Between each pair of adjacent layers is an interface. The interface defines wh ich primitives operations and services the lower layer makes available to the upper one.解释这句话

时间: 2024-01-24 12:18:52 浏览: 121
这句话的意思是,在每一对相邻的层之间都有一个接口。这个接口定义了下层向上层提供的原始操作和服务。换句话说,每个层次都有自己独特的功能和责任,而接口则允许不同的层次之间相互通信和交换信息。在这种情况下,每个层次可以使用下层提供的功能,同时也可以向上层提供自己的服务。这种分层的结构使得系统更加灵活,因为它允许各个层次之间相互独立,并且可以通过接口进行协作和交互。
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用中文翻译:The anode side coolant channels W1 ~ W6 at the upper (solid lines) and lower (dashed lines) layers overlap and interweave, as shown in Fig. 11 (c). In the present model, coolant of the lower layer contact anode metallic plate directly with higher temperature than that of the upper layer. The temperature distribution results show apparent heat exchange effect between the adjacent channels at intersections of the wavy channels, for example, the W1 channel at lower layer and the W2 channel at upper layer. The heat exchange effect of the intercrossed wavy flow field is mainly caused by two thermal processes: the thermal conduction through the coolant material in contact and the thermal convection by the secondary flow induced mass transfer indicated with yellow dashed circles. Fig. 11 (d) and (e) show local temperature of the anode and cathode metallic plates. Due to the fine thermal conductivity, the maximum temperature deviation is less than 1 K. The ribs which contact the MEA directly present apparently higher temperature than the channel compartments. The edge areas show the highest temperature where the local portion of MEA generates reaction heat without coverage by the wavy coolant channels. From the thermal management point of view, the edge area of the wavy flow fields with local hot spot behavior should be concerned when designing the MBPP fuel cell stack.

热液流道W1~W6上下层(实线和虚线)重叠交错,如图11(c)所示。在当前模型中,下层冷却剂与阳极金属板直接接触,温度比上层高。温度分布结果表明,波浪流道交叉点处相邻流道之间存在明显的热交换效应,例如下层的W1流道和上层的W2流道。交错的波浪流场的热交换效应主要由两个热过程引起:接触介质中的热传导和由黄色虚线圆圈所标示的辅助流引起的热对流。图11(d)和(e)显示了阳极和阴极金属板的局部温度。由于热导率较高,最大温差不超过1 K。直接与MEA接触的肋骨表现出明显高于流道舱的温度。边缘区域显示最高温度,其中MEA的局部部分产生反应热而没有被波浪冷却流道覆盖。从热管理的角度来看,在设计MBPP燃料电池堆时,应关注具有局部热点行为的波浪流场的边缘区域。

WishingBone plans to decorate his hexagon-shaped room with unique flooring patterns. First he divides it into triangles, and colors them with different colors. Now that he has got a satisfactory pattern, he uses small rugs to cover the floor. The rugs sold at shops are diamond-shaped; exact the shape of two adjacent triangles. He is not willing to cut any of them. Different colored rugs are labeled different price. And he does care about the price - he cares about the most expensive rugs that he should use to finish the covering. Input There are multiple tests, for each test: The first line is an integer specifying the side of his room (1 <= n <= 50). The following 2n lines with the pattern of the room, with upper-case letters denoting different colors. So there are at most 26 colors. Since the room is considerably large and sometimes needs pillars to support the ceiling, some triangles are marked with '.' rather than a letter. Such triangles are not to be covered. The following line contains one integer m (1 <= m <= 351) - number of rugs in the shop. The following m lines contains one type of rug (with two upper-case letters) and its price p (1 <= p <= 100000). There will not be duplicate items listed, that is, if AB is listed, BA will not appear. You may assume the number of each type to be infinite. Output One line for each test, the price of the most expensive rugs used. Of course, you have to minimize this value. If there is no solution to the problem, print a 0.

这是一道图论和最小权匹配问题,需要使用匈牙利算法来解决。 首先,将六边形的每个三角形看作一个节点,将相邻的三角形之间的边看作图中的边。这样就将问题转化为了求解这个图的最小权完美匹配问题。 接下来,对于每个给定的小地毯,将其两个顶点所对应的三角形之间的边标记为有权值的边,权值为该小地毯的价格。 最后,使用匈牙利算法求解这个图的最小权完美匹配,即可得到使用的最贵地毯的价格。 需要注意的是,如果无法完美匹配,则说明无法覆盖整个六边形的地面,此时输出0。
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用c++和segment tree解决下述问题Doing Exercises 描述 As we all know, the lines of students doing exercises between classes are always unsatisfactory to teachers. Today, a teacher wants to require something new. Firstly, he lets some students of N classes correspondingly form n lines. Then, he randomly selects a class to add some of its remaining students to its line, or selects a class to let some students leave its line, or lets the monitors from some adjacent classes report the total number of students in all these classes. This is very annoying for the monitors. Can you write a program to help them complete the reporting task? 输入 The first line is an integer T (T<50), the number of test cases. For each test case, its first line is an integer N (1<=N<=50000), representing the number of classes, and its second line include N integers (a1, a2, a3, ... , an), and ai (1<=ai<=100) means the number of students in the line of class i at the beginning. Then, each next line will be an order. There are 4 kinds of orders: (1) "Add x i" means adding x students to the line of class i; (2) "Sub x i" means that x students leave the line of class i; (3) "Query i j" means that the monitors from class i to class j report the teacher the total number (sum) of students in their classes at that moment (i<j); (4) "End" means ending the exercises, which will only appear at the end of each test case. The number of orders will not exceed 40000. The number of students in any line will never below 0. 输出 For each test case, you must print "Case i:" in the first line. Then for each Query, print the result in one line.

Create a function pixel_flip(lst, orig_lst, budget, results, i=0) that uses recursion to generate all possible new unique images from the input orig_lst, following these rules: • The input lst is the current list being processed. Initially, this will be the same as orig_lst which is the original flattened image. • The input budget represents the number of pixels that can still be flipped. When the budget reaches 0, no more pixels can be flipped. • The input results is a list of resulting flattened images with flipped pixels. Initially, this will be an empty list. • The input i represents the index of the pixel being processed, by default set to 0, which is used to drive the recursive function towards its base case (i.e., initially starting from i=0). At termination of the function, the argument results should contain all possibilities of the input orig_lst by only flipping pixels from 0 to 1 under both the budget and the adjacency constraints. fill code at #TODO def pixel_flip(lst: list[int], orig_lst: list[int], budget: int, results: list, i: int = 0) -> None: """ Uses recursion to generate all possibilities of flipped arrays where a pixel was a 0 and there was an adjacent pixel with the value of 1. :param lst: 1D list of integers representing a flattened image . :param orig_lst: 1D list of integers representing the original flattened image. :param budget: Integer representing the number of pixels that can be flipped . :param results: List of 1D lists of integers representing all possibilities of flipped arrays, initially empty. :param i: Integer representing the index of the pixel in question. :return: None. """ #TODO def check_adjacent_for_one(flat_image: list[int], flat_pixel: int) -> bool: """ Checks if a pixel has an adjacent pixel with the value of 1. :param flat_image: 1D list of integers representing a flattened image . :param flat_pixel: Integer representing the index of the pixel in question. :return: Boolean. """ #TODO

The starting configuration of this puzzle is a row of cells, with disks located on cells through . The goal is to move the disks to the end of the row using a constrained set of actions. At each step, a disk can only be moved to an adjacent empty cell, or to an empty cell two spaces away if another disk is located on the intervening square. Given these restrictions, it can be seen that in many cases, no movements will be possible for the majority of the disks. For example, from the starting position, the only two options are to move the last disk from cell to cell , or to move the second-to-last disk from cell to cell . 1. [15 points] Write a function solve_identical_disks(length, n) that returns an optimal solution to the above problem as a list of moves, where length is the number of cells in the row and n is the number of disks. Each move in the solution should be a twoelement tuple of the form (from, to) indicating a disk movement from the cell from to the cell to. As suggested by its name, this function should treat all disks as being identical. Your solver for this problem should be implemented using a breadth-first graph search. The exact solution produced is not important, as long as it is of minimal length. Unlike in the previous two sections, no requirement is made with regards to the manner in which puzzle configurations are represented. Before you begin, think carefully about which data structures might be best suited for the problem, as this choice may affect the efficiency of your search

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