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[Other] 图形学直线的生成算法 DL : 0: 圆心在原点，半径为R的第一个4分圆的Bresenham画圆算法程序，将该算法程序推广到任一四分圆，从而形成一般的Bresenham画圆算法。并利用Bresenham画圆算法画出一个圆心在点(xc,yc), 半径为R, 圆周颜色为color的圆-Center in the basics of radius R for the first four hours round the Bresenham algorithm Circle procedures, procedures for the promotion of this algorithm, arrived on January 4 pm yen, thus forming a general Circle Bresenham algorithm. Circle and the use of Bresenham algorithm to identify a point in the center of a circle (xc, yc), the radius R, Circle color color circle
Date : 2025-12-31 Size : 1kb User :
[Other] p219__3__2 DL : 0: #include<iostream> using namespace std class Point {public: Point(int a){x=a } ~Point(){cout< "execuing Point destructor"<<endl } private: int x } class Circle:public Point {public: Circle(int m,int n):Point(m){radius=m } ~Circle(){cout<<"execuing Circle destructor"<<radius<<endl } private: int radius } int main() {Point *p=new Circle(5,12) delete p system("pause") return 0 } -# Include <iostream> using namespace std class Point (public: Point (int a) (x = a) ~ Point () (cout < execuing Point destructor <<endl) private: int x) class Circle: public Point (public: Circle (int m, int n): Point (m) (radius = m) ~ Circle () (cout << execuing Circle destructor <<radius <<endl) private: int radius) int main () (Point* p = new Circle (5,12) delete p system ( pause ) return 0)
Date : 2025-12-31 Size : 1kb User : 华盛
[Other] dv DL : 0: #include<iostream> using namespace std class Point {public: Point(int a){x=a } ~Point(){cout< "execuing Point destructor"<<endl } private: int x } class Circle:public Point {public: Circle(int m,int n):Point(m){radius=m } ~Circle(){cout<<"execuing Circle destructor"<<radius<<endl } private: int radius } int main() {Point *p=new Circle(5,12) delete p system("pause") return 0 } -# Include <iostream> using namespace std class Point (public: Point (int a) (x = a) ~ Point () (cout < execuing Point destructor <<endl) private: int x) class Circle: public Point (public: Circle (int m, int n): Point (m) (radius = m) ~ Circle () (cout << execuing Circle destructor <<radius <<endl) private: int radius) int main () (Point* p = new Circle (5,12) delete p system ( pause ) return 0)
Date : 2025-12-31 Size : 2kb User : 华盛
[Other] 3 DL : 0: 心形线，是一个圆上的固定一点在它绕着与其相切且半径相同的另外一个圆周滚动时所形成的轨迹，该程序可以动态的演示心形线的生成过程，-Heart-shaped line, is a fixed point circle around it with its tangent and the radius of a circle the same rolling track being formed, the program can be a demonstration of the dynamic heart-shaped line generation process,
Date : 2025-12-31 Size : 51kb User : 张校
[Other] snake DL : 0: 在数字图像中虹膜位置的有效定位是虹膜识别的关键问题。用一种基于主动轮廓线模型的方法定位虹膜的位置，先用灰度投影法检测出瞳孔内的一点作为瞳孔的伪圆心，该圆心只要能落在瞳孔内部即可。然后以该伪圆心为中心，在其周围等角度间隔地取N个点作为初始的snake基准点，按照snake 的运行机制不断进化，直到虹膜的内边界为止。最后，计算进化后的snake形心和snake上的控制点与该形心的距离，取其平均值作为瞳孔的半径，动态轮廓模型的形心作为瞳孔的圆心，即可准确定位出虹膜内边界的位置。实验表明，与常见的定位方法相比，文中的方法速度快、精度高，而且，对瞳孔初始的伪圆心要求不高，鲁棒性更强。-Digital image of the iris in the location of the effective positioning of the key issues of iris recognition. With a model based on the active contour method of positioning the location of the iris, the first gray projection method used to detect the pupil as the point of the pseudo-pupil center, the center of a circle as long as it can fall within the pupil. And then to the pseudo-center as the center, in terms of their spacing around the N points to get the snake as the initial reference point, the operating mechanism in accordance with the snake continued to evolve until the border until the inner iris. Finally, after calculating the evolution of snake and snake-shaped heart of the control point on the centroid distance, choose the average as the radius of the pupil, the dynamic contour model of the centroid as the center of pupil can be an accurate positioning within the borders of the iris position. The experimental results show that the position with common methods, the methods o
Date : 2025-12-31 Size : 1.68mb User : 侯汶杉
[Other] BEM2 DL : 0: 基于matlab边界元的经典算法。垂直接地极半径r=0.05m，长度l=0.5m,垂直接地极上电位V0=30V,边界元法求解径向向外8个点的电势.-Classical algorithm matlab Boundary Element. Vertical ground electrode radius r = 0.05m, length l = 0.5m vertical grounding on potential V0 = 30V, the Boundary Element Method radially outwardly eight-point potential.
Date : 2025-12-31 Size : 86kb User : 凌厉
[Other] Hough DL : 0: 用HOUGH检测直线的时候是两个未知量,因为轴半径和相角可以确定一条直线，在HOUGH域出现累计最大点就可能是原域中的直线。现在回到圆的检测上来，确定一个圆需要什么：X坐标，Y坐标，半径三个未知量吧，好了，现在，你做一个三维空间的HOUGH域，以这三个未知量作为三个轴，现在按照一定步长进行三重循环，在最内层循环是这样的，X,Y 确定，以不同的半径进行搜索，如果你的X,Y刚好就是实际图像的X,Y处，半径又搜索到实际真实的半径-The HOUGH detect straight when two unknown amount of axis radius and phase angle can be determined in a straight line, in the HOUGH domain cumulative maximum point may be the straight line of the original domain. Now back to the circle detection up to determine what a circle: X coordinate, Y coordinate, radius three unknowns, now, you do the HOUGH domain of a three-dimensional space, three to three unknown quantity as axis, in accordance with a certain step size triple loop, is such that in the innermost loop, X, and Y determined search for a different radius, If you X, Y, exactly is the actual image of the X, Y, at the radius radius of the search to the actual real
Date : 2025-12-31 Size : 45kb User : chen
[Other] PolarCo DL : 0: 测绘类计算，利用已知点，以及极角与极半径，计算另一点的坐标-The mapping class calculated using a known point, and the polar angle and polar radius, calculate Another point coordinates
Date : 2025-12-31 Size : 42kb User : 后腾辉
[Other] Circle DL : 0: Matlab圆弧插补，只要给定起点与终点，半径，设点圆心方向以及顺逆时针方向即可实现插补-Matlab circular interpolation, given the starting point and end point, radius, set the direction of the point of the center as well as Shun counterclockwise to achieve interpolation
Date : 2025-12-31 Size : 1kb User : xiaozhi
[Other] exp1 DL : 0: 理解和掌握C++抽象类和具体类的实现，了解运算符重载。实验内容：　　（1）从Point、Circle类中抽象出基类Shape，研究抽象类和具体类的接口和实现。　　（2）从中派生出一个正方形类和圆柱体类，然后写一个测试程序，输出正方形的面积和圆柱体的体积。　　（3）在Circle类中实现关系运算符重载（<、<=、==、!=、>、>=），实现按半径对Circle对象排序。-Understand and grasp the abstract and concrete classes C++ realization understand operator overloading. Experiment: (a) From the Point, Circle class abstract base class Shape, research abstract and concrete classes of interface and implementation. (2) derived from a square cylinder class and the class, and then write a test program, the output square area and volume of a cylinder. (3) In the Circle class operator overloading of relations (< , < =, ==,! =,> ,> =), To achieve by the radius of the right sort Circle object.
Date : 2025-12-31 Size : 1.36mb User : wu
[Other] point DL : 0: （1）在二维平面空间上，使用x、y坐标可以确定一个点；确定了圆心坐标和半径后可以确定一个圆。声明一个点类，并使用这个点类的对象为数据成员声明圆类。（2）设计并测试一个名为Ellipse的椭圆类，其属性为外接矩形的左上角与右下角两个点的坐标，并能计算出椭圆的面积 -(1) in two-dimensional space, using the x, y coordinates of a point can be determined determine the center coordinates and the radius of a circle can be determined later. Declaring a Point class, and use this point as a data object of the class member declarations round classes. (2) to design and test a class named Ellipse ellipse whose property is the upper left and lower right corner of bounding rectangle coordinates of two points, and to calculate the area of the ellipse
Date : 2025-12-31 Size : 1.16mb User : 李光炜
[Other] TranCoordinate DL : 0: 輸入點位於WGS84之大地經、緯度，可於命令窗顯示此點之卯酉圈曲率半徑N、子午圈曲率半徑M及平均曲率半徑R及此點之空間直角坐標X、Y及Z。-Input point is located in the WGS84 geodetic latitude and longitude can be displayed in the command window of the prime vertical radius of curvature point N, M meridian radius of curvature and the average radius of curvature R and the point of the space rectangular coordinates X, Y and Z.
Date : 2025-12-31 Size : 1kb User : 源
[Other] cover DL : 0: 覆盖问题 ★问题描述：通信公司准备在n 个村庄中的某些村庄安装基站,对于村庄i ，在其中建立基站的花费是Ci。我们把村庄看成是一个二维平面上的点。对于每一个基站，他们的规格，配置都是相同的，一个基站发射的信号可以覆盖以它为中心的半径为R 米的圆形区域(边界上的也算被覆盖)。现在公司希望用最小的花费，使得这n 个村庄都能被信号覆盖。 ★实验任务：对于给定的村庄的位置和修建基站的信息，求出公司覆盖所有村庄的最小花费。 ★数据输入：第一行是一个整数n（1<=n<=20）代表给定的村庄个数。接下来n 行，每行有3 个整数xi yi ci, 分别表示村庄的坐标以及在第i 个村庄修建基站的花费。(0<= xi, yi, ci <= 100) 最后一行是一个整数R , 表示基站的覆盖范围。(0 <= L <= 100) ★结果输出：公司覆盖所有村庄的最小花费。输入: 3 0 0 1 1 0 2 1 1 4 1 输出: 2-Cover problem ★ Problem Description : Communications ready n villages installation of base stations in certain villages , the village i, in which the base station to establish spending Is Ci. We village as a point on a two-dimensional plane . For each base station , their specifications , configurations are Same , the base station transmits a signal which can be covered with the center of the circular area of radius R m ( also the boundary Operators covered) . Now the company hopes to use the smallest cost , making this n villages can be overwritten signal. ★ experimental tasks : For information given the location and construction of the village station , find the minimum cost companies covering all the villages. ★ Data input : The first line is an integer n (1 <= n <= 20) representative of a given number of villages. The next n lines, each line has three integers xi yi ci, denote the coordinates of the village and the village built in the i-th Takes the base station.
Date : 2025-12-31 Size : 1kb User : 迷若烟雨
[Other] huofu DL : 0: matlab中运用hough变换，设定恰当的阈值，将超过阈值的点在三维空间标示出来&将三维参数矩阵根据半径的不同做成不同的帧，制作成视频-hough transform matlab in use, set the appropriate threshold, will exceed the threshold point in three-dimensional space marked out & three-dimensional parameter matrix made of different frames depending on the radius, made into video
Date : 2025-12-31 Size : 1kb User : 唐心
[Other] yuan DL : 0: 问题描述给定圆的半径r，求圆的面积。输入格式输入包含一个整数r，表示圆的半径。输出格式输出一行，包含一个实数，四舍五入保留小数点后7位，表示圆的面积。说明：在本题中，输入是一个整数，但是输出是一个实数。对于实数输出的问题，请一定看清楚实数输出的要求，比如本题中要求保留小数点后7位，则你的程序必须严格的输出7位小数，输出过多或者过少的小数位数都是不行的，都会被认为错误。实数输出的问题如果没有特别说明，舍入都是按四舍五入进行。样例输入 4 样例输出 50.2654825 数据规模与约定 1 <= r <= 10000。提示本题对精度要求较高，请注意π的值应该取较精确的值。你可以使用常量来表示π，比如PI=3.14159265358979323，也可以使用数学公式来求π，比如PI=atan(1.0)*4。-Problem description A given circle radius r, and the area of a circle. Input format Input contains an integer r, said the radius of the circle. The output format Output a line contains a real number, rounded after the decimal point 7, said the area of a circle. Note: in the ontology, the input is an integer, but the output is a real number. Answer to the question of the real output please see real output requirements, such as the subject of seven after decimal point, is the output of your application must be strictly 7 decimal places, the output of decimal digits too little or too much is not enough, will be considered. The problem of real output if there are no special instructions, rounding is each round. The sample input 4 Sample output 50.2654825 The data scale and conventions 1 < = r < = 10000. prompt Subject to the precision demand is higher, please pay attention to the value of PI should take more accurate values. You can use a constant to represent the PI
Date : 2025-12-31 Size : 9kb User : 童话Bu说话
[Other] huawujiaoxing DL : 0: 通过点击鼠标确定五角星中心，拖动鼠标后松开可绘制出五角星，也可直接点出固定半径的五角星-Determined by the click of a mouse pentagram center, after dragging the mouse to draw a pentagram release may also be directly fixed radius point out the five-pointed star
Date : 2025-12-31 Size : 3kb User : 郑洪振
[Other] YJS DL : 0: 机器人圆弧轨迹离散，需要的可以看看，三点确定圆坐标和半径，算出三点角度-Robot arc track discrete, need to look at the circle coordinates and radius determined by three points, three point
Date : 2025-12-31 Size : 19kb User : 迟锐
[Other] 1 DL : 0: 定义点类，然后由点类生成圆类，再由圆类生成圆柱体类。圆类的属性有圆心、半径；成员函数有构造函数、析构函数、显示、计算面积。圆柱体类的新增属性有高度；方法有构造函数、析构函数、显示、计算体积。类的应用：创建一个圆柱体对象，显示其信息，求它的体积。-Defined point category, then generates round classes the point of class, then class is generated by the circular cylinder class. Property round class are center, radius member functions have constructors, destructors, display, calculate the area. New property cylinder class of highly methods are constructors, destructors, display, calculate the volume. Type of application: Create a cylinder object to display its information, find its volume.
Date : 2025-12-31 Size : 1kb User : 何嘉荣
[Other] solve-centerial-point DL : 0: 在直角坐标系中，已知两点坐标和半径求圆心坐标程序-In the Cartesian coordinate system, the coordinates of known points and a radius of center coordinates program requirements
Date : 2025-12-31 Size : 925kb User : 梅跃
[Other] LATTICE DL : 0: 格子手绢问题描述久城送给维多利加一块苏格兰格子手绢，可是手绢不小心被咖啡弄脏了，久城不得不负起清洗的责任。手绢的格子可以用二维坐标系来描述，手绢左下角坐标为(0,0)，右上角坐标为(N,M)，也就是说，手绢上一共有N*M个格子。咖啡在手绢上形成了一个圆形的污点，圆心位于整点( X,Y)处，半径为R。久城想知道，有多少个格子被完全污染了，多少个格子被部分污染了。这里，某个格子被“完全污染”的意思为该格内被污点覆盖面积＝1，被“部分污染”的意思为0＜该格内被污点覆盖面积＜1。输入格式一行5个整数N,M,X,Y,R。输出格式一行两个整数，由一个空格隔开，表示被完全污染的格子数量和被部分污染的格子数量。样例输入 2 2 0 0 2 样例输出 1 3 数据范围对于20 的数据，N,M< 10；对于60 的数据，N,M< 1000；对于100 的数据，N,M< 1000000，0< X< N,0< Y< M，R< 1000000.-Plaid handkerchief Problem description For a long time the city to Victoria and a piece of Scotland Plaid handkerchief, but the handkerchief was accidentally dirty coffee, for a long time the city had negative Responsibility for cleaning. The handkerchief lattice can be used to describe the two-dimensional coordinates of the lower left corner of the handkerchief, coordinates (0,0), the upper right corner coordinates (N, M), That is to say, a handkerchief and a total of N*M lattice. Coffee has formed a circular spot in a handkerchief, a center at the whole point ( X, Y), the radius is R. For a long time the city wanted to know, how many squares were completely polluted, and how many checks were partly polluted. Here, a square was completely pollution means the box is spot coverage area 1, is pollution . The box is for 0 < < 1 spot covered area. Input format One line of 5 integers N, M, X, Y, R. Output format A line of two integers, separated by a space, that is co
Date : 2025-12-31 Size : 1kb User : zzz

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