 Same or not

Problem Description
Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation.
Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times.
Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie.Input
On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree.Output
For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth.Sample Input
2
010011
001101
010011
01010011Sample Output
True
False
Same or not _course
20171125Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. ![](http://acm.hdu.edu.cn/data/images/C13810051.jpg) Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
Same or not 代码实现的技术_course
20190921Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
Same or not 怎么做的呢_course
20190901Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
C语言实现 Same or not_course
20190812Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
判断一个tree是否存在相似性的问题，怎么用C语言的程序的编写来实现计算的_course
20190630Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
难易程度的判断、打分的问题，要求使用C语言的程序的设计的思想原理来解决，怎么去实现的_course
20190613Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
判断二叉数的相似的遍历的问题，要求使用C语言的程序的设计的办法的解决的过程怎么做_course
20190613Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
相似性的判断的算法，怎么判断图的节点上的相似性用C语言的程序的代码设计的程序怎么实现_course
20190607Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
相等性和相似性的判断的算法，如何才能利用C程序的编写的技术来实现的呢_course
20190405Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
二叉树的相等性的判断算法问题，运用C语言具体怎么实现的一个思路？_course
20190316Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
分型二叉树的相似性的判断算法，如何利用C语言编程技术的实现的_course
20190323Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
关于二叉树的路径的相似行的判断的问题，如何使用C语言解决啊_course
20190207Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
C语言如何输入两个二叉树的二进制判断二叉树是否是相等的_course
20190105Problem Description Everyone konws how to traveling a tree using the DFS strategy. But we also know that there are many ways to do so. For example, giving a tree as the following picture, we may get three ways: 010011, 001101, 01010011. 0 stands for the down operation while 1 means the up operation. Now we make a constraint: if one node has k direct childs, you can visit a node at most 2*k times, if k == 0, you can visit it only once, in the example, the root has two direct child. Like the example, you can only get two ways: 010011, 001101. Because the way 01010011 will visit the node in yellow four times. Here is the problem: ACboy drawed a tree, but is not very nice, so he won't show you the picture. Instread he will give you two strings indicating that the ways to travel the tree. Of cource, the strings will only contain 0 and 1. And your mission is to tell whether ACboy is telling the truth. For example, he drawed a picture as the following, if he give you 010011 and 001101, then he is telling the truth, but if he give you 010011 and 01010011, you konw that he is telling a lie. Input On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test case consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a way to travel the tree. Output For each test case output a line containing the word "True" or the word "False", depending on whether ACboy is telling the truth. Sample Input 2 010011 001101 010011 01010011 Sample Output True False
算法的，自动的Ajax_course
20110801<div class="posttext" itemprop="text"> <p>I have made a framework that generates a HTML "DOM" tree on the server, as a tree of python objects, and then spits it out as a string to be sent to the client. The way it does this is via a recursive depthfirst traversal of the tree: for example a div would spit out the opening "div", spit out all it's children's html and then spit out the closing "/div".</p> <p>This tree is broken down into conceptual components, as shown below:</p> <p><a href="http://lhy.mit.edu/media/Flow_Chart.png" rel="nofollow noreferrer">graph http://lhy.mit.edu/media/Flow_Chart.png</a></p> <p>This only shows the first two levels of hierarchy; the actual site has many more: for example each comment in the comment bar is a self contained component, each button on the menu bar is a self contained component. As you can see, the various components do not need to be on the same depth in the tree. What constitutes a "component" is decided by me.</p> <p>What I want is the <em>complete</em> html string for each component (everything from the root node of that component downwards), as well as the <em>partial</em> HTML string for every component (The HTML of that component, minus the HTML of its children). The <em>partial</em> HTML of <strong>main section</strong>, for example, would be the html, head and two div tags <em>only</em>. The <em>complete</em> html of <strong>main section</strong>, on the other hand, would be every node on the page. </p> <p>How would i do this? I could just find the <em>complete</em> HTML string of every component and subcomponent, mark the boundaries of each subcomponent with some string and do RegexRemovals in order to find the <em>partial</em> HTML string for every component, but that feels clunky and inefficient. </p> <p>I could do an iterativedeepening DFS, halting at the boundary between a component and its subcomponents until every node in that component has been explored. I would then have the <em>partial</em> HTML for every component but then i would need to do a similarly hacky RegexInserts to later build up the <em>complete</em> HTML for every component.</p> <p>I could do both, but that would take two passes and would be expensive, though maybe not as expensive as the above Regex gymnastics.</p> <p>I could do a priorityqueue Dijkstra's, having each component be strictly higher priority than its children. It would traverse the tree in the correct order, finishing each component before moving on to its children, but i have no idea how i would get the final wellformed HTML string out of it.</p> <p>The purpose of all this is so the server can intelligently and completely autonomously determine the minimal set of components on the client's page that need to change on a pagetransition between two arbitrary pages.</p> <p>If i create a new page on my site, I should need no more than <strong>Zero</strong> extra lines of code to have it ajax smoothly with any existing page.</p> <p>But first i need to get my graphtraversing htmlspewing algorithms in order. Any ideas?</p> </div>
玩转Linux：常用命令实例指南
20190928人工智能、物联网、大数据时代，Linux正有着一统天下的趋势，几乎每个程序员岗位，都要求掌握Linux。本课程零基础也能轻松入门。 本课程以简洁易懂的语言手把手教你系统掌握日常所需的Linux知识，每个知识点都会配合案例实战让你融汇贯通。课程通俗易懂，简洁流畅，适合0基础以及对Linux掌握不熟练的人学习； 【限时福利】 1）购课后按提示添加小助手，进答疑群，还可获得价值300元的编程大礼包！ 2）本月购买此套餐加入老师答疑交流群，可参加老师的免费分享活动，学习最新技术项目经验。  29元=掌握Linux必修知识+社群答疑+讲师社群分享会+700元编程礼包。
专为程序员设计的数学课_course
20191111<p> 限时福利限时福利，<span>15000+程序员的选择！</span> </p> <p> 购课后添加学习助手（微信号：csdn590），按提示消息领取编程大礼包！并获取讲师答疑服务！ </p> <p> <br> </p> <p> 套餐中一共包含5门程序员必学的数学课程（共47讲） </p> <p> 课程1：《零基础入门微积分》 </p> <p> 课程2：《数理统计与概率论》 </p> <p> 课程3：《代码学习线性代数》 </p> <p> 课程4：《数据处理的最优化》 </p> <p> 课程5：《马尔可夫随机过程》 </p> <p> <br> </p> <p> 哪些人适合学习这门课程？ </p> <p> 1）大学生，平时只学习了数学理论，并未接触如何应用数学解决编程问题； </p> <p> 2）对算法、数据结构掌握程度薄弱的人，数学可以让你更好的理解算法、数据结构原理及应用； </p> <p> 3）看不懂大牛代码设计思想的人，因为所有的程序设计底层逻辑都是数学； </p> <p> 4）想学习新技术，如：人工智能、机器学习、深度学习等，这门课程是你的必修课程； </p> <p> 5）想修炼更好的编程内功，在遇到问题时可以灵活的应用数学思维解决问题。 </p> <p> <br> </p> <p> 在这门「专为程序员设计的数学课」系列课中，我们保证你能收获到这些:<br> <br> <span> </span> </p> <p class="qllong24357476"> <span class="qlauthor24357476">①价值300元编程课程大礼包</span> </p> <p class="qllong24357476"> <span class="qlauthor24357476">②应用数学优化代码的实操方法</span> </p> <p class="qllong24357476"> <span class="qlauthor24357476">③数学理论在编程实战中的应用</span> </p> <p class="qllong24357476"> <span class="qlauthor24357476">④程序员必学的5大数学知识</span> </p> <p class="qllong24357476"> <span class="qlauthor24357476">⑤人工智能领域必修数学课</span> </p> <p> <br> 备注：此课程只讲程序员所需要的数学，即使你数学基础薄弱，也能听懂，只需要初中的数学知识就足矣。<br> <br> 如何听课？ </p> <p> 1、登录CSDN学院 APP 在我的课程中进行学习； </p> <p> 2、登录CSDN学院官网。 </p> <p> <br> </p> <p> 购课后如何领取免费赠送的编程大礼包和加入答疑群？ </p> <p> 购课后，添加助教微信：<span> csdn590</span>，按提示领取编程大礼包，或观看付费视频的第一节内容扫码进群答疑交流！ </p> <p> <img src="https://imgbss.csdn.net/201912251155398753.jpg" alt=""> </p>
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20190516【为什么学爬虫？】 1、爬虫入手容易，但是深入较难，如何写出高效率的爬虫，如何写出灵活性高可扩展的爬虫都是一项技术活。另外在爬虫过程中，经常容易遇到被反爬虫，比如字体反爬、IP识别、验证码等，如何层层攻克难点拿到想要的数据，这门课程，你都能学到！ 2、如果是作为一个其他行业的开发者，比如app开发，web开发，学习爬虫能让你加强对技术的认知，能够开发出更加安全的软件和网站 【课程设计】 一个完整的爬虫程序，无论大小，总体来说可以分成三个步骤，分别是： 网络请求：模拟浏览器的行为从网上抓取数据。 数据解析：将请求下来的数据进行过滤，提取我们想要的数据。 数据存储：将提取到的数据存储到硬盘或者内存中。比如用mysql数据库或者redis等。 那么本课程也是按照这几个步骤循序渐进的进行讲解，带领学生完整的掌握每个步骤的技术。另外，因为爬虫的多样性，在爬取的过程中可能会发生被反爬、效率低下等。因此我们又增加了两个章节用来提高爬虫程序的灵活性，分别是： 爬虫进阶：包括IP代理，多线程爬虫，图形验证码识别、JS加密解密、动态网页爬虫、字体反爬识别等。 Scrapy和分布式爬虫：Scrapy框架、Scrapyredis组件、分布式爬虫等。 通过爬虫进阶的知识点我们能应付大量的反爬网站，而Scrapy框架作为一个专业的爬虫框架，使用他可以快速提高我们编写爬虫程序的效率和速度。另外如果一台机器不能满足你的需求，我们可以用分布式爬虫让多台机器帮助你快速爬取数据。 从基础爬虫到商业化应用爬虫，本套课程满足您的所有需求！ 【课程服务】 专属付费社群+每周三讨论会+1v1答疑
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20181010本课程内容系统、全面、简洁、通俗易懂，通过2个多小时的介绍，让大家对软件测试有个系统的理解和认识，具备基本的软件测试理论基础。 主要内容分为5个部分： 1 软件测试概述，了解测试是什么、测试的对象、原则、流程、方法、模型； 2.常用的黑盒测试用例设计方法及示例演示； 3 常用白盒测试用例设计方法及示例演示； 4.自动化测试优缺点、使用范围及示例‘； 5.测试经验谈。
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