Your top-level group Ali-partners is over the 5 user limit and has been placed in a read-only state. To remove the read-only state and regain write access, ask your top-level group owner(s) to reduce the number of users in your top-level group to 5 users or less, or to upgrade to a paid tier which do not have user limits.

时间: 2024-03-19 13:40:13 浏览: 127
RAR

CRC循环校验的具体算法.rar_CRC 校验_To Be or Not To Be_bch_crc-8_filename—ma

这是GitLab的限制规则之一，即免费版的顶级组织（top-level group）最多只能添加5个用户，超过限制则会将该组织设置为只读状态。要解除只读状态并恢复写访问权限，需请组织拥有者减少组织中用户的数量至5个或以下，或者升级到不受用户限制的付费版。付费版提供了更多功能和更高级别的支持，可以根据具体需求选择适合的版本。
阅读全文

相关推荐

pdf

Test and analysis of the halo in low-light-level image intensifiers

A halo tester is designed to analyze the halo formation in low-light-level (LLL) image intensifiers and the influencing factors on halo size. The tester is used to collect a 0.1922-mm hole image ...

Please write a simply code in python, the requirement is as followed: 1. use the list type describe rod A, B and C 2. to The Tower of Hanoi is a mathematical game or puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape. The following figure shows the initial state of the Tower of Hanoi with 5 disks.The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules: 1. Only one disk can be moved at a time. 2. Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack i.e. a disk can only be moved if it is the uppermost disk on a stack. 3. No disk may be placed on top of a smaller disk. Assume that initially all the disks are placed on rod A. Write a non-recursive Python function to print out the steps to move all the disks from rod A to rod C via rod B (Hint: a recursive algorithm can be converted into a non-recursive algorithm using stack). The header of the function is: def HanoiTower(n) Here n represents the number of disks.

首先，我们可以使用 Python 的列表类型来描述棒 A，B 和 C： rod_A = [] rod_B = [] rod_C = [] 然后，我们可以定义一个非递归的 Python 函数来打印汉诺塔的移动步骤： def HanoiTower(n): stack = ...

Description Consider the following 5 picture frames placed on an 9 x 8 array. ........ ........ ........ ........ .CCC.... EEEEEE.. ........ ........ ..BBBB.. .C.C.... E....E.. DDDDDD.. ........ ..B..B.. .C.C.... E....E.. D....D.. ........ ..B..B.. .CCC.... E....E.. D....D.. ....AAAA ..B..B.. ........ E....E.. D....D.. ....A..A ..BBBB.. ........ E....E.. DDDDDD.. ....A..A ........ ........ E....E.. ........ ....AAAA ........ ........ EEEEEE.. ........ ........ ........ ........ 1 2 3 4 5 Now place them on top of one another starting with 1 at the bottom and ending up with 5 on top. If any part of a frame covers another it hides that part of the frame below. Viewing the stack of 5 frames we see the following. .CCC.... ECBCBB.. DCBCDB.. DCCC.B.. D.B.ABAA D.BBBB.A DDDDAD.A E...AAAA EEEEEE.. In what order are the frames stacked from bottom to top? The answer is EDABC. Your problem is to determine the order in which the frames are stacked from bottom to top given a picture of the stacked frames. Here are the rules: 1. The width of the frame is always exactly 1 character and the sides are never shorter than 3 characters. 2. It is possible to see at least one part of each of the four sides of a frame. A corner shows two sides. 3. The frames will be lettered with capital letters, and no two frames will be assigned the same letter. Input Each input block contains the height, h (h<=30) on the first line and the width w (w<=30) on the second. A picture of the stacked frames is then given as h strings with w characters each. Your input may contain multiple blocks of the format described above, without any blank lines in between. All blocks in the input must be processed sequentially. Output Write the solution to the standard output. Give the letters of the frames in the order they were stacked from bottom to top. If there are multiple possibilities for an ordering, list all such possibilities in alphabetical order, each one on a separate line. There will always be at least one legal ordering for each input block. List the output for all blocks in the input sequentially, without any blank lines (not even between blocks). Sample Input 9 8 .CCC.... ECBCBB.. DCBCDB.. DCCC.B.. D.B.ABAA D.BBBB.A DDDDAD.A E...AAAA EEEEEE.. Sample Output EDABC

这是一道经典的拓扑排序问题。我们可以将每个图形看作一个节点，如果一个图形可以覆盖另一个图形，那就从覆盖的图形指向被覆盖的图形。最后得到的图是一个 DAG，我们可以对其进行拓扑排序。拓扑排序的结果就是图形的...

翻译Device configuration register The device has various configuration settings that are global in nature. The configuration settings are as follows: • When the 33978 is in the overvoltage region, a Logic [0] on the VBATP OV bit limits the wetting current on all input channels to 2 mA and the 33978 will not be able to enter into the Low-power mode. A Logic [1] allows the device to operate normally even in the overvoltage region. The OV flag will be set when the device enters in the OV region, regardless the value of the VBATP OV bit. • WAKE_B can be used to enable an external power supply regulator to supply the VDDQ voltage rail. When the WAKE_B VDDQ check bit is a Logic [0], the WAKE_B pin is expected to be pulled-up internally or externally to VDDQ and VDDQ is expected to go low, therefore the 33978 does not wake-up on the falling edge of WAKE_B. A Logic [1], assumes the user is using an external pull-up to VBATP or VDDQ (when VDDQ is not expected to be off) and the IC wakes up on a falling edge of WAKE_B. • INT_B out is used to select how the INT_B pin operates when an interrupt occurs. The IC is able to pulse low [1] or latch low [0]. • Aconfig[1-0] is used to determine the method of selecting the AMUX output, either a SPI command or using a hardwired setup using SG[3-1]. • Inputs SP0-7 may be programmable for switch-to-battery or switch-to-ground. These inputs types are defined using the settings command. To set a SPn input for switch-to-battery, a logic [1] for the appropriate bit must be set. To set a SPn input for switch-toground, a logic [0] for the appropriate bit must be set. The MCU may change or update the programmable switch register via software at any time in Normal mode. Regardless of the setting, when the SPn input switch is closed a logic [1] is placed in the serial output response register.

• Aconfig [1-0]用于确定选择AMUX输出的方法，可以是SPI命令或使用硬连线设置使用SG [3-1]。 • 输入SP0-7可以编程为切换至电池或切换至地。这些输入类型使用设置命令进行定义。要将SPn输入设置为切换至电池，必须...

It is said that in ancient Indian holy temples, there was a game called Hanoi. The game is on a copper plate device with three rods (numbered a, B and C). there are many plates are placed in order from bottom to top and from large to small in rod A. The goal of the game: move all the gold plates on pole A to pole C, and keep the original order. Operating rules: only one plate can be moved at a time, and the large plate is always kept under the three rods and the small plate is on the top. During the operation, the plate can be placed on any rod A, B and C. Solve the above problems with programs, try to propose an optimized algorithm.

The Hanoi game is a classic problem in computer science and can be solved using recursion. Here is a Python implementation of the Hanoi game: python def move(n, source, target, auxiliary): if n >...

a vagrantfile has been placed in this directory. you are now ready to vagrant up your first virtual environment! please read the comments in the vagrantfile as well as documentation on vagrantup.com for more information on using vagrant.

在这个目录下已经放置了一个叫做 vagrantfile 的文件。你现在可以准备启动你的第一个虚拟环境了！请阅读 vagrantfile 中的注释，以及在 vagrantup.com 上的文档，获取更多有关使用 vagrant 的信息。

When all of a vertex’s input channels become ready a new execution record is created for the vertex and placed in a scheduling queue. A disk-based channel is considered to be ready when the entire file is present. A channel that is a TCP pipe or shared-memory FIFO is ready when the predecessor vertex has at least one running execution record. A vertex and any of its channels may each specify a “hardconstraint” or a “preference” listing the set of computers on which it would like to run. The constraints are combined and attached to the execution record when it is added to the scheduling queue and they allow the application writer to require that a vertex be co-located with a large input file, and in general let the scheduler preferentially run computations close to their data.数据库中这些具体能做什么

这段文本描述了一个分布式计算系统中的调度和执行过程，其中包括了对输入通道的就绪状态判断、执行记录的创建和调度队列的管理，以及对计算节点的硬性约束和偏好设置等。如果将这个系统应用到数据库中，可以实现...

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L). To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river. Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N). FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks. Input Line 1: Three space-separated integers: L, N, and M Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position. Output Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks Sample Inputcopy Outputcopy 25 5 2 2 14 11 21 17 4 Hint Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).

To determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks, we can follow these steps: 1. Read the input values for L, N, and M. 2. Read the ...

Robert is a famous engineer. One day he was given a task by his boss. The background of the task was the following: Given a map consisting of square blocks. There were three kinds of blocks: Wall, Grass, and Empty. His boss wanted to place as many robots as possible in the map. Each robot held a laser weapon which could shoot to four directions (north, east, south, west) simultaneously. A robot had to stay at the block where it was initially placed all the time and to keep firing all the time. The laser beams certainly could pass the grid of Grass, but could not pass the grid of Wall. A robot could only be placed in an Empty block. Surely the boss would not want to see one robot hurting another. In other words, two robots must not be placed in one line (horizontally or vertically) unless there is a Wall between them. Now that you are such a smart programmer and one of Robert's best friends, He is asking you to help him solving this problem. That is, given the description of a map, compute the maximum number of robots that can be placed in the map. Input The first line contains an integer T (<= 11) which is the number of test cases. For each test case, the first line contains two integers m and n (1<= m, n <=50) which are the row and column sizes of the map. Then m lines follow, each contains n characters of '#', '', or 'o' which represent Wall, Grass, and Empty, respectively. Output For each test case, first output the case number in one line, in the format: "Case :id" where id is the test case number, counting from 1. In the second line just output the maximum number of robots that can be placed in that map.

To solve it, we can use a simple algorithm that iterates over all empty cells in the map and checks whether a robot can be placed there. For each empty cell, we can simulate a robot firing its laser ...

For the following C statement,write the corresponding RISC-V assembly code.Assume that the C variables f，g，and h,have already been placed in registers x5，x6，and x7 respectively.Use a minimal number of RISC-V assemblyinstructions.

C statement: c f = (g + h) ; RISC-V assembly code: assembly addw x5, x6, x7 # 将 g 和 h 相加并将结果放入 x5 寄存器 slli x5, x5, 2 # 将 x5 寄存器的值左移 2 位，相当于乘以 4 ...

Write a Java program using JavaFX to create a graphical user interface for a microwave similar to the one presented below. The numbers are placed on buttons. The node displaying the text “Time to be displayed here” is a TextField and the node with the text “Place Food Here” is a Label.

我可以帮你编写一个使用JavaFX创建一个图形用户界面的Java程序，用于创建下面...按钮上放置的是数字，显示文本“Time to be displayed here”的节点是一个TextField，显示文本“Place Food Here”的节点是一个Label。

[Place 30-575] Sub-optimal placement for a clock-capable IO pin and MMCM pair. If this sub optimal condition is acceptable for this design, you may use the CLOCK_DEDICATED_ROUTE constraint in the .xdc file to demote this message to a WARNING. However, the use of this override is highly discouraged. These examples can be used directly in the .xdc file to override this clock rule. < set_property CLOCK_DEDICATED_ROUTE BACKBONE [get_nets sys_clk_IBUF] > sys_clk_IBUF_inst (IBUF.O) is locked to IOB_X1Y74 instance_name/inst/mmcm_adv_inst (MMCME2_ADV.CLKIN1) is provisionally placed by clockplacer on MMCME2_ADV_X1Y2 The above error could possibly be related to other connected instances. Following is a list of all the related clock rules and their respective instances. Clock Rule: rule_mmcm_bufg Status: PASS Rule Description: An MMCM driving a BUFG must be placed on the same half side (top/bottom) of the device instance_name/inst/mmcm_adv_inst (MMCME2_ADV.CLKFBOUT) is provisionally placed by clockplacer on MMCME2_ADV_X1Y2 and instance_name/inst/clkf_buf (BUFG.I) is provisionally placed by clockplacer on BUFGCTRL_X0Y31

Based on the provided information, it seems that there is a sub-optimal placement for a clock-capable IO pin and MMCM pair in your design. If this condition is acceptable, you can use the CLOCK_...

2. (2 points) Please find the dates without any placed order between the time range. (Hint: write a recursive query to create date sequence and use the select statement in Q1 to assign user-defined variables so that you can have a neat recursive query.)

id, order_date, and customer_id, we can use the following recursive query to generate a sequence of dates between a specified time range and then find the dates without any placed order: ...

Please rewrite the following paragraphs to make them more logical and highlight the keyword: Photocatalytic degradation tests were performed in a quartz reactor. Ten mg of PCN-134/PCN-134 was weighed and dispersed in 100 mL of 40 mg/L NZT solution, stirred at 400 rpm on a magnetic stirrer and placed under a 500 W xenon lamp (Beijing Pulinsaisi, China) before the reaction started. At regular intervals, 3 mL of the mixed solution was removed and filtered through a 0.22μm nylon filter tip and subsequently measured by UV-Vis spectrophotometer at 314 nm. The photocatalytic performance of PCN-134/PCN-134 for NZT in the presence of different pH, anions and dissolved organic matter (DOM) was then investigated. The photocatalytic performance of PCN-134/PCN-134 was investigated after 5 times of repeated use.

which was then filtered through a 0.22μm nylon filter tip and analyzed using a UV-Vis spectrophotometer at 314 nm to evaluate the photocatalytic performance of PCN-134/PCN-134 for NZT in various ...
zip

office2john的源文件

office2john的源文件
rp

青海省各市、县区及街镇SVG图

青海省各市、县区及街镇SVG图
zip

(源码)基于JavaNIO框架的网络IO处理系统.zip

# 基于Java NIO框架的网络IO处理系统 ## 项目简介 本项目是一个基于Java NIO（非阻塞IO）框架的网络IO处理系统。通过使用Java NIO的多路复用技术，系统能够高效地处理多个客户端的连接和IO操作。项目涵盖了从基本的Socket编程到复杂的NIO多路复用实现，适合学习和理解Java网络编程的高级特性。 ## 项目的主要特性和功能 1. 多路复用技术使用Java NIO的Selector机制，实现单线程处理多个客户端的连接和IO操作。 2. 多线程处理通过多线程技术，系统能够并发处理多个客户端的请求，提高服务器的处理能力。 3. 文件IO操作包括基本的文件写入、缓冲文件IO、随机访问文件写入等操作，展示了Java文件IO的多种实现方式。 4. 网络IO操作包括Socket客户端和服务器的实现，展示了如何使用Java进行网络通信。
zip

数据结构- C语言版-树的遍历代码

二叉树先、中、后序遍历及层次遍历代码 C90版

最新推荐

recommend-type

office2john的源文件

office2john的源文件
recommend-type

青海省各市、县区及街镇SVG图

青海省各市、县区及街镇SVG图
recommend-type

(源码)基于JavaNIO框架的网络IO处理系统.zip

# 基于Java NIO框架的网络IO处理系统 ## 项目简介 本项目是一个基于Java NIO（非阻塞IO）框架的网络IO处理系统。通过使用Java NIO的多路复用技术，系统能够高效地处理多个客户端的连接和IO操作。项目涵盖了从基本的Socket编程到复杂的NIO多路复用实现，适合学习和理解Java网络编程的高级特性。 ## 项目的主要特性和功能 1. 多路复用技术使用Java NIO的Selector机制，实现单线程处理多个客户端的连接和IO操作。 2. 多线程处理通过多线程技术，系统能够并发处理多个客户端的请求，提高服务器的处理能力。 3. 文件IO操作包括基本的文件写入、缓冲文件IO、随机访问文件写入等操作，展示了Java文件IO的多种实现方式。 4. 网络IO操作包括Socket客户端和服务器的实现，展示了如何使用Java进行网络通信。
recommend-type

Java集合ArrayList实现字符串管理及效果展示

资源摘要信息:"Java集合框架中的ArrayList是一个可以动态增长和减少的数组实现。它继承了AbstractList类，并且实现了List接口。ArrayList内部使用数组来存储添加到集合中的元素，且允许其中存储重复的元素，也可以包含null元素。由于ArrayList实现了List接口，它支持一系列的列表操作，包括添加、删除、获取和设置特定位置的元素，以及迭代器遍历等。 当使用ArrayList存储元素时，它的容量会自动增加以适应需要，因此无需在创建ArrayList实例时指定其大小。当ArrayList中的元素数量超过当前容量时，其内部数组会重新分配更大的空间以容纳更多的元素。这个过程是自动完成的，但它可能导致在列表变大时会有性能上的损失，因为需要创建一个新的更大的数组，并将所有旧元素复制到新数组中。 在Java代码中，使用ArrayList通常需要导入java.util.ArrayList包。例如： ```java import java.util.ArrayList; public class Main { public static void main(String[] args) { ArrayList<String> list = new ArrayList<String>(); list.add("Hello"); list.add("World"); // 运行效果图将显示包含"Hello"和"World"的列表 } } ``` 上述代码创建了一个名为list的ArrayList实例，并向其中添加了两个字符串元素。在运行效果图中，可以直观地看到这个列表的内容。ArrayList提供了多种方法来操作集合中的元素，比如get(int index)用于获取指定位置的元素，set(int index, E element)用于更新指定位置的元素，remove(int index)或remove(Object o)用于删除元素，size()用于获取集合中元素的个数等。 为了演示如何使用ArrayList进行字符串的存储和管理，以下是更加详细的代码示例，以及一个简单的运行效果图展示： ```java import java.util.ArrayList; import java.util.Iterator; public class Main { public static void main(String[] args) { // 创建一个存储字符串的ArrayList ArrayList<String> list = new ArrayList<String>(); // 向ArrayList中添加字符串元素 list.add("Apple"); list.add("Banana"); list.add("Cherry"); list.add("Date"); // 使用增强for循环遍历ArrayList System.out.println("遍历ArrayList:"); for (String fruit : list) { System.out.println(fruit); } // 使用迭代器进行遍历 System.out.println("使用迭代器遍历:"); Iterator<String> iterator = list.iterator(); while (iterator.hasNext()) { String fruit = iterator.next(); System.out.println(fruit); } // 更新***List中的元素 list.set(1, "Blueberry"); // 移除ArrayList中的元素 list.remove(2); // 再次遍历ArrayList以展示更改效果 System.out.println("修改后的ArrayList:"); for (String fruit : list) { System.out.println(fruit); } // 获取ArrayList的大小 System.out.println("ArrayList的大小为: " + list.size()); } } ``` 在运行上述代码后，控制台会输出以下效果图： ``` 遍历ArrayList: Apple Banana Cherry Date 使用迭代器遍历: Apple Banana Cherry Date 修改后的ArrayList: Apple Blueberry Date ArrayList的大小为: 3 ``` 此代码段首先创建并初始化了一个包含几个水果名称的ArrayList，然后展示了如何遍历这个列表，更新和移除元素，最终再次遍历列表以展示所做的更改，并输出列表的当前大小。在这个过程中，可以看到ArrayList是如何灵活地管理字符串集合的。 此外，ArrayList的实现是基于数组的，因此它允许快速的随机访问，但对元素的插入和删除操作通常需要移动后续元素以保持数组的连续性，所以这些操作的性能开销会相对较大。如果频繁进行插入或删除操作，可以考虑使用LinkedList，它基于链表实现，更适合于这类操作。 在开发中使用ArrayList时，应当注意避免过度使用，特别是当知道集合中的元素数量将非常大时，因为这样可能会导致较高的内存消耗。针对特定的业务场景，选择合适的集合类是非常重要的，以确保程序性能和资源的最优化利用。"
recommend-type

管理建模和仿真的文件

管理Boualem Benatallah引用此版本：布阿利姆·贝纳塔拉。管理建模和仿真。约瑟夫-傅立叶大学-格勒诺布尔第一大学，1996年。法语。NNT：电话：00345357HAL ID：电话：00345357https://theses.hal.science/tel-003453572008年12月9日提交HAL是一个多学科的开放存取档案馆，用于存放和传播科学研究论文，无论它们是否被公开。论文可以来自法国或国外的教学和研究机构，也可以来自公共或私人研究中心。L’archive ouverte pluridisciplinaire
recommend-type

【MATLAB信号处理优化】：算法实现与问题解决的实战指南

![【MATLAB信号处理优化】：算法实现与问题解决的实战指南](https://i0.hdslb.com/bfs/archive/e393ed87b10f9ae78435997437e40b0bf0326e7a.png@960w_540h_1c.webp) # 1. MATLAB信号处理基础 MATLAB，作为工程计算和算法开发中广泛使用的高级数学软件，为信号处理提供了强大的工具箱。本章将介绍MATLAB信号处理的基础知识，包括信号的类型、特性以及MATLAB处理信号的基本方法和步骤。 ## 1.1 信号的种类与特性 信号是信息的物理表示，可以是时间、空间或者其它形式的函数。信号可以被分
recommend-type

在西门子S120驱动系统中，更换SMI20编码器时应如何确保数据的正确备份和配置？

在西门子S120驱动系统中更换SMI20编码器是一个需要谨慎操作的过程，以确保数据的正确备份和配置。这里是一些详细步骤： 参考资源链接：[西门子Drive_CLIQ编码器SMI20数据在线读写步骤](https://wenku.csdn.net/doc/39x7cis876?spm=1055.2569.3001.10343) 1. 在进行任何操作之前，首先确保已经备份了当前工作的SMI20编码器的数据。这通常需要使用STARTER软件，并连接CU320控制器和电脑。 2. 从拓扑结构中移除旧编码器，下载当前拓扑结构，然后删除旧的SMI
recommend-type

实现2D3D相机拾取射线的关键技术

资源摘要信息: "camera-picking-ray:为2D/3D相机创建拾取射线" 本文介绍了一个名为"camera-picking-ray"的工具，该工具用于在2D和3D环境中，通过相机视角进行鼠标交互时创建拾取射线。拾取射线是指从相机（或视点）出发，通过鼠标点击位置指向场景中某一点的虚拟光线。这种技术广泛应用于游戏开发中，允许用户通过鼠标操作来选择、激活或互动场景中的对象。为了实现拾取射线，需要相机的投影矩阵（projection matrix）和视图矩阵（view matrix），这两个矩阵结合后可以逆变换得到拾取射线的起点和方向。 ### 知识点详解 1. **拾取射线（Picking Ray）**: - 拾取射线是3D图形学中的一个概念，它是从相机出发穿过视口（viewport）上某个特定点（通常是鼠标点击位置）的射线。 - 在游戏和虚拟现实应用中，拾取射线用于检测用户选择的对象、触发事件、进行命中测试（hit testing）等。 2. **投影矩阵（Projection Matrix）与视图矩阵（View Matrix）**: - 投影矩阵负责将3D场景中的点映射到2D视口上，通常包括透视投影（perspective projection）和平面投影（orthographic projection）。 - 视图矩阵定义了相机在场景中的位置和方向，它将物体从世界坐标系变换到相机坐标系。 - 将投影矩阵和视图矩阵结合起来得到的invProjView矩阵用于从视口坐标转换到相机空间坐标。 3. **实现拾取射线的过程**: - 首先需要计算相机的invProjView矩阵，这是投影矩阵和视图矩阵的逆矩阵。 - 使用鼠标点击位置的视口坐标作为输入，通过invProjView矩阵逆变换，计算出射线在世界坐标系中的起点（origin）和方向（direction）。 - 射线的起点一般为相机位置或相机前方某个位置，方向则是从相机位置指向鼠标点击位置的方向向量。 - 通过编程语言（如JavaScript）的矩阵库（例如gl-mat4）来执行这些矩阵运算。 4. **命中测试（Hit Testing）**: - 使用拾取射线进行命中测试是一种检测射线与场景中物体相交的技术。 - 在3D游戏开发中，通过计算射线与物体表面的交点来确定用户是否选中了一个物体。 - 此过程中可能需要考虑射线与不同物体类型的交互，例如球体、平面、多边形网格等。 5. **JavaScript与矩阵操作库**: - JavaScript是一种广泛用于网页开发的编程语言，在WebGL项目中用于处理图形渲染逻辑。 - gl-mat4是一个矩阵操作库，它提供了创建和操作4x4矩阵的函数，这些矩阵用于WebGL场景中的各种变换。 - 通过gl-mat4库，开发者可以更容易地执行矩阵运算，而无需手动编写复杂的数学公式。 6. **模块化编程**: - camera-picking-ray看起来是一个独立的模块或库，它封装了拾取射线生成的算法，让开发者能够通过简单的函数调用来实现复杂的3D拾取逻辑。 - 模块化编程允许开发者将拾取射线功能集成到更大的项目中，同时保持代码的清晰和可维护性。 7. **文件名称列表**: - 提供的文件名称列表是"camera-picking-ray-master"，表明这是一个包含多个文件和子目录的模块或项目，通常在GitHub等源代码托管平台上使用master分支来标识主分支。 - 开发者可以通过检查此项目源代码来更深入地理解拾取射线的实现细节，并根据需要进行修改或扩展功能。 ### 结论 "camera-picking-ray"作为一个技术工具，为开发者提供了一种高效生成和使用拾取射线的方法。它通过组合和逆变换相机矩阵，允许对3D场景中的物体进行精准选择和交互。此技术在游戏开发、虚拟现实、计算机辅助设计（CAD）等领域具有重要应用价值。通过了解和应用拾取射线，开发者可以显著提升用户的交互体验和操作精度。
recommend-type

"互动学习：行动中的多样性与论文攻读经历"

多样性她- 事实上SCI NCES你的时间表ECOLEDO C Tora SC和NCESPOUR l’Ingén学习互动，互动学习以行动为中心的强化学习学会互动，互动学习，以行动为中心的强化学习计算机科学博士论文于2021年9月28日在Villeneuve d'Asq公开支持马修·瑟林评审团主席法布里斯·勒菲弗尔阿维尼翁大学教授论文指导奥利维尔·皮耶昆谷歌研究教授：智囊团论文联合主任菲利普·普雷教授，大学。里尔/CRISTAL/因里亚报告员奥利维耶·西格德索邦大学报告员卢多维奇·德诺耶教授，Facebook /索邦大学审查员越南圣迈IMT Atlantic高级讲师邀请弗洛里安·斯特鲁布博士，Deepmind对于那些及时看到自己错误的人...3谢谢你首先，我要感谢我的两位博士生导师Olivier和Philippe。奥利维尔，"站在巨人的肩膀上"这句话对你来说完全有意义了。从科学上讲，你知道在这篇论文的（许多）错误中，你是我可以依
recommend-type

【MATLAB时间序列分析】：预测与识别的高效技巧

![MATLAB](https://img-blog.csdnimg.cn/direct/8652af2d537643edbb7c0dd964458672.png) # 1. 时间序列分析基础概念 在数据分析和预测领域，时间序列分析是一个关键的工具，尤其在经济学、金融学、信号处理、环境科学等多个领域都有广泛的应用。时间序列分析是指一系列按照时间顺序排列的数据点的统计分析方法，用于从过去的数据中发现潜在的趋势、季节性变化、周期性等信息，并用这些信息来预测未来的数据走向。 时间序列通常被分为四种主要的成分：趋势（长期方向）、季节性（周期性）、循环（非固定周期）、和不规则性（随机波动）。这些成分