 Magic Number

Problem Description
There are many magic numbers whose lengths are less than 10. Given some queries, each contains a single number, if the Levenshtein distance (see below) between the number in the query and a magic number is no more than a threshold, we call the magic number is the lucky number for that query. Could you find out how many luck numbers are there for each query?Levenshtein distance (from Wikipedia http://en.wikipedia.org/wiki/Levenshtein_distance):
In information theory and computer science, the Levenshtein distance is a string metric for measuring the amount of difference between two sequences. The term edit distance is often used to refer specifically to Levenshtein distance.
The Levenshtein distance between two strings is defined as the minimum number of edits needed to transform one string into the other, with the allowable edit operations being insertion, deletion, or substitution of a single character. It is named after Vladimir Levenshtein, who considered this distance in 1965.
For example, the Levenshtein distance between "kitten" and "sitting" is 3, since the following three edits change one into the other, and there is no way to do it with fewer than three edits:
1.kitten → sitten (substitution of 's' for 'k')
2.sitten → sittin (substitution of 'i' for 'e')
3.sittin → sitting (insertion of 'g' at the end).Input
There are several test cases. The first line contains a single number T shows that there are T cases. For each test case, there are 2 numbers in the first line: n (n <= 1500) m (m <= 1000) where n is the number of magic numbers and m is the number of queries.
In the next n lines, each line has a magic number. You can assume that each magic number is distinctive.
In the next m lines, each line has a query and a threshold. The length of each query is no more than 10 and the threshold is no more than 3.Output
For each test case, the first line is "Case #id:", where id is the case number. Then output m lines. For each line, there is a number shows the answer of the corresponding query.Sample Input
1
5 2
656
67
9313
1178
38
87 1
9509 1Sample Output
Case #1:
1
0
错误的幻数  将shapefile上传到GeoServer_course
20171011<div class="posttext" itemprop="text"> <p>I successfully created workspace using the GeoServer REST API. But I am having problem uploading shape file.</p> <p>I followed <a href="http://docs.geoserver.org/2.0.0/user/extensions/rest/restconfigexamplescurl.html#uploadingashapefile" rel="nofollow noreferrer">this</a> guide.</p> <pre><code>curl u admin:geoserver XPUT H 'Contenttype: application/zip' databinary "@D:/trash.zip" http://localhost:9090/geoserver/rest/workspaces/string/datastores/trash/file.shp </code></pre> <p>PHP  Equivalent</p> <pre><code>$ch = curl_init(); curl_setopt($ch, CURLOPT_URL, "http://localhost:9090/geoserver/rest/workspaces/string/datastores/trash/file.shp"); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); $post = array( "file" => "@" .realpath("D:/trash.zip") ); curl_setopt($ch, CURLOPT_POSTFIELDS, $post); curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "PUT"); curl_setopt($ch, CURLOPT_USERPWD, "admin" . ":" . "geoserver"); $headers = array(); $headers[] = "ContentType: application/zip"; curl_setopt($ch, CURLOPT_HTTPHEADER, $headers); $result = curl_exec($ch); echo $result; if (curl_errno($ch)) { echo 'Error:' . curl_error($ch); } curl_close ($ch); </code></pre> <p>When I run it I get:</p> <blockquote> <p>java.lang.RuntimeException: java.io.IOException: Wrong magic number, expected 9994, got 757935405</p> </blockquote> <p>Have no clue how to upload shapefile to geoserver. Thank you!</p> </div>
在linux中执行java.jar时报错_course
20110408[devafs:/home/devafs/batch/sbin] sh start.sh Usage: java [options] class [args...] (to execute a class) or java [options] jar jarfile [args...] (to execute a jar file) where options include: d32 use a 32bit data model if available d64 use a 64bit data model if available client to select the "client" VM server to select the "server" VM hotspot is a synonym for the "client" VM [deprecated] The default VM is server, because you are running on a serverclass machine. cp <class search path of directories and zip/jar files> classpath <class search path of directories and zip/jar files> A : separated list of directories, JAR archives, and ZIP archives to search for class files. D<name>=<value> set a system property verbose[:classgcjni] enable verbose output version print product version and exit version:<value> require the specified version to run showversion print product version and continue jrerestrictsearch  jrenorestrictsearch include/exclude user private JREs in the version search ? help print this help message X print help on nonstandard options ea[:<packagename>...:<classname>] enableassertions[:<packagename>...:<classname>] enable assertions da[:<packagename>...:<classname>] disableassertions[:<packagename>...:<classname>] disable assertions esa  enablesystemassertions enable system assertions dsa  disablesystemassertions disable system assertions agentlib:<libname>[=<options>] load native agent library <libname>, e.g. agentlib:hprof see also, agentlib:jdwp=help and agentlib:hprof=help agentpath:<pathname>[=<options>] load native agent library by full pathname javaagent:<jarpath>[=<options>] load Java programming language agent, see java.lang.instrument splash:<imagepath> show splash screen with specified image invalid file (bad magic number): Exec format error invalid file (bad magic number): Exec format error invalid file (bad magic number): Exec format error invalid file (bad magic number): Exec format error invalid file (bad magic number): Exec format error Failed to load MainClass manifest attribute from /home/devafs/batch/lib/db2jcc.jar start.sh #!/bin/sh export args="" while [ $# ne 0 ] do export args="$args""$1 " shift done java Xms512M Xmx512M Dfile.encoding=GBK classpath ../lib/c3p00.9.1.2.jar;../lib/commonsbeanutils1.8.0.jar;../lib/commonsdigester2.0.jar;../lib/commonsio1.4.jar;../lib/commonslang2.5.jar;../lib/commonslogging1.1.1.jar;../lib/db2jcc.jar;../lib/freemarker2.3.8.jar;../lib/junit4.8.1.jar;../lib/log4j1.2.16.jar;../lib/slf4japi1.6.1.jar;../lib/slf4jlog4j121.6.1.jar;../lib/springasm3.0.4.RELEASE.jar;../lib/springcore3.0.4.RELEASE.jar;../lib/topbatch1.2.jar com.huateng.topafs.frame.BatchRunner $args 描述：代码在doc界面中执行成功，但是在linux下执行则报此错误，请知道的朋友指点。
用python的pcapfile解析pcapng文件时出错：pcapfile.UnknownMagicNumber: No supported Magic Number found_course
20190826在做攻防世界crypto时看到一个题是有关pcapng文件的，我想用调用python的pcapfile库去解析pacpng文件里的内容，但是出错了，显示UnknownMagicNumber: No supported Magic Number found ![图片说明](https://imgask.csdn.net/upload/201908/26/1566804985_691013.png) 本人新手一枚。。。求大佬出手相助
tensorflow模型载入失败_course
20180521载入模型的时候不知道怎么出现这种报错 DataLossError (see above for traceback): Unable to open table file .\eyemodel\eye_kaggle.ckpt192.meta: Data loss: not an sstable (bad magic number): perhaps your file is in a different file format and you need to use a different restore operator?
Happy Matt Friends _course
20170825Problem Description Matt has N friends. They are playing a game together. Each of Matt’s friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusiveor) sum of the selected friends’magic numbers is no less than M , Matt wins. Matt wants to know the number of ways to win. Input The first line contains only one integer T , which indicates the number of test cases. For each test case, the first line contains two integers N, M (1 ≤ N ≤ 40, 0 ≤ M ≤ 106). In the second line, there are N integers ki (0 ≤ ki ≤ 106), indicating the ith friend’s magic number. Output For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y indicates the number of ways where Matt can win. Sample Input 2 3 2 1 2 3 3 3 1 2 3 Sample Output Case #1: 4 Case #2: 2
在c＃中解压缩由php的gzcompress（）压缩的字符串_course
20101102<div class="posttext" itemprop="text"> <p>i'm querying a webservice in c# 4.0, which provides me with a string compressed by php's <a href="http://php.net/gzcompress" rel="nofollow noreferrer">gzcompress()</a>. Now i need to decompress this string in c#. I tried several ways including</p> <ul> <li><a href="https://stackoverflow.com/questions/1522040/gzipstreamdecompressionperformanceispoor">GZipStream decompression performance is poor</a></li> <li><a href="https://stackoverflow.com/questions/1408267/ctojavabase64stringmemorystreamgzipstream">C# to Java: Base64String, MemoryStream, GZipStream</a></li> <li><a href="https://stackoverflow.com/questions/1025777/howtosolvegzipmagicnumbermissing">How to solve Gzip Magic Number Missing</a></li> </ul> <p>but everytime i get an "Missing Magic Number" exception.</p> <p>Can someone provide me with some hints?</p> <p>Thank you</p> <p><strong>Edit 1:</strong></p> <p>My latest try:</p> <pre><code>public static string Decompress(string compressed) { byte[] compressedBytes = Encoding.ASCII.GetBytes(compressed); MemoryStream mem = new MemoryStream(compressedBytes); GZipStream gzip = new GZipStream(mem, CompressionMode.Decompress); StreamReader reader = new StreamReader(gzip); return reader.ReadToEnd(); } </code></pre> </div>
无法启动进程：字节0x0处的记录中无效的幻数Delve Golang_course
20160610<div class="posttext" itemprop="text"> <p>Just got a:</p> <pre><code>could not launch process: invalid magic number in record at byte 0x0 </code></pre> <p>When running on OSX after working fine for a while:</p> <pre><code>dlv debug main.go </code></pre> <p>reinstalled Delve, but still persists. What can it be?</p> </div>
从小到大输出魔法数字，用C语言的程序编写设计的代码的代码编写的思想方法来做_course
20190610Problem Description On New Year Festival, Liu Qian’s magic impressed on little WisKey’s heart and he wants to learn some magic to make himself stronger. One day, he met a cowman named LinLe. Linle is very nice, he told little WisKey the mysteries of magic. Now, little WisKey began to perform the magic to you. “Hello, Everybody. I have five decimal numbers named a, b, c, d, e, (0 <= a, b, c, d, e <= 9) and I rearranged them, and then combined them into a number, for example <a, b, c, d, e> = a*10000 + b*1000 + c*100 + d*10 + e*1. You know the number of permutations is 5! = 120. So you have 120 numbers in your hands, you can pick a number N from the 120 numbers and calculate the sum S of remain 119 numbers. AHA~, If you tell me the S, I can guess the N~!” It’s easy? Okay, you can challenge this. Input Each line will contain an integer S. Process to end of file. Output For each case, output all possible N, print the number N with 5 digits, including the leading zeros, one line per case. I promise every case have one solution at least. If have many N, please output them from small to large in one line, separate them with a blank space. Sample Input 2933266 6392217 4245386 Sample Output 00038 07719 21238
Magic Square _course
20170327In recreational mathematics, a magic square of ndegree is an arrangement of n2 numbers, distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant. For example, the picture below shows a 3degree magic square using the integers of 1 to 9. Given a finished number square, we need you to judge whether it is a magic square. Input The input contains multiple test cases. The first line of each case stands an only integer N (0 < N < 10), indicating the degree of the number square and then N lines follows, with N positive integers in each line to describe the number square. All the numbers in the input do not exceed 1000. A case with N = 0 denotes the end of input, which should not be processed. Output For each test case, print "Yes" if it's a magic square in a single line, otherwise print "No". Sample Input 2 1 2 3 4 2 4 4 4 4 3 8 1 6 3 5 7 4 9 2 4 16 9 6 3 5 4 15 10 11 14 1 8 2 7 12 13 0 Sample Output No No Yes Yes
1/8圆的贝塞尔曲线的m值怎么换算？_course
20180218一般的情形是用贝塞尔曲线画一个1/4的直径是1的圆，![图片说明](https://imgask.csdn.net/upload/201802/18/1518961288_614959.png)但是我想划分成1/8的圆，那图片中的M值是多少？怎么换算
Magic Trick _course
20161202Background Warning! This problem statement contains a serious spoiler. It shows the trick behind a magic trick. So if you still want to be amazed in case somebody shows this trick to you then do NOT read the rest of this problem statement. Stop reading… NOW! Problem Well, you’re still reading, so obviously you have no respect for magic tricks. Be ashamed, please. Ok,here’s what happens. The magician shows you a text with three paragraphs like this one: It was a horribly dark night. The moon was shining, but not much. A suspicious stranger entered the bar and went straight to John Doe. “I’m searching for aliens, can I borrow your computer?”, he said. He then asks you to secretly pick a word in the first paragraph. Then you shall do this: 1. Count the number of characters in your word (call that number X). 2. From your word move on X words. Repeat these two steps until you reach the third paragraph. Then tell the magician that you’re done.After some hocus pocus he tells you the word you ended up with. For our purposes, a “word” is defined as consecutive letters (AZ,az). For example, “I’m” is regarded as two separate words. For example, let’s say you choose “night” in the above example. It has 5 characters, so you move on five words: “The”, “moon”, “was”, “shining”, “but”. Our new word is “but”. You move on 3 words to “A”,then 1 to “suspicious”, then 10 to “Doe” and then 3 to “searching”. Now you tell the magician that you’re ready. He says that you’ve reached “searching”. How can he know? Well, it doesn’t matter where you start in the first paragraph, you’ll always end up at “searching”. The magician needs new texts and asks you to help him to find all possible outcomes (in the above example, “searching” is the only one). Apart from words, a possible outcome is “outside”,which means it’s possible to jump behind the third paragraph. Also, he’s not interested if more than three outcomes are possible. 输入: The first line contains the number of scenarios. For each scenario, three lines are given, representing the three paragraphs. No line is longer than 100000 characters. Every paragraph will contain at least one word. 输出: The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Then print the possible outcomes (possibly including “outside”) in alphabetical/lexicographical order, one word per line. Write words in lower case. Don’t list outcomes more than once. If however there are more than three possible outcomes, then print “too many” and do *not* print any of them. Terminate the output for the scenario with a blank line. 样例输入: 4 It was a horribly dark night. The moon was shining, but not much. A suspicious stranger entered the bar and went straight to John Doe. "I'm searching for aliens, can I borrow your computer?", he said. !pablo espanol! !pablo espanol! !pablo espanol! c'mon howLongOrShortCanASingleWordBe? a b c d e f g f e d c b a 54254#@%$^%^@4626^#^%^$hahaha#$@%#$@63456326 Hello buddy dance tango! PleaseOhPleaseJumpOverMe This is too much for me... 样例输出: Scenario #1: searching Scenario #2: outside espanol Scenario #3: outside hahaha Scenario #4: too many
aidl 向远程服务注册listener问题_course
20160405在android开发中遇到一个aidl问题。 向远程服务注册监听器，但是报一个Bad magic number for Bundle错误 在普通用法是没有问题的，这个错误是因为我有个需求是在静态方法中调用 有没有人遇到过这样的问题？
Magic Bitstrings _course
20171016Description A bitstring, whose length is one less than a prime, might be magic. 1001 is one such string. In order to see the magic in the string let us append a nonbit x to it, regard the new thingy as a cyclic string, and make this square matrix of bits each bit 1001 every 2nd bit 0110 every 3rd bit 0110 every 4th bit 1001 This matrix has the same number of rows as the length of the original bitstring. The mth row of the matrix has every mth bit of the original string starting with the mth bit. Because the enlarged thingy has prime length, the appended x never gets used. If each row of the matrix is either the original bitstring or its complement, the original bitstring is magic. Input Each line of input (except last) contains a prime number p <= 100000. The last line contains 0 and this line should not be processed. Output For each prime number from the input produce one line of output containing the lexicographically smallest, nonconstant magic bitstring of length p1, if such a string exists, otherwise output Impossible. Sample Input 5 3 17 47 2 79 0 Sample Output 0110 01 0010111001110100 0000100001101010001101100100111010100111101111 Impossible 001001100001011010000001001111001110101010100011000011011111101001011110011011
Magic WisKey _course
20171127Problem Description On New Year Festival, Liu Qian’s magic impressed on little WisKey’s heart and he wants to learn some magic to make himself stronger. One day, he met a cowman named LinLe. Linle is very nice, he told little WisKey the mysteries of magic. Now, little WisKey began to perform the magic to you. “Hello, Everybody. I have five decimal numbers named a, b, c, d, e, (0 <= a, b, c, d, e <= 9) and I rearranged them, and then combined them into a number, for example <a, b, c, d, e> = a*10000 + b*1000 + c*100 + d*10 + e*1. You know the number of permutations is 5! = 120. So you have 120 numbers in your hands, you can pick a number N from the 120 numbers and calculate the sum S of remain 119 numbers. AHA~, If you tell me the S, I can guess the N~!” It’s easy? Okay, you can challenge this. Input Each line will contain an integer S. Process to end of file. Output For each case, output all possible N, print the number N with 5 digits, including the leading zeros, one line per case. I promise every case have one solution at least. If have many N, please output them from small to large in one line, separate them with a blank space. Sample Input 2933266 6392217 4245386 Sample Output 00038 07719 21238
如何在golang中实现“ tar cvfz xxx.tar.gz”？_course
20140513<div class="posttext" itemprop="text"> <p>I would like to decompress targz file by golang.</p> <blockquote> <p>err := DecompressTarGz('xxx.tar.gz', '/Users/foobarbuzz/')</p> </blockquote> </div>
Bit Magic _course
20170418Yesterday, my teacher taught me about bit operators: and (&), or (), xor (^). I generated a number table a[N], and wrote a program to calculate the matrix table b[N][N] using three kinds of bit operator. I thought my achievement would get teacher's attention. The key function is the code showed below. void calculate(int a[N], int b[N][N]) { for (int i = 0; i < N; ++i) { for (int j = 0; j < N; ++j) { if (i == j) b[i][j] = 0; else if (i % 2 == 1 && j % 2 == 1) b[i][j] = a[i]  a[j]; else if (i % 2 == 0 && j % 2 == 0) b[i][j] = a[i] & a[j]; else b[i][j] = a[i] ^ a[j]; } } } There is no doubt that my teacher raised lots of interests in my work and was surprised to my talented programming skills. After deeply thinking, he came up with another problem: if we have the matrix table b[N][N] at first, can you check whether corresponding number table a[N] exists? Input There are multiple test cases. For each test case, the first line contains an integer N, indicating the size of the matrix. (1 ≤ N ≤ 500). The next N lines, each line contains N integers, the jth integer in ith line indicating the element b[i][j] of matrix. (0 ≤ b[i][j] ≤ 2 ^ 31  1) Output For each test case, output "YES" if corresponding number table a[N] exists; otherwise output "NO". Sample Input 2 0 4 4 0 3 0 1 24 1 0 86 24 86 0 Sample Output YES NO
Magic Board _course
20170930Problem Description Sths is a happy boy~ Sths has got a Magic Board for his birthday gift! A Magic board is an N*M sized grids which were painted by black and white. According to the parity of N and M, the Magic Board would be a little different, but it can be guaranteed that any two adjacent grids are painted by different colors, and the amount of black grid is not less than white ones. The Magic Board is called MAGIC because it is formed by a Chain. A Chain is a set of grids in which a grid has at most 2 adjacent grids and there are only 2 grids which has only 1 adjacent grid. Those two special grids are called the endpoints of the Chain. The joint between two adjacent grids is very flexible. It can be rotate by any angle. Here’s an example of transforming a Chain into a Magic Board The Chain was connected by a Magic String. The existence of Magic String relies on the power of fengshui. But recently the fengshui in Beijing was ruined because there is a university installing airconditionor (which also cause the HUGE RAIN in Beijing).So the Magic String disappears, and the Magic Board is totally fell apart. Sths feels upset, because he really likes the Magic Board (since it can form a lot of things). So he is thinking about how to reconstruct it. The only thing Sths has got now is the separated grids. But surprisingly, Sths finds out that there are differences between these grids. 1. There are black grids and white grids. 2. There are three different grids in the same color because the Magic String goes through it in 3 different ways shown below: So there are 6 different kinds of grids. Now Sths has counted the amount of each kind of grids, he wants to know: by using the grids in his hand, how many kinds of legal Chains (which can form an N*M sized Magic Board) can be constructed. We shall say two Chains is the same if and only if the standard expression of these two Chains is the same. The standard expression is a set of numbers which decided by following method: 1. Starting from one of a Chain's endpoint. 2. Write down the color of the grids (1 for black and 0 for white) before direction changing. 3. Write down 2 then change direction and repeat Step 2 until reaching another endpoint of the Chain. 4. Choose the expression which lexicographical lower between the two expressions just generated since there are two endpoints. For example, the standard expression of the example of “N=M=3” is “10120120120212” (another expression is “10212012012012”, which is lexicographical greater than standard expression). And the standard expression of the example of “N=M=4” is “01012010210120120120212”. Input There are Multiple Test Cases For each case, there will be six integer numbers in one line, N, M, BO, BA, WO, and WA, indicating the number of rows and columns, the amount of “Black and Opposite” grids, the amount of “Black and Adjacent” grids, the amount of “White and Opposite” grids, the amount of “White and Adjacent” grids. 2<=N*M<=30 The input end with a line of 0 0 0 0 0 0. Output For each case, output the kinds of legal Chains that can be constructed by given grids. Sample Input 3 3 1 2 2 2 3 3 0 3 3 1 5 5 5 6 5 7 0 0 0 0 0 0 Sample Output 1 1 4
Magic Sticks _course
20171210Problem Description Magic was accepted by all ancient peoples as a technique to compel the help of divine powers. In a wellknown story, one group of sorcerers threw their walking sticks on the floor where they magically appeared to turn into live serpents. In opposition, another person threw his stick on the floor, where it turned into a serpent which then consumed the sorcerers’ serpents! The only magic required for this problem is its solution. You are given a magic stick that has several straight segments, with joints between the segments that allow the stick to be folded. Depending on the segment lengths and how they are folded, the segments of the stick can be arranged to produce a number of polygons. You are to determine the maximum area that could be enclosed by the polygons formed by folding the stick, using each segment in at most one polygon. Segments can touch only at their endpoints. For example, the stick shown below on the left has five segments and four joints. It can be folded to produce a polygon as shown on the right. ![](http://acm.hdu.edu.cn/data/images/38431.jpg) Input The input contains several test cases. Each test case describes a magic stick. The first line in each test case contains an integer n (1 <= n <= 500) which indicates the number of the segments in the magic stick. The next line contains n integers S1,S2, . . . , Sn (1 <= Si <= 1000) which indicate the lengths of the segments in the order they appear in the stick. The last test case is followed by a line containing a single zero. Output For each case, display its case number followed by the maximum total enclosed area that can be obtained by folding the magic stick at the given points. Answers within an absolute or relative error of 10^4 will be accepted. Follow the format of the sample output. Sample Input 4 1 2 3 4 8 3 4 5 33 3 4 3 5 0 Sample Output Case 1: 4.898979 Case 2: 19.311
 11.19MB
Android支付宝支付demo
20181016快捷简洁的集成了安卓端支付宝支付demo只需三个参数，
21天通关Python（仅视频课）
20190521本页面购买不发书！！！仅为视频课购买！！！ 请务必到https://edu.csdn.net/bundled/detail/49下单购买课+书。 本页面，仅为观看视频页面，如需一并购买图书，请务必到https://edu.csdn.net/bundled/detail/49下单购买课程+图书！！！ 疯狂Python精讲课程覆盖《疯狂Python讲义》全书的主体内容。 内容包括Python基本数据类型、Python列表、元组和字典、流程控制、函数式编程、面向对象编程、文件读写、异常控制、数据库编程、并发编程与网络编程、数据可视化分析、Python爬虫等。 全套课程从Python基础开始介绍，逐步步入当前就业热点。将会带着大家从Python基础语法开始学习，为每个知识点都提供对应的代码实操、代码练习，逐步过渡到文件IO、数据库编程、并发编程、网络编程、数据分 析和网络爬虫等内容，本课程会从小案例起，至爬虫、数据分析案例终、以Python知识体系作为内在逻辑，以Python案例作为学习方式，最终达到“知行合一”。
 373KB
2020数学建模A题
202009112020数学建模国赛A题及其数据 2020数学建模国赛A题及其数据2020数学建模国赛A题及其数据 2020数学建模国赛A题及其数据 2020数学建模国赛A题及其数据 2020数学建模国赛A题及其数据
Java学习指南（Java入门与进阶）
20170809这是Java学习指南系列课程的第1篇，介绍Java语言的入门语法，引领希望学习Java语言编程的初学者进入Java大门。 本课程不需要其他语言作为基础，可以直接学习。 课程从Java开发平台的下载和安装开始，从浅到深、从易到难，循序渐进地进行语法讲解。 为了让学员更好的掌握Java语言，本课程配套在线的Java题库及答案解析。 相比于其他语言，Java语言更科学、更容易掌握，快来和大家一起学习Java吧。
 42KB
81个Python爬虫源代码
2018121381个Python爬虫源代码，内容包含新闻、视频、中介、招聘、图片资源等网站的爬虫资源
Google开发专家带你入门神经网络
20200609限时秒杀，先到先得！！！ 购买后添加“助手小姐姐”微信，还可有不定期直播活动和老师答疑哦！ 【授课老师】 彭靖田 Google Developer Experts。 曾为 TensorFlow Top 40 的贡献者，著书《深入理解TensorFlow》，是国内第一本深度剖析 Google AI 框架的畅销书。 曾从0到1深入参与了华为 2012 实验室深度学习平台和华为深度学习云服务的设计与研发工作。 【课程会讲哪些知识？】 整个课程以理论加实战为核心，通过从基础原理、代码案例带你手把手入门神经网络。下面是课程的知识概览思维导图。
 71.49MB
jboss4.2.3.GA
20190327此版本在其余地方已经下载不到了哟，但是这个版本还是很好的，jboss4.2.3.GA,此jboss支持jboss 4.x 开发，jdk需满足5.0及以上的版本，不然会运行出错的，需要的自行下载哈，无
 博客 FCM服务器 adminSDK接入指引
 学院 GIT版本管理精讲
 博客 SWIFT与新加坡资讯通信媒体发展局携手推动全球贸易数字化
 博客 html href与src的区别
 博客 过拟合与欠拟合
 博客 锐浪报表(实现打印机打印配货单功能)
 博客 1.17循环细节
 学院 企业级SpringBoot使用
 博客 viewui
 学院 Java+FFmpeg开发音视频流媒体直播等
 学院 朱老师鸿蒙系列课程第1期1.鸿蒙系统HarmonyOS打通开发实践
 下载 手把手教你成为架构师课程资料包大放送
 下载 正则测试小工具，可以保存一些常用正则（附源码）
 下载 ITK4.13.2源码压缩包
 学院 Spring从入门到精通
 下载 板材开料软件 荣朗木工优化开料软件 v2.0
 学院 Python基础特训课
 下载 javascript仿XP关机效果的弹出窗口功能
 博客 飞凌FET3399C核心板揭开“刷脸时代”人脸识别背后的奥秘
 下载 axure rp 8.0视频从入门到精通
 学院 uniapp实战在线教育项目/小程序/H5/APP(第二季)
 博客 linux基本操作
 学院 AI工程师深度学习与计算机视觉实战
 学院 Cocos Creator 微信小游戏实战开发
 下载 在ASP.Net中实现flv视频转换的代码
 博客 python005集合
 学院 SpringBoot2实战教程
 博客 六种查看卫星历史影像数据方法，别怪我没跟你说
 下载 Newtonsoft.Json .net framework4.0版
 学院 极简数据分析入门