成功 镜像也换过了 代理也试过了 都无法解决该问题
 4年前回答 2 已采纳 Background If thou doest well, shalt thou not be accepted? and if thou doest not well, sin lieth at the door. And unto thee shall be his desire, and thou shalt rule over him. And Cain talked with Abel his brother: and it came to pass, when they were in the field, that Cain rose up against Abel his brother, and slew him. And the LORD said unto Cain, Where is Abel thy brother? And he said, I know not: Am I my brother's keeper? And he said, What hast thou done? the voice of thy brother's blood crieth unto me from the ground. And now art thou cursed from the earth, which hath opened her mouth to receive thy brother's blood from thy hand; When thou tillest the ground, it shall not henceforth yield unto thee her strength; a fugitive and a vagabond shalt thou be in the earth. —— Bible Chapter 4 Now Cain is unexpectedly trapped in a cave with N paths. Due to LORD's punishment, all the paths are zigzag and dangerous. The difficulty of the ith path is ci. Then we define f as the fighting capacity of Cain. Every day, Cain will be sent to one of the N paths randomly. Suppose Cain is in front of the ith path. He can successfully take ti days to escape from the cave as long as his fighting capacity f is larger than ci. Otherwise, he has to keep trying day after day. However, if Cain failed to escape, his fighting capacity would increase ci as the result of actual combat. (A kindly reminder: Cain will never died.) As for ti, we can easily draw a conclusion that ti is closely related to ci. Let's use the following function to describe their relationship: After D days, Cain finally escapes from the cave. Please output the expectation of D. Input The input consists of several cases. In each case, two positive integers N and f (n ≤ 100, f ≤ 10000) are given in the first line. The second line includes N positive integers ci (ci ≤ 10000, 1 ≤ i ≤ N) Output For each case, you should output the expectation(3 digits after the decimal point). Sample Input 3 1 1 2 3 Sample Output 6.889
 回答 1 已采纳 I'm learning Go and I have some problem. I following go install But an error: no install location for directory /Users/skan/documents/study/golang/src/section1 outside GOPATH For more details see: ’go help gopath’ so I return to my .bash/profile this is my .bash/profile # added by Anaconda3 2018.12 installer # >>> conda init >>> # !! Contents within this block are managed by ’conda init’ !! __conda_setup=”$(CONDA_REPORT_ERRORS=false ’/anaconda3/bin/conda’ shell.bash hook 2> /dev/null)” if [ $? eq 0 ]; then \eval ”$__conda_setup” else if [ f ”/anaconda3/etc/profile.d/conda.sh” ]; then . ”/anaconda3/etc/profile.d/conda.sh” CONDA_CHANGEPS1=false conda activate base else \export PATH=”/anaconda3/bin:$PATH:$GOPATH:$GOBIN” fi fi unset __conda_setup # <<< conda init <<< export GOPATH="/Users/skan/Documetns/study/golang" export GOBIN=$GOPATH/bin
 4年前回答 1 已采纳 Problem Description “Help! Help!” While walking in the park, you suddenly hear someone shouting for help. You immediately realize that a person has fallen into the lake. As a brave man, you decide to save him. You are really familiar with the terrain of the park. The park can be regarded as a 2D plane. And the lake is a convex polygon. At current, you are on (Xo, Yo), and the person is on (Xp, Yp). You also know that you can run Vr per second on the land, or swim Vs per second in the lake. Notice that you are allowed to run along the edge of the lake. You are not good at swimming. You cannot stay in the lake longer than Ts second. And carrying another person will cut down your swimming speed by half. Can you save the poor guy? What is the minimum time for you to reach him, and carry him back to the border of the lake? Input There are several test cases in the input. The first line contains an integer T (1 <= T <= 20)  the number of test cases. For each case: The first line contains three real numbers Ts, Vr, Vs. 0 < Ts < 108, 0 < Vs < Vr < 108. The second line contains two real numbers Xo, Yo, indicate the position (Xo, Yo) of you at current. The third line contains two real numbers Xp, Yp, indicate the position (Xp, Yp) of the person you are going to save. The forth line contains only one integer N  the number of vertices of the lake. 3 <= N <= 50000. The follow N lines, each line contains two real numbers x, y, indicating one of the vertex (x, y) of the lake. The vertices of lake are listed in either clockwise or counterclockwise order. Each coordinate in the input does not exceed 106 by its absolute value. Your position is on the land and the person’s is in the lake. Output For each test case, output the minimum time(in seconds) to save the poor person, rounded to two digits after the decimal point. If you cannot save he, output “1” instead. Sample Input 2 100 2 1 0 10 0 0 3 1 1 1 1 0 1 1 2 1 0 10 0 0 3 1 1 1 1 0 1 Sample Output 6.39 1
 回答 1 已采纳 Problem Description A doll vendor has just received a shipment of dolls from the factory. The vendor can only sell dolls in pairs, and have to ensure that in each pair one doll is K times as large as the other. What is the maximum number of pairs that the vendor can assemble? Each doll can only be sold one time. Now the vendor wants your help to solve this problem. Of course, he will give you 30% of the profit, so, what are you waiting for? Input There are multiple test cases.For each case the input contains two integers N (1<=50<=N) and K (1<=K<=25) in first line and N doll sizes (integers between 1 and 100000, inclusive) below, where N is the number of dolls. The doll sizes may separated by white spaces or newline.Input ends when N and K is 0. Output For each case, output the maximum number of pairss that the vendor can assemble in one line. Sample Input 5 2 1 2 1 2 4 0 0 Sample Output 2 Hint: you can sell two pairs at most, {(1,2),(1,2)} or {(1,2),(2,4)}.
 回答 1 已采纳 Description Preparing a problem for a programming contest takes a lot of time. Not only do you have to write the problem description and write a solution, but you also have to create difficult input files. In this problem, you get the chance to help the problem setter to create some input for a certain problem. For this purpose, let us select the problem which was not solved during last year's local contest. The problem was about finding the optimal binary search tree, given the probabilities that certain nodes are accessed. Your job will be: given the desired optimal binary search tree, find some access probabilities for which this binary search tree is the unique optimal binary search tree. Don't worry if you have not read last year's problem, all required definitions are provided in the following. Let us define a binary search tree inductively as follows: The empty tree which has no node at all is a binary search tree Each nonempty binary search tree has a root, which is a node labelled with an integer, and two binary search trees as left and right subtree of the root A left subtree contains no node with a label >= than the label of the root A right subtree contains no node with a label <= than the label of the root Given such a binary search tree, the following search procedure can be used to locate a node in the tree: Start with the root node. Compare the label of the current node with the desired label. If it is the same, you have found the right node. Otherwise, if the desired label is smaller, search in the left subtree, otherwise search in the right subtree. The access cost to locate a node is the number of nodes you have to visit until you find the right node. An optimal binary search tree is a binary search tree with the minimum expected access cost. Input The input contains several test cases. Each test case starts with an integer n (1 <= n <= 50), the number of nodes in the optimal binary search tree. For simplicity, the labels of the nodes will be integers from 1 to n. The following n lines describe the structure of the tree. The ith line contains the labels of the roots of the left and right subtree of the node with label i (or 1 for an empty subtree). You can assume that the input always defines a valid binary search tree. The last test case is followed by a zero. Output For each test case, write one line containing the access frequency for each node in increasing order of the labels of the nodes. To avoid problems with floating point precision, the frequencies should be written as integers, meaning the access probability for a node will be the frequency divided by the sum of all frequencies. Make sure that you do not write any integer bigger than 263  1 (the maximum value fitting in the GCC/G++ data type long long , the Java data type long or VC data type __int64). Otherwise, you may produce any solution ensuring that there is exactly one optimal binary search tree: the binary search tree given in the input. Sample Input 3 1 1 1 3 1 1 10 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 1 1 0 Sample Output 1 1 1 512 256 128 64 32 16 8 4 2 1