drawn area

时间: 2024-06-15 22:05:56 浏览: 66
"drawn area"这个词组在不同的上下文中可能有不同的含义。请问你是指绘制区域的概念吗?如果是的话,绘制区域通常指的是在计算机图形学中,用于绘制图形或图像的特定区域。这个区域可以是屏幕上的特定位置,也可以是图像或画布上的特定区域。 绘制区域可以通过指定坐标、尺寸和其他属性来定义。在绘制区域内,可以使用不同的绘图算法和技术来创建各种图形效果,如线条、形状、颜色等。
相关问题

原理图右下角的drawn by

原理图右下角的drawn by通常是指该原理图的绘制者或者设计师。在电路设计过程中,每个原理图都需要有一个负责绘制和设计的人员,这个人员通常会将自己的名字或者工号填写在原理图右下角的drawn by字段中。这样可以方便其他人员了解该原理图的绘制者,并在需要时与其进行沟通和协作。此外,drawn by字段还可以用于质量控制和责任追溯,确保原理图的准确性和可靠性。

give me a function,finds and returns list of corners and new image with corners drawn

As an AI language model, I cannot provide you with any specific programming language or library. However, I can provide you with a general algorithm that you can implement in any language or library. Algorithm: 1. Import the necessary libraries for image processing and corner detection. 2. Load the image. 3. Convert the image to grayscale. 4. Apply a Gaussian blur to smoothen the image and remove any noise. 5. Apply a corner detection algorithm (e.g., Harris corner detector) to the image. 6. Extract the coordinates of the corners detected. 7. Draw circles around the corners in the original image. 8. Return the list of corner coordinates and the new image with corners drawn. Pseudo-code: ``` import cv2 def find_corners(image): # Convert the image to grayscale. gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY) # Apply a Gaussian blur to the image. blurred = cv2.GaussianBlur(gray, (5, 5), 0) # Apply a corner detection algorithm. corners = cv2.cornerHarris(blurred, 2, 3, 0.04) # Normalize the corner response. cv2.normalize(corners, corners, 0, 255, cv2.NORM_MINMAX) # Extract the coordinates of the corners. corner_coords = [] threshold = 0.1 * corners.max() for i in range(corners.shape[0]): for j in range(corners.shape[1]): if corners[i,j] > threshold: corner_coords.append((j, i)) # Draw circles around the corners in the original image. for coord in corner_coords: cv2.circle(image, coord, 5, (0, 0, 255), 2) # Return the list of corner coordinates and the new image with corners drawn. return corner_coords, image ``` Note: This is just a rough algorithm and may need to be adapted to your specific use case.

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Dog Card is a card game. In the game, there are a total of 2n cards in the deck, each card has a value, and the values of these 2n cards form a permutation of 1 ~ 2n. There is a skill that works as follows: 1. Draw a card from the top of the deck. 2. If the deck is empty, then skip to step 3, otherwise you guess whether the card on the top of the deck has a higher value than your last drawn card and draw a card from the top of the deck. If your guess is correct, then repeat this step, otherwise skip to step 3. 3. End this process. Nana enjoys playing this game, although she may not be skilled at it. Therefore, her guessing strategy when using this skill is simple: if the value of the last drawn card is less than or equal to n, then she guesses that the next oard's valve is higher, ther wse, she guedses thet the next card's vaue s lomler she wârns tb dmokt tor anfafrhlm decks of cards (Obviously, there are (2n)! cases), how many cards she can draw in total if she uses the skill only once in each case. Since this number can be very large,please provide the answer modulo a given value.Dog Card is a card game. In the game, there are a total of 2n cards in the deck, each card has a value, and the values of these 2n cards form a permutation of 1 ~ 2n. There is a skill that works as follows: 1. Draw a card from the top of the deck. 2. If the deck is empty, then skip to step 3, otherwise you guess whether the card on the top of the deck has a higher value than your last drawn card and draw a card from the top of the deck. If your guess is correct, then repeat this step, otherwise skip to step 3. 3. End this process. Nana enjoys playing this game, although she may not be skilled at it. Therefore, her guessing strategy when using this skill is simple: if the value of the last drawn card is less than or equal to n, then she guesses that the next oard's valve is higher, ther wse, she guedses thet the next card's vaue s lomler she wârns tb dmokt tor anfafrhlm decks of cards (Obviously, there are (2n)! cases), how many cards she can draw in total if she uses the skill only once in each case. Since this number can be very large,please provide the answer modulo a given value.给出c++代码及中文解释

代码We now want to always redraw all the points that have ever been drawn in the panel, not just the last point. To do this, we must save the coordinates of all these points so that we can redraw them all one by one in the paintComponent method every time this method is called. To save the coordinates of the various mouse positions we click, replace the x and y instance variables of the MyPanel class with a single private instance variable called points of type ArrayList. The Point class is provided to you by Swing. In the constructor of MyPanel, initialize the points instance variable with a new arraylist object of the same type. In the mouseClicked method of the mouse listener, use the getPoint method of the mouse event object to get a Point object representing the position of the mouse click (that Point object internally stores both the x and y coordinates of the mouse click event). Then add this Point object to the arraylist using the arraylist’s add method. Then, in the paintComponent method, add a loop to draw in the panel all the points of the arraylist. You can get the number of elements in the arraylist by using the size method of the arraylist; you can access a specific element of the arraylist at index i by using the get(i) method of the arraylist (element indexes start at zero in an arraylist). The Point class has getX and getY methods to get the coordinates of the point (these two methods return values of type double so you need to cast the returned values into the int type before you can use them to draw a point).

④ 模拟C/S通信,要求如下: a.模拟客户端(Client端)程序client,其功能是: 1.显示下列服务功能菜单: ************************** * 1:Query balance * * 2:Draw money * * 3:Save money * * 4:Change password * * 0:Exit * ************************* * 2.接收用户键入的功能号选择; 3.将用户键入的功能号作为一条信息发送到消息队列,然后结束。 b.模拟服务器端(Server端)程序server,其功能是: 1.从消息队列接收Client端发来的一条消息; 2.父进程创建一个子进程; 3.根据消息作如下处理: 若消息为"1",子进程1加载服务模块query,该模块的内容为显示以下信息:You have $10000! 若消息为"2",子进程1加载服务模块draw,该模块的内容为显示以下信息:You have drawn $10000! 若消息为"3",子进程1加载服务模块save,该模块的内容为显示以下信息:You have saved $10000! 若消息为"4",子进程1加载服务模块change,该模块的内容为显示以下信息:Your password has changed! 若消息为"0",退出子进程。 4.等待子进程终止后,Server进程删除消息缓冲队列,然后结束。 注意: I).各个子模块query、draw、save和change要事先编译连接好,放在你认为合适的子目录下; II).采用先运行客户端进程,然后运行服务器端进程的方式实现同步; III).注意子进程的加载方法 在linux系统下,根据要求给出代码要求

用C++编写程序,实现以下问题2、题目ID Codes(POJ1146) Time Limit: 1000MS Memory Limit: 10000K 描述: It is 2084 and the year of Big Brother has finally arrived, albeit a century late. In order to exercise greater control over its citizens and thereby to counter a chronic breakdown in law and order, the Government decides on a radical measure--all citizens are to have a tiny microcomputer surgically implanted in their left wrists. This computer will contains all sorts of personal information as well as a transmitter which will allow people's movements to be logged and monitored by a central computer. (A desirable side effect of this process is that it will shorten the dole queue for plastic surgeons.) An essential component of each computer will be a unique identification code, consisting of up to 50 characters drawn from the 26 lower case letters. The set of characters for any given code is chosen somewhat haphazardly. The complicated way in which the code is imprinted into the chip makes it much easier for the manufacturer to produce codes which are rearrangements of other codes than to produce new codes with a different selection of letters. Thus, once a set of letters has been chosen all possible codes derivable from it are used before changing the set. For example, suppose it is decided that a code will contain exactly 3 occurrences of a', 2 of b' and 1 of c', then three of the allowable 60 codes under these conditions are: abaabc abaacb ababac These three codes are listed from top to bottom in alphabetic order. Among all codes generated with this set of characters, these codes appear consecutively in this order. Write a program to assist in the issuing of these identification codes. Your program will accept a sequence of no more than 50 lower case letters (which may contain repeated characters) and print the successor code if one exists or the message No Successor' if the given code is the last in the sequence for that set of characters. 输入: Input will consist of a series of lines each containing a string representing a code. The entire file will be terminated by a line consisting of a single #. 输出: Output will consist of one line for each code read containing the successor code or the words 'No Successor'. 样例输入 abaacb cbbaa # 样例输出 ababac No Successor

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