下面句子的语法错误：Hu (2005) considered that it is much impossible for the author to make a wonderful classification of the English business correspondences in this paper because there are so many various categories in the communicative process. Therefore, sometimes it is of great difficulty to clearly tell one type of correspondences from another. More specifically, such business correspondences as those of establishing business relationship, orders and confirmation to orders, routine inquiries, affirmative response to the requests of claims and adjustment, affirmative replies to credit inquiries, etc. can be divided into good-news and neutral-news correspondences, that is, everyday correspondences; on the other hand, those correspondences, for example, declining invitations and requests for favors, cancelling the orders, replacements, non-confirming orders, negative answers to routine requests, and refusals of adjustment of claims and complaints etc. pertain to bad-news correspondences. The third type persuasive correspondences are referred to those whose aim is to convince the readers to conduct some performance that is not considered before or do something that might be inconvenient. This type includes sales correspondences, claims correspondences and correspondences of requesting special favor or information taking collection correspondences for example.

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英语语法焦点：过去分词构成与环境项目参与

本资源是一份关于英语语法教学的PPT模板，主要讲解了过去分词的构成，并通过对话形式展示了现在进行时和现在完成时的用法。 ### 过去分词的构成过去分词是动词的一种非限定形式，通常用于构成完成时态、被动语态...

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The veri¯cation of latency time for distributed hard real-time systems is considered to be the most important performance metrics. It is important for safety critical applications and for performance evaluation of the transmission systems. The integration of a middleware in such application, can induce some communication delays. Therefore, we focused our interest to the computation of end to end latency. Some existing works, adopt optimization or scheduling methods in network computation.35,36 In our work, we propose the WCRT to take into account all possible delays. Latency is considered as the period of time since a message is written by a DataWriter until it is received by a DataReader. It consists of delay caused by signal processing, propagation, queuing and transmission. We detail the computation of latency period of a packet on its entire network path, and we verify its validity with DDS QoS:

对于分布式硬实时系统的延迟时间的验证被认为是最重要的性能指标。对于安全关键应用程序和传输系统的性能评估而言，它非常重要。在这种应用程序中集成中间件可能会引入一些通信延迟。因此，我们将重点放在计算端到端...

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...

帮我写出以下java代码：Add a class Bubble that extends Shape. The Bubble class has an instance variable called radius of type double that represents the radius of the bubble. The constructor of the Bubble class takes an x and a y as arguments, which represent the position of the new bubble. The radius of a new bubble is always 10 and never changes after that. The isVisible method indicates whether the bubble is currently visible inside a window of width w and height h (position (0, 0) is in the upper-left corner of the window). The bubble is considered visible if at least one pixel of the bubble is visible. Therefore a bubble might be visible even when its center is outside the window, as long as the edge of the bubble is still visible inside the window. The code of the isVisible method is a little bit complex, mostly because of the case where the center of the circle is just outside one of the corners of the window. So here is the code of the isVisible method, which you can directly copy-paste into your assignment: // Find the point (wx, wy) inside the window which is closest to the // center (x, y) of the circle. In other words, find the wx in the // interval [0, w - 1] which is closest to x, and find the wy in the // interval [0, h - 1] which is closest to y. // If the distance between (wx, wy) and (x, y) is less than the radius // of the circle (using Pythagoras's theorem) then at least part of // the circle is visible in the window. // Note: if the center of the circle is inside the window, then (wx, wy) // is the same as (x, y), and the distance is 0. public boolean isVisible(int w, int h) { double x = getX(); double y = getY(); double wx = (x < 0 ? 0 : (x > w - 1 ? w - 1 : x)); double wy = (y < 0 ? 0 : (y > h - 1 ? h - 1 : y)); double dx = wx - x; double dy = wy - y; return dx * dx + dy * dy <= radius * radius; } The isIn method indicates whether the point at coordinates (x, y) (which are the arguments of the method) is currently inside the bubble or not. The edge of the bubble counts as being inside of the bubble. HINT: use Pythagoras's theorem to compute the distance from the center of the bubble to the point (x, y). The draw method uses the graphics object g to draw the bubble. HINT: remember that the color of the graphics object g is changed in the draw method of the superclass of Bubble. Also add a testBubble method to test all your methods (including inherited methods, but excluding the isVisible method, which I provide, and excluding the draw method since it requires as argument a graphics object g that you

You will need to implement the isIn and draw methods yourself, using Pythagoras's theorem to calculate the distance between the center of the bubble and the given point, and the Graphics object ...

继续理解以下的数据库中的操作When a vertex execution fails for any reason the job manager is informed. If the vertex reported an error cleanly the process forwards it via the daemon before exiting; if the process crashes the daemon notifies the job manager; and if the daemon fails for any reason the job manager receives a heartbeat timeout. If the failure was due to a read error on an input channel (which is be reported cleanly) the default policy also marks the execution record that generated that version of the channel as failed and terminates its process if it is running. This will cause the vertex that created the failed input channel to be re-executed, and will lead in the end to the offending channel being re-created. Though a newly-failed execution record may have non-failed successor records, errors need not be propagated forwards: since vertices are deterministic two successors may safely compute using the outputs of different execution versions. Note however that under this policy an entire connected component of vertices connected by pipes or shared-memory FIFOs will fail as a unit since killing a running vertex will cause it to close its pipes, propagating errors in both directions along those edges. Any vertex whose execution record is set to failed is immediately considered for re-execution.

如果顶点干净地报告了错误，进程会在退出前通过守护程序将错误信息转发给作业管理器。如果进程崩溃，守护程序会通知作业管理器。如果守护程序由于任何原因失败，作业管理器会接收到心跳超时。如果失败是由于输入通道...

The objective of the program you are going to produce is to evaluate boolean expressions as the one shown next: Expression: ( V | V ) & F & ( F | V ) where V is for True, and F is for False. The expressions may include the following operators: ! for not, & for and, | for or, the use of parenthesis for operations grouping is also allowed. To perform the evaluation of an expression, it will be considered the priority of the operators, the not having the highest, and the or the lowest. The program must yield V or F, as the result for each expression in the input file. Input The expressions are of a variable length, although will never exceed 100 symbols. Symbols may be separated by any number of spaces or no spaces at all, therefore, the total length of an expression, as a number of characters, is unknown. The number of expressions in the input file is variable and will never be greater than 20. Each expression is presented in a new line, as shown below. Output For each test expression, print "Expression " followed by its sequence number, ": ", and the resulting value of the corresponding test expression. Separate the output for consecutive test expressions with a new line. Use the same format as that shown in the sample output shown below. Sample Input ( V | V ) & F & ( F| V) !V | V & V & !F & (F | V ) & (!F | F | !V & V) (F&F|V|!V&!F&!(F|F&V)) Sample Output Expression 1: F Expression 2: V Expression 3: V

Note that the evaluate() function uses a stack to keep track of the operands and operators in the expression. It starts by iterating over each symbol in the expression, and for each symbol, it ...

You are the author of a Codeforces round and have prepared n problems you are going to set, problem i having difficulty ai . You will do the following process: remove some (possibly zero) problems from the list; rearrange the remaining problems in any order you wish. A round is considered balanced if and only if the absolute difference between the difficulty of any two consecutive problems is at most k (less or equal than k ). What is the minimum number of problems you have to remove so that an arrangement of problems is balanced? Input The first line contains a single integer t (1≤t≤1000 ) — the number of test cases. The first line of each test case contains two positive integers n (1≤n≤2⋅105 ) and k (1≤k≤109 ) — the number of problems, and the maximum allowed absolute difference between consecutive problems. The second line of each test case contains n space-separated integers ai (1≤ai≤109 ) — the difficulty of each problem. Note that the sum of n over all test cases doesn't exceed 2⋅105 . Output For each test case, output a single integer — the minimum number of problems you have to remove so that an arrangement of problems is balanced.用cpp怎么写

for (int i = 1; i ; i++) { int diff = difficulties[i] - difficulties[i-1]; int problemsToRemove = (diff - 1) / k; minProblemsToRemove += problemsToRemove; } return minProblemsToRemove; } int ...

用C语言解决下列问题：Kirill wants to weave the very beautiful blanket consisting of n×m of the same size square patches of some colors. He matched some non-negative integer to each color. Thus, in our problem, the blanket can be considered a B matrix of size n×m consisting of non-negative integers. Kirill considers that the blanket is very beautiful, if for each submatrix A of size 4×4 of the matrix B is true: A11⊕A12⊕A21⊕A22=A33⊕A34⊕A43⊕A44, A13⊕A14⊕A23⊕A24=A31⊕A32⊕A41⊕A42, where ⊕ means bitwise exclusive OR Kirill asks you to help her weave a very beautiful blanket, and as colorful as possible! He gives you two integers n and m . Your task is to generate a matrix B of size n×m , which corresponds to a very beautiful blanket and in which the number of different numbers maximized. Input The first line of input data contains one integer number t (1≤t≤1000 ) — the number of test cases. The single line of each test case contains two integers n and m (4≤n,m≤200) — the size of matrix B . It is guaranteed that the sum of n⋅m does not exceed 2⋅105 . Output For each test case, in first line output one integer cnt (1≤cnt≤n⋅m) — the maximum number of different numbers in the matrix. Then output the matrix B (0≤Bij<263) of size n×m . If there are several correct matrices, it is allowed to output any one. It can be shown that if there exists a matrix with an optimal number of distinct numbers, then there exists among suitable matrices such a B that (0≤Bij<263) .

我可以回答这个问题。这是一个矩阵问题，可以使用C语言解决。...对于每个4×4的子矩阵A，可以使用条件判断来判断是否满足美丽的条件，如果满足，则在矩阵B中填入不同的数字。最后输出矩阵B和其中不同数字的数量即可。

Develop a NetBeans application that illustrates the applicability of one sorting algorithm on a collection of 10 items. The algorithm should sort the items in descending order. The algorithm must work based on the principle of one of the four sorting algorithms studied during the semester. The collection of 10 items is defined and populated in the application code. You can decide the type and values of items stored in the collection. The algorithm must be implemented as part of a class called MyArrayList that extends the ArrayList class functionality with your sorting algorithm. [20 marks] 5. Extend the application implemented for Question 4 above to be able to measure the efficiency of the implemented sorting algorithm in term of number of comparisons performed by the algorithm. The application should display the total number of comparisons performed by the algorithm for a given collection of items. [10 marks] 6. In an Excel graph, plot the number of comparisons measured by the sorting algorithm when three different sizes of collection are considered. Submit your Excel file on Moodle. [20 marks]

You want me to develop a NetBeans application that demonstrates the use of one sorting algorithm on a collection of 10 items, sorting them in descending order. The algorithm should be based on one of...

c++Bessie and friends have been captured and are trapped in a secret compound in a location far from their farm, and it is up to Bessie to plan their escape! The compound consists of � � NK holding cells arranged in an � × � N×K rectangular grid, where there are gates between horizontally and vertically adjacent cells. Each cell houses exactly one cow. Bessie has hacked into the system, and is able to unlock any subset of the gates, but each gate has a cost. For the cows to escape, Bessie must open enough gates that all the cows can gather in a single cell (so that they have enough cow-power to tunnel to the surface!). Bessie wants to minimize the total unlocking cost. But the stakes are higher than ever, and Bessie cannot be content with just one escape plan: she needs back-ups. Help her count the number of minimum-cost escape plans; two plans are considered different if some gate needs to be unlocked in one of the plans but not the other. Since this number may be very large, only output its remainder modulo 1 0 9 + 7 10 9 +7.

下面是对应的 C++ 代码实现： cpp #include #include using namespace std; const int MOD = 1e9 + 7; int countEscapePlans(vector<vector<int>>& gates) { int N = gates.size(); int K = gates[0]....

用中文翻译：A coupled three-dimensional model is developed to study the internal parameter distributions of the MBPP fuel cell stack, considering fluid dynamics, electro-chemical reactions, multi-species mass transfer, twophase flow of water and thermal dynamics. The model geometry domains include anode MBPP, anode gas wavy flow field (5 parallel flow channels), anode GDL, anode catalyst layer (CL), membrane, cathode CL, cathode GDL, cathode gas wavy flow field (5 parallel flow channels), cathode MBPP and the two-layered coolant wavy flow fields at anode/cathode sides. According to the stack design, the design parameters of wavy flow fields for anode and cathode sides are the same but the phase deviation between their wave cycles presents 180◦. The two wavy flow fields of coolant, at the respective back sides of the anode and cathode plates, form the intercrossed two-layered coolant flow fields inside the MBPP, due to the phase difference of 180◦ between the wave cycles (Fig. 3). The mismatched flow field patterns between the neighbored fluid flows lead to complicated geometry and mesh building. The presented model geometry is divided into several layers (xz plane) according to the different domain materials so that the thin metallic plate and fluid domains with complicated 3D morphologies could be finely meshed layer by layer. As the real geometry of the experimental stack is too large for calculation, the modeled flow field consists of 5 parallel wavy channels, each of which includes 2 wave periods and corresponding inlet/outlet portions as well. To study the detailed thermal behavior of the presented design, the two-layered coolant fluid flow at the back side of the anode plate is considered and so is for the cathode plate. The counter flow operation is conducted where the air flows at the same direction with coolant but the opposite with hydrogen, shown in Fig. 3 (b).

研究MBPP燃料电池堆内部参数分布的三维耦合模型被建立，考虑流体力学、电化学反应、多物种质量转移、水的两相流动和热力学。模型几何域包括阳极MBPP、阳极气体波浪流场（5个平行流道）、阳极GDL、阳极催化层（CL）、...

Standard SPI Mode Standard SPI mode is selected when the Mode option in the Vivado IDE is set to Standard. The relevant parameters in this mode are: • Mode • Enable STARTUPE2 Primitive • Transaction Width • No. of Slaves • Frequency Ratio Send Feedback AXI Quad SPI v3.2 7 PG153 April 4, 2018 www.xilinx.com Chapter 1: Overview • Enable FIFO The properties of the core in standard SPI mode, including or excluding a FIFO, are described as: • The choice of inclusion of FIFO is based on the Enable FIFO parameter. FIFO Depth parameter is linked to Enable FIFO parameter. FIFO Depth limits the transmit and receive FIFO depth to 16 or 256 when FIFO is enabled. When FIFO is not enabled, the value of FIFO depth parameter is considered to be 0. A FIFO depth of 256 should be used because this is the most suitable depth in relation to the flash memory page size. • The valid values for the FIFO Depth option in this mode are 16 or 256 when FIFO is enabled through Enable FIFO parameter. When Enable FIFO is 0 and no FIFO is included in the core. Data transmission occurs through the single transmit and receive register. When FIFO Depth is 16 or 256, the transmit or receive FIFO is included in the design with a depth of 16 or 256 elements. The width of the transmit and receive FIFO is configured with the Transaction Width option. The AXI Quad SPI core supports continuous transfer mode. When configured as master, the transfer continues until the data is available in the transmit register/FIFO. This capability is provided in both manual and automatic slave select modes. As an example, during the page read command, the command, address, and number of data beats in the DTR must be set equal to the same number of data bytes intended to be read by the SPI memory. When the core is configured as a slave, if the slave select line (SPISEL) goes High (inactive state) during the data element transfer, the current transfer is aborted. If the slave select line goes Low, the aborted data element is transmitted again. The slave mode of the core is allowed only in the standard SPI mode.

of Slaves（从设备数量） • Frequency Ratio（频率比例） • Enable FIFO（启用FIFO）在标准SPI模式下，核心的属性包括是否包含FIFO等，具体描述如下： • 是否包含FIFO的选择取决于Enable FIFO参数。FIFO Depth...

You are given two arrays a and b each consisting of n integers. All elements of a are pairwise distinct. Find the number of ways to reorder a such that ai>bi for all 1≤i≤n , modulo 109+7 . Two ways of reordering are considered different if the resulting arrays are different. Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤104 ). The description of the test cases follows. The first line of each test case contains a single integer n (2≤n≤2⋅105 ) — the length of the array a and b . The second line of each test case contains n distinct integers a1 , a2 , … , an (1≤ai≤109 ) — the array a . It is guaranteed that all elements of a are pairwise distinct. The second line of each test case contains n integers b1 , b2 , … , bn (1≤bi≤109 ) — the array b . It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 . Output For each test case, output the number of ways to reorder array a such that ai>bi for all 1≤i≤n , modulo 109+7 .

题意简述：给定两个长度为n的数组a和b，每个数组都由n个不同的正整数组成。找出重新排列数组a的方式，使得对于所有1≤i≤n，ai>bi。如果得到的数组不同，则认为两种重新排列方式是不同的。求满足条件的方案数，对109...

This function first checks if the input n is less than or equal to 1, and returns FALSE in that case, because 1 is not considered a prime number. Next, the function uses a for loop to check if n is evenly divisible by any number between 2 and n-1. If n is evenly divisible by any of these numbers, the function returns FALSE, as n is not a prime number. If none of the numbers between 2 and n-1 divide n evenly, the function returns TRUE, indicating that n is a prime number.

接下来，该函数使用for循环检查n是否能够被2到n-1之间的任何数字整除。如果n能够被这些数字之一整除，则函数返回FALSE，表示n不是质数。如果2到n-1之间的数字都不能整除n，则函数返回TRUE，表示n是质数。

Follow the instructions below to construct a complete program. We are going to write a program that reads a textfile with keywords and their sentiment value, and calculates the sentiment of a tweet. The text file has some commonly used words and their sentiment values. If we group these keywords into categories, we can check any tweet by comparing the words in it against the different sentiment categories and their values and coming up with a sentiment value for the tweet. We will group the keywords from our text file that have a 20 sentiment value into a “positive” group, the ones with a value of 0 into a “neutral” group and the ones with a -10 sentiment value into a “negative group”. Note that if you copy and paste code from this document, the minus signs will likely cause issues with the code. It is better to type the minus signs rather than copy and paste them in to avoid these issues.These are the contents of our text file:

To clarify, you want to group the keywords based on their sentiment value, where keywords with a value of 20 are considered positive, those with a value of 0 are neutral, and those with a value of -10...

请用C语言实现Once upon a time in the mystical land of Draconis, there existed two powerful arrays: M and N . These arrays were filled with positive integers, each carrying its own magical essence. The inhabitants of the land were intrigued by the concept of similarity between arrays. They discovered that two arrays, M and N , could be considered similar if it was possible to transform a subarray of N into M by adding or subtracting a constant value to each element. You are now summoned to solve a puzzle. Given two arrays, M and N , your task is to determine the number of subarrays of N that are similar to M . Will you be able to unravel this mystical connection? Input The input consists of multiple lines. The first line contains two integers M and N (1≤M≤N≤106) , representing the lengths of arrays M and N respectively. The second line contains M space-separated positive integers m1,m2,…,mM (1≤mi≤109) , representing the magical elements of array M . The third line contains N space-separated positive integers n1,n2,…,nN (1≤ni≤109) , representing the mystical elements of array N . Output Output a single integer, the number of subarrays of N that are similar to M .

int compare(const void *a, const void *b) { return *(int*)a - *(int*)b; } int count_similar() { int i = 0, j = 0; int count = 0; while (i ) { if (M[i] == N[j]) { // 找到相同的元素 int k = j; ...

翻译这段英语并把里面用到的语法讲解一下In Polling mode, the HAL functions return the process status when the data processing in blocking mode is complete. The operation is considered complete when the function returns the HAL_OK status, otherwise an error status is returned. The user can get more information through the HAL_PPP_GetState() function. The data processing is handled internally in a loop. A timeout (expressed in ms) is used to prevent process hanging. The example below shows the typical Polling mode processing sequence :

2. 从句"when the data processing in blocking mode is complete"表示当阻塞模式的数据处理完成时； 3. 主语"the operation"表示操作； 4. 从句"otherwise an error status is returned"表示否则返回错误状态； 5. ...

用C++写Mr. Wor's router at home is broken again. After some repairs, he needs to reconnect the Ethernet cables to the router. It is known that Mr. Wor's router has n ports, and there are n Ethernet cables. Then, Mr. Wor will sequentially insert the i -th Ethernet cable to the i -th port, for all i from 1 to n . However, due to Mr. Wor's mysterious actions, the first m cables were not inserted into their designated ports, but were inserted into m ports uniformly at random. For the remaining cables, when cable i is being inserted, if port i is not occupied, then cable i will be inserted into port i . Otherwise, cable i will be inserted into a randomly chosen unoccupied port. Mr. Wor wants to know the probability that the last cable, cable n , is plugged into port n . Input Only one line contains two integers n,m(1≤n≤10,0≤m≤n) , denoting the number of cables and the number of cables which was inserted randomly. Output Print the probability in one line. Your answer is considered correct if the relative or absolute error is less than or equal to 10−6 .

c++ #include ... for (int i = 2; i ; i++) { for (int j = 0; j ; j++) { f[i][j] = (i-j)*f[i-1][j]/i + (j+1)*f[i-1][j+1]/i; } } printf("%.8lf", f[n][n-m]); return 0; }

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