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Description: 计算子午圈半径M
计算卯酉圈半径N
计算玮圈半径r
计算经线弧长Sm
计算纬线弧长Sn
计算球面梯形面积
计算最大角度变形
计算球面极坐标-calculated Meridian circle radius circle conveniently calculated M N calculation Wei radius circle radius r calculated meridian arc Sm calculated latitude arc Sn calculated spherical long trapezoidal area of the greatest deformation angle spherical polar coordinates
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Size: 16124 |
Author: 刘黎 |
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Description: 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
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Size: 878 |
Author: 成龙 |
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Description: 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
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Size: 1024 |
Author: |
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Description: 算法实现题1-2 平面几何问题
.问题描述：
试用C++ 的类来定义表示简单平面几何对象的抽象数据类型Geometry 。
(1) 用浮点数定义2 个几何量m1 和m2；m2 的缺省值为0.0。
(2) 实现2 个构造函数：允许用1 个参数或2 个参数声明几何对象。用1 个参数r 声明
的几何对象为半径为r 的圆；用2 个参数r1 和r2 声明的几何对象为r1′r2 的矩形。所声明
的几何对象的中心在原点。
(3) 定义计算几何对象的面积，周长和对角线长度的成员函数。
(4) 对于给定的实数r，先声明一个半径为r 的圆c1；然后声明一个c1 的外切矩形s，
接着再声明s 的外接圆c2。
(5) 输出(4) 中几何对象的面积，周长和对角线长度。
.编程任务：
给定实数r，计算几何对象c1,s 和c2 的面积，周长和对角线长度。
.数据输入：
由文件input.txt 给出输入数据。第1 行有1 个实数r，表示圆c1 的半径。
.结果输出:
将计算出的几何对象c1,s 和c2 的面积，周长和对角线长度分别输出到文件output.txt 。
文件的第1 行是计算得到的圆c1 的面积和周长；第2 行是矩形s 的面积，周长和对角线长
度；第3 行是圆c2 的面积和周长。-algorithm that 1-2 plane geometry problem. Problem description : probationary category C to the definition simple plane geometry objects to the abstract data type Geometry. (1) The definition of a float with two geometric m1 and m2; M2, the default value of 0.0. (2) to achieve two Constructors : permits the use of a parameter or two parameters statement geometric objects. Using a parameter r statement geometric objects to a radius of the circle r; With two parameters r1 and r2 statement to the geometric objects r1'r2 rectangular. Statement by the geometric objects in the center of origin. (3) calculate the geometric definition of the target area, perimeter and length of the diagonal members function. (4) For a given set of real r, first declare a radius of the circle r c1; Then a state
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Size: 66560 |
Author: stzxxxstz |
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Description: 计算子午圈半径M
计算卯酉圈半径N
计算玮圈半径r
计算经线弧长Sm
计算纬线弧长Sn
计算球面梯形面积
计算最大角度变形
计算球面极坐标-calculated Meridian circle radius circle conveniently calculated M N calculation Wei radius circle radius r calculated meridian arc Sm calculated latitude arc Sn calculated spherical long trapezoidal area of the greatest deformation angle spherical polar coordinates
Platform: |
Size: 15360 |
Author: 刘黎 |
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Description: 圆心(a,b),半径r 圆外一点坐标(m,n) 求其切点的坐标(x,y)-Center of a circle (a, b), outer radius r circle point coordinates (m, n) for its cut-point coordinates (x, y)
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Size: 1024 |
Author: 李智信 |
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Description: linux系统下的文件操作实验程序，完成的功能如下。
在一个半径为R的监测区域，随机分布有M个节点，每个节点的感应半径为r，请完成：
（1）建立节点的位置信息文件
（2）建立网络的拓扑信息文件
-linux operating system files under the experimental procedures, the completion of the function as follows. Radius of R in a monitoring region, a random distribution of M nodes, each node of the sensor radius r, please complete: (1) to establish the location of the node information file (2) the establishment of the network topology information file
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Size: 1024 |
Author: menchael |
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Description: 在一个半径为R的监测区域，随机分布有M个节点，每个节点的感应半径为r，请完成：
（1）建立节点的位置信息文件
（2）建立网络的拓扑信息文件
（3）输出最佳路由即最小生成树
-Radius of R in a monitoring region, a random distribution of M nodes, each node of the sensor radius r, please complete: (1) to establish the location of the node information file (2) the establishment of the network topology information file (3) output the best route that is minimum spanning tree
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Size: 35840 |
Author: zouna |
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Description: DDA的模拟算法。指定的显示窗口的屏幕划分成64X64的区域；每个区域有一个象素，是位于区域中心半径为r的圆。进行DDA直线段绘制时，算法中选中的象素就用窗口中相应的圆代替。-DDA simulation algorithm. The specified display window of the screen is divided into 64X64 areas each region has a pixel is located in the Regional Center radius r circle. DDA to draw a straight line segment, the algorithm used to select the pixel on the window, replaced by the corresponding circle.
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Size: 1024 |
Author: 枵 |
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Description: radius相关rfc（全）+ 华为radius培训pdf-rfc documents about radius protocol and authenticate protocol.
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Size: 1188864 |
Author: redchair |
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Description: 知道圆上两点A(x1,y1),B(x2,y2)和半径r，求出圆心坐标-Know the circle two points A (x1, y1), B (x2, y2) and radius r, find the center of a circle coordinates
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Size: 1024 |
Author: 天河 |
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Description: 从键盘输入两个矩阵,输出这两个矩阵和这两个矩阵相乘的矩阵,图书管理系统的一小部分 要求能 登陆 注册 推出 登陆要有次数限制（失败三次退出程序等），注册信息要求写入到文本文件里保存,已知圆半径r=2.5 圆柱高h=4,求圆的周长，面积，圆球体积，圆柱体积，用cin输入要计算的项目，然后输出计算结果，输入输出时要有文字提示,求1！+2！。。。+10！c语言代码,贪吃蛇游戏的C语言代码,求几个带权字符的哈夫曼编码,本打字游戏可进行英文，其它字符和全部字符的打字训练，每次训练若干个字符并进行打字时间，速度，正确率的统计显示,数值分析的龙贝格求积代码(C语言),以足球,篮球,羽毛球.性别.为内容设置checkbox,raidobutton.并且设置确定跟退出按钮,确定时候弹出消息框显示出你所选的checkbox,raidobutton内容,建立数据库，创建表，进行简单的查询，删除，修改即可，这些事件是点击按钮时发生的,Checkbox ,RADIONBUTTON,的简单事例代码,要求确定返回一些关于这些对象的内容 ,DialogBoxSquareFigureTranscriptExitframe的实例源代码.
-From the keyboard input two matrices, the output of these two matrices and a matrix of these two matrices, library management system, a small portion of requests can log landing should be introduced limiting the number of registration (failed three out of Procedure, etc.), registration information requirements write to a text file saved, known radius r = 2.5 cylindrical high-h = 4, seeking circumference of a circle, area, sphere volume, cylinder volume, with the cin input to calculate the project, and then output the calculation results, input output should be text prompts, seek a!+2! . . .+10! c language code, Snake game C language code, the right to seek a few characters with Huffman encoding, this game can be typed in English, other characters and all characters typing training, each training session and conduct a number of characters typed time, speed, accuracy statistics, numerical analysis of the Romberg quadrature codes (C language), to football, basketball, badminton. sex. for
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Size: 16384 |
Author: wangdongdong |
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Description: 安全可靠的网络互联和基于 VPN 隧道的加密数据传输
华为 Quidway 系列多协议路由器，支持 PPP、FR、X25、SLIP、HDLC 等网络协议，用户可通过各种广域网线路接入 Internet 及互联互访。
路由器内置安全防火墙、NAT、RADIUS 安全认证，以及 CALL BACK 等多项互联安全防护功能；并通过 CA 认证协议的验证和 IPSec 隧道协议，在 Internet 上构建安全可靠的网络通道，基于硬件的数据加密，将保密数据加密后在隧道上传输，加密数据有可靠的保密性但牺牲了效率；非保密数据可以通过一般性安全处理在隧道上传输，保证较高的转发速率。硬件加密的算法采用专门的芯片（经过国家相关部门批准）进行加密，密钥长度大于 128 位，并提供高强度的一次一密功能。配合完全自主知识产权的软硬件平台，提供网络的安全互联和数据的安全传输的安全解决方案。
而 "备份中心"的可靠性备份解决方案，实现多种线路之间的任意备份和业务分担，为网络提供业界最坚固的保护屏障，保证网络的可靠运行。
-主要设备
Quidway TA128 ISDN 适配器
Quidway R1603/1604 路由器
Quidway R2501 路由器
Quidway R2509/2511 路由器
Quidway R3640/4508 模块化路由器
Quidway A8010 Refiner 接入服务器
Quidway S2403 以太网交换机
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Size: 81920 |
Author: 程教育 |
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Description: 题意:已知一个圆的弦长l0及这条弦所在的弧长l1,求弦的中心点到弧的中心点的距离
思想:这是一个列方程然后利用二分法解方程的题目,令该疑弧所对的圆心角为anlg,
半径为r,根据题意有两个方程:l1=anlg*r l0=2*r*sin(anlg/2) 两个方程两个未知数,
通过化简有:2*l1*sin(anlg/2)-anlg*l0=0 因为角度的值是从0到2*pi,题目中讲到过
弧的长度不可能大于弦的两倍,所以角度不可能取到2*pi,但是有可能为0,把零特殊考虑,
再从0到2*pi间二分找解就不会出错了!但是要注意精度问题,取七位小数才能得出正确解.-Italian title: Known l0 chord of a circle and arc length of this string where l1, seeking the center of the arc chord distance of the center of thought: This is a solution out equations and then use the dichotomy of the title equation, so that The arc of the doubt on the central angle for the anlg, radius r, according to the meaning of problems with two equations: l1 = anlg* r l0 = 2* r* sin (anlg/2) two equations two unknowns, by simplification are: 2* l1* sin (anlg/2)-anlg* l0 = 0 because the value of the angle is from 0 to 2* pi, the title referred to had not greater than the arc length of the string twice, So can not get to the point of 2* pi, but there may be 0, to zero special consideration, and then from 0 to 2* pi to find solution between the two points will not be wrong! but to pay attention to accuracy problems, were taken to seven decimal the correct solution.
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Size: 1024 |
Author: yangxiuyi |
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Description: vc2008 dos 程序
已知AB两点坐标，半径R。
求2个圆心。
公式使用MATLAb推导的，做了一点简化，和小的改动。
都在函数 double jisuanYanxin(double Ax,double Ay,double Bx,double By,double R)
里面-vc2008 dos program known AB two coordinates, radius R. Seeking two center of the circle. Formula derived using MATLAb do a little simplification, and small changes. In the function double jisuanYanxin (double Ax, double Ay, double Bx, double By, double R) which
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Size: 20480 |
Author: 王伟 |
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Description: Input: XY(n,2) is the array of coordinates of n points x(i)=XY(i,1), y(i)=XY(i,2)
Output: Par = [a b R] is the fitting circle:
center (a,b) and radius R
Note: this fit does not use built-in matrix functions (except "mean"),
so it can be easily programmed in any programming language-Input: XY (n, 2) is the array of coordinates of n points x (i) = XY (i, 1), y (i) = XY (i, 2) Output: Par = [ab R] is the fitting circle: center (a, b) and radius R Note: this fit does not use built-in matrix functions (except " mean" ), so it can be easily programmed in any programming language
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Size: 1024 |
Author: 田雨 |
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Description: Draws a circle with center at (x1,y1) and radius r
in matlab
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Size: 1024 |
Author: farzan |
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Description: Draws a circle with center at (x1,y1) and radius r
in space
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Size: 2048 |
Author: farzan |
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Description: 编写JAVA程序求园柱体的表面积和体积，已知底面圆心p为（0，0），半径r为10，圆柱体高为5.-JAVA program written request park area and volume of cylinder, known as the bottom center of the circle p (0,0), radius r is 10, a cylinder height of 5.
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Size: 1024 |
Author: william |
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Description: 在windows界面输入圆心坐标（x,y）及半径r任意画圆，并自动生成一个椭圆-The center coordinates of the windows interface of the input (x, y) and radius r arbitrary circle, and automatically generates an ellipse
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Size: 2510848 |
Author: 李倩 |
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