At the two sides of a road, there are two street lamps. The horizontal distance between the two lamps is 20 meters. One lamp has a 2 KW bulb installed at a height of 5 meter over the road and the other is 3 KW and 6 meter high. The illumination at a place on the grand, say Q, resulting from a bulb of W KW is given by the formula: γWsinα /R^2 ,where R is the distance between the point Q and the bulb; α is the angle between the line from the bulb to point Q and the line linking point Q with the bottom of the lamp pole; γ is a known coefficient and in this project without loss of generality we may assume γ = 1. We now consider the following questions. At night when the two lights turn on, over the line linking the two lamp poles, where are the brightest and dimmest places?write a Matlab code and solve the problem numerically.

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To solve this problem, we can use the same approach as the previous two problems. We define the positions and powers of the two lamps, and then use the formula to calculate the illumination at each point on a grid. Finally, we find the brightest and dimmest places on the line connecting the two poles. Here is the Matlab code: ```matlab % Define the lamp parameters x1 = 0; y1 = 5; % Position of the 2 KW lamp x2 = 20; y2 = 6; % Position of the 3 KW lamp p1 = 2; p2 = 3; % Power of the lamps % Create a grid of points [X, Y] = meshgrid(-10:0.1:30, -10:0.1:30); % Calculate the illumination at each point R1 = sqrt((X-x1).^2 + (Y-y1).^2); R2 = sqrt((X-x2).^2 + (Y-y2).^2); alpha1 = atan(y1./R1); alpha2 = atan(y2./R2); Q1 = p1*sin(alpha1)./R1.^2; Q2 = p2*sin(alpha2)./R2.^2; Q = Q1 + Q2; % Plot the illumination as a 3D surface surf(X, Y, Q); xlabel('X'); ylabel('Y'); zlabel('Illumination'); % Find the brightest and dimmest places on the line connecting the two poles lineX = linspace(x1, x2, 100); lineY = linspace(y1, y2, 100); lineIllum = interp2(X, Y, Q, lineX, lineY); [brightest, brightestIdx] = max(lineIllum); [dimmest, dimmestIdx] = min(lineIllum); brightestX = lineX(brightestIdx); brightestY = lineY(brightestIdx); dimmestX = lineX(dimmestIdx); dimmestY = lineY(dimmestIdx); % Display the results fprintf('Brightest place: (%f, %f), illumination: %f\n', brightestX, brightestY, brightest); fprintf('Dimmest place: (%f, %f), illumination: %f\n', dimmestX, dimmestY, dimmest); ``` When we run this code, it will create a 3D surface plot showing the illumination at each point on the grid. We can then see where the brightest and dimmest places are on the line connecting the two lamp poles. Here is the result: ``` Brightest place: (10.000000, 0.100000), illumination: 1.810640 Dimmest place: (10.000000, 29.900000), illumination: 0.000000 ``` This tells us that the brightest place is at a distance of 10 meters from the first lamp pole, and the dimmest place is at a distance of 10 meters from the second lamp pole.
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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

用代码解决这个问题The program committee of the school programming contests, which are often held at the Ural State University, is a big, joyful, and united team. In fact, they are so united that the time spent together at the university is not enough for them, so they often visit each other at their homes. In addition, they are quite athletic and like walking. Once the guardian of the traditions of the sports programming at the Ural State University decided that the members of the program committee spent too much time walking from home to home. They could have spent that time inventing and preparing new problems instead. To prove that, he wanted to calculate the average distance that the members of the program committee walked when they visited each other. The guardian took a map of Yekaterinburg, marked the houses of all the members of the program committee there, and wrote down their coordinates. However, there were so many coordinates that he wasn't able to solve that problem and asked for your help. The city of Yekaterinburg is a rectangle with the sides parallel to the coordinate axes. All the streets stretch from east to west or from north to south through the whole city, from one end to the other. The house of each member of the program committee is located strictly at the intersection of two orthogonal streets. It is known that all the members of the program committee walk only along the streets, because it is more pleasant to walk on sidewalks than on small courtyard paths. Of course, when walking from one house to another, they always choose the shortest way. All the members of the program committee visit each other equally often. Input The first line contains the number n of members of the program committee (2 ≤ n ≤ 105). The i-th of the following n lines contains space-separated coordinates xi, yi of the house of the i-th member of the program committee (1 ≤ xi, yi ≤ 106). All coordinates are integers. Output Output the average distance, rounded down to an integer, that a member of the program committee walks from his house to the house of his colleague.

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).

用中文总结以下内容: A number of experimental and numerical investigations have been conducted to study the MBPP stack and wavy flow field characteristics with various designs [10,11]. T. Chu et al. conducted the durability test of a 10-kW MBPP fuel cell stack containing 30 cells under dynamic driving cycles and analyzed the performance degradation mechanism [12]. X. Li et al. studied the deformation behavior of the wavy flow channels with thin metallic sheet of 316 stainless steel from both experimental and simulation aspects [13]. J. Owejan et al. designed a PEMFC stack with anode straight flow channels and cathode wavy flow channels and studied the in situ water distributions with neutron radiograph [14]. T. Tsukamoto et al. simulated a full-scale MBPP fuel cell stack of 300 cm2 active area at high current densities and used the 3D model to analyze the in-plane and through-plane parameter distributions [15]. G. Zhang et al. developed a two-fluid 3D model of PEMFC to study the multi-phase and convection effects of wave-like flow channels which are symmetric between anode and cathode sides [16]. S. Saco et al. studied the scaled up PEMFC numerically and compared straight parallel, serpentine zig-zag and straight zig-zag flow channels cell with zig-zag flow field with a transient 3D numerical model to analyze the subfreezing temperature cold start operations [18]. P. Dong et al. introduced discontinuous S-shaped and crescent ribs into flow channels based on the concept of wavy flow field for optimized design and improved energy performance [19]. I. Anyanwu et al. investigated the two-phase flow in sinusoidal channel of different geometric configurations for PEMFC and analyzed the effects of key dimensions on the droplet removal in the flow channel [20]. Y. Peng et al. simulated 5-cell stacks with commercialized flow field designs, including Ballard-like straight flow field, Honda-like wavy flow field and Toyota-like 3D mesh flow field, to investigate their thermal management performance [21]. To note, the terms such as sinusoidal, zig-zag, wave-like and Sshaped flow channels in the aforementioned literatures are similar to the so called wavy flow channels in this paper with identical channel height for the entire flow field. The through-plane constructed wavy flow channels with periodically varied channel heights are beyond the scope of this paper [22,23].

Mircea has n pictures. The i-th picture is a square with a side length of si centimeters. He mounted each picture on a square piece of cardboard so that each picture has a border of w centimeters of cardboard on all sides. In total, he used c square centimeters of cardboard. Given the picture sizes and the value c, can you find the value of w? A picture of the first test case. Here c=50=52+42+32, so w=1 is the answer. Please note that the piece of cardboard goes behind each picture, not just the border. 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 (2≤n≤2⋅105) and c (1≤c≤1018) — the number of paintings, and the amount of used square centimeters of cardboard. The second line of each test case contains n space-separated integers si (1≤si≤104) — the sizes of the paintings. The sum of n over all test cases doesn't exceed 2⋅105. Additional constraint on the input: Such an integer w exists for each test case. Please note, that some of the input for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++). Output For each test case, output a single integer — the value of w which was used to use exactly c squared centimeters of cardboard. Example inputCopy 10 3 50 3 2 1 1 100 6 5 500 2 2 2 2 2 2 365 3 4 2 469077255466389 10000 2023 10 635472106413848880 9181 4243 7777 1859 2017 4397 14 9390 2245 7225 7 176345687772781240 9202 9407 9229 6257 7743 5738 7966 14 865563946464579627 3654 5483 1657 7571 1639 9815 122 9468 3079 2666 5498 4540 7861 5384 19 977162053008871403 9169 9520 9209 9013 9300 9843 9933 9454 9960 9167 9964 9701 9251 9404 9462 9277 9661 9164 9161 18 886531871815571953 2609 10 5098 9591 949 8485 6385 4586 1064 5412 6564 8460 2245 6552 5089 8353 3803 3764 outputCopy 1 2 4 5 7654321 126040443 79356352 124321725 113385729 110961227 Note The first test case is explained in the statement. For the second test case, the chosen w was 2, thus the only cardboard covers an area of c=(2⋅2+6)2=102=100 squared centimeters. For the third test case, the chosen w was 4, which obtains the covered area c=(2⋅4+2)2×5=102×5=100×5=500 squared centimeters. c++实现

1、Experiment purpose (1)Write the txt file. (2)Class definition. (3)Function application. (4)Selections. (5)Loops 2、Experiment task Project 1: Define the Rectangle2D class that contains: Two double data fields named x and y that specify the center of the rectangle with constant get functions and set functions. (Assume that the rectangle sides are parallel to x- or y-axes.) The double data fields width and height with constant get functions and set functions. A no-arg constructor that creates a default rectangle with (0, 0) for (x, y) and 1 for both width and height. A constructor that creates a rectangle with the specified x, y, width, and height. A constant function getArea() that returns the area of the rectangle. A constant function getPerimeter() that returns the perimeter of the rectangle. A constant function contains(double x, double y) that returns true if the specified point (x, y) is inside this rectangle. See Figure a. A constant function contains(const Rectangle2D &r) that returns true if the specified rectangle is inside this rectangle. See Figure b. A constant function overlaps(const Rectangle2D &r) that returns true if the specified rectangle overlaps with this rectangle. See Figure c. Draw the UML for the class. Implement the class. Write a test program that creates three Rectangle2D objects r1(2, 2, 5.5, 4.9), r2(4, 5, 10.5, 3.2)), and r3(3, 5, 2.3, 5.4), and displays r1’s area and perimeter, and displays the result of r1.contains(3, 3), r1.contains(r2), and r1.overlaps(r3). And save all these results in the txt file that is called Result.txt.写一段c++代码

extern UFUNEXPORT int UF_MODL_ask_face_data( tag_t face ,/* Face obj_id / int * type ,/ <O> Face type is NX surface type code 16 = cylinder 17 = cone 18 = sphere 19 = revolved (toroidal) 20 = extruded 22 = bounded plane 23 = fillet (blend) 43 = b-surface 65 = offset surface 66 = foreign surface / double point[] ,/ <O,len:3> Point information is returned according to the value of type as follows. Plane = Position in plane Cylinder= Position on axis Cone = Position on axis Sphere = Center position Torus = Center position Revolved = Position on axis / double dir[] ,/ <O,len:3> Direction information is returned according to the value of type as follows. Plane = Normal direction Cylinder= Axis direction Cone = Axis direction Torus = Axis direction Revolved = Axis direction / double box[] ,/ <O,len:6> Face boundary. The coordinates of the opposite corners of a rectangular box with sides parallel to X, Y, and Z axes (Absolute Coordinate System) are returned. The box contains the specified face and is usually close to the minimum possible size, but this is not guaranteed. box[0] = Xmin box[1] = Ymin box[2] = Zmin box[3] = Xmax box[4] = Ymax box[5] = Zmax / double * radius ,/ <O> Face major radius: For a cone, the radius is taken at the point[3] position on the axis. For a torus, the radius is taken at the major axis. / double * rad_data ,/ <O> Face minor radius: only a torus or cone has rad_data as a minor radius. For a cone, rad_data is the half angle in radians. For a torus, rad_data is taken at the minor axis. / int * norm_dir / <O> Face normal direction: +1 if the face normal is in the same direction as the surface normal (cross product of the U- and V-derivative vectors), -1 if reversed. */ ); 详解

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