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[Graph Drawingfigure

Description: 有100个数据点,知道x,y,z的坐标,用命令把这些点拟合成空间的曲线?看点的分布应该是螺旋线,得出一个表达式来表达曲线,有程序-Has 100 data points, know that x, y, z coordinates, using the order of these points to be synthetic curve space? Highlight the distribution should be a spiral line, derived an expression to express the curve, the procedure has
Platform: | Size: 1024 | Author: hujik | Hits:

[SCMsensor

Description: MC9S08AW60的3轴加速度传感器实验 本实验利用MCU内部ADC不停的采集3轴加速度传感器的X,Y,Z方向的输出-MC9S08AW60 the 3-axis acceleration sensor experiments in this experiment using the internal ADC clock MCU collection of 3-axis acceleration sensor X, Y, Z direction output
Platform: | Size: 51200 | Author: 李鹏 | Hits:

[ARM-PowerPC-ColdFire-MIPSARM28

Description: 替代加密: A B C D E F G H I J K L M N O P Q R S T U V W 密文 Y Z D M R N H X J L I O Q U W A C B E G F K P 明文 X Y Z T S V I HAVE A DREAM!# 密文?? 用ARM编程实现替代加密。-Alternative encryption: ABCDEFGHIJKLMNOPQRSTU VW ciphertext YZDMRNHXJLIOQUWACBEGF KP expressly XYZTSVI HAVE A DREAM!# Ciphertext? ? Programming using ARM alternative encryption.
Platform: | Size: 18432 | Author: 富洋 | Hits:

[Software EngineeringMatrixBible

Description: OpenGL中的各种转换是通过矩阵运算实现的,具体的说,就是当发出一个转换命令时,该命令会生成一个4X4阶的转换矩阵(OpenGL中的物体坐标一律采用齐次坐标,即(x, y, z, w),故所有变换矩阵都采用4X4矩阵),当前矩阵与这个转换矩阵相乘,从而生成新的当前矩阵。例如,对于顶点坐标v ,转换命令通常在顶点坐标命令之前发出,若当前矩阵为C,转换命令构成的矩阵为M,则发出转换命令后,生成的新的当前矩阵为CM,这个矩阵再乘以顶点坐标v,从而构成新的顶点坐标CMv。上述过程说明,程序中绘制顶点前的最后一个变换命令最先作用于顶点之上。这同时也说明,OpenGL编程中,实际的变换顺序与指定的顺序是相反的。文档对其进行了详细的分析。
Platform: | Size: 17408 | Author: | Hits:

[CSharpRS232Comm

Description: 采集X,Y,Z坐标系显示位移值,数据来源与位移传感器采集的信号,经MPU处理,经RS-232接口传送与电脑-Acquisition X, Y, Z coordinate system shows displacement values, data sources and the displacement sensor signal acquisition, processing by the MPU via RS-232 interface to send and computer
Platform: | Size: 62464 | Author: 黄泽俭 | Hits:

[assembly languagehuibianshiyan1

Description: 汇编语言程序上机过程DEBUG程序的使用 1.编制程序计算Z=X+Y,其中X、Y、Z均为字型无符号数。2.将数据区中的100个‘A’字符组成的字符串传送到附加段中。-Assembly language program on the machine the process of using one of the DEBUG program. Programming calculation Z = X+ Y, which X, Y, Z are the number of fonts unsigned. 2. Data area of 100
Platform: | Size: 38912 | Author: 卢雪 | Hits:

[Compress-Decompress algrithmssmd380_program

Description: 三轴加速度传感器读数据,计算x,y,z,三个倾角度数,可以测速度。-Three-axis acceleration sensor data to calculate the x, y, z, three degrees inclination, you can test the speed.
Platform: | Size: 26624 | Author: 徐贤 | Hits:

[matlabpca

Description: 在MATLAB上所使用的PCA程序,主要應用於過濾相對較不重要的特徵值(dimension),例如在三度空間的某些點具有(x,y,z)值,因為這些點有共同的一個持徵,就是z值相對於x,y值來得小很多(不明顯),所以就以X,Y軸來表示這些點。此時就達成去除掉z的特徵值(dimension)。 -MATLAB used in the PCA procedure, mainly used in filtering relatively unimportant eigenvalue (dimension), for example, in some points with three-dimensional (x, y, z) values, because these points have a common one sign holders, that is, z values compared with the x, y values much smaller than (not obvious), so on to X, Y axis to express these points. Reached at this time to get rid of z eigenvalues (dimension).
Platform: | Size: 3072 | Author: iceryu01 | Hits:

[OtherC4

Description: 两个乒乓球队进行比赛,各出3人。甲队为a,b,c三人,乙队为x,y,z三人,已抽签决定比赛名单。有人向队员打听比赛的名单,a说他不和x比,c说他不和x,z比,请编程序找出三队赛手的名单。-Two table tennis team competition, each person 3. A team for a, b, c three B teams for the x, y, z three, has decided to match the list of drawing. It was asked to match the list of members, a said that he did not and the x ratio, c said that he did not, and x, z ratio, the procedure is requested for three teams racing to find the list.
Platform: | Size: 1024 | Author: likai | Hits:

[Other4

Description: 输入某年某月某日,判断是不是闰年? 输入三个整数x,y,z,请把这三个数由小到大输出。 -Importation of a certain period of a day, to judge is not a leap year? Enter three integers x, y, z, invited these three the number of small to large output.
Platform: | Size: 1024 | Author: sunyan | Hits:

[Windows Developplotdata

Description: 绘散点图,x,y,z,x和y确定坐标, z确定颜色级别-Painted scatter, x, y, z, x and y determine the coordinates, z-level to determine color
Platform: | Size: 3665920 | Author: xiaozhao | Hits:

[Windows Developkuaisufuzhi

Description: entity/ent(1000),obj(1000) number/mat(12) l1: ident/ select objects to transform ,ent,cnt,num,rsp jump/l1:,trm:,,rsp l2: gpos/ select reference point ,x1,y1,z1,rsp jump/l1:,trm:,,rsp l3: gpos/ select next point ,x2,y2,z2,rsp jump/l2:,trm:,,rsp l4: x=x2-x1 y=y2-y1 z=z2-z1 mat=matrix/transl,x,y,z obj=transf/mat,ent(1..num) jump/l3: trm: halt-entity/ent (1000), obj (1000) number/mat (12) l1: ident/select objects to transform, ent, cnt, num, rspjump/l1:, trm:,, rspl2: gpos/select reference point, x1 , y1, z1, rspjump/l1:, trm:,, rspl3: gpos/select next point, x2, y2, z2, rspjump/l2:, trm:,, rspl4: x = x2-x1y = y2-y1z = z2-z1mat = matrix/transl, x, y, zobj = transf/mat, ent (1 .. num) jump/l3: trm: halt
Platform: | Size: 2048 | Author: 云飞 | Hits:

[Linux-UnixfirstGTK

Description: 一个基于GTK+的单词数值计算器,1、 按照规则计算单词的值,如果 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 26个字母(全部用大写)的值分别为 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26,如: WINJACK这个单词的值就为:W+I+N+J+A+C+K=23+9+14+1+3+11=71% HARDWORK=H+A+R+D+W+O+R+D=8+1+18+4+23+15+18+11=98% LOVE=L+O+V+E=12+15+22+5=54% LUCK=L+U+C+K=12+21+3+11=47% ATTITUDE= A+T+T+I+T+U+D+E=1+20+20+9+20+24+4+5=100% 2、对程序的界面布局参考如下图所示,在第一个单行文本框输入一个单词,点击“计算”按钮,按照以上算法计算出该单词的值。 3、如果在最下面的单行文本框输入一个文件路径,此文件每行记录一个单词,那么经过程序计算出各个单词的值,并把结果输出到当前目录下result.txt文件中。如果文件不存在,应该提示错误。 -err
Platform: | Size: 1024 | Author: | Hits:

[Algorithmgenetic_algorithm

Description: 基于求函数f(x,y,z)=xyz*sin(xyz)最大值问题的演示程序-Based on the demand function f (x, y, z) = xyz* sin (xyz) max demo issues
Platform: | Size: 7441408 | Author: lty | Hits:

[matlab4dofrobot

Description: 四自由度的简易机器人(转载),通过输入四个关节的旋转角度,显示出实体在X、Y、Z三轴上的投影,并可计算出位置坐标-Simple robot four degrees of freedom (reprint), by entering the four joint rotation angle, showing that the entity in the X, Y, Z axis on the projection, and calculate the location coordinates
Platform: | Size: 44032 | Author: yn | Hits:

[JSP/Java11

Description: 1. 在No.1图形窗口中绘制 y=sin(x)在[0,2*pi]内的曲线。要求曲线的颜色为绿色,线型为 点划线,用*标示坐标点,在x轴的附近用 黑体 标注 ‘x轴’字样,在图形的上方加上标题 ‘正弦函数’,严格控制x,y轴分度相等,并开启网格。 2. 在No.2图形窗口中创建四个子窗口,在第一、二子窗口中用不同的方法同时绘制 y=x^2,y=-x^2,y=x^2*sin(x) 在[0,2*pi]内的曲线,并要给出标注 在第三个子窗口中绘制 三维曲线 3. 把No.3图形窗口分成五个子窗口,分别用plot3 mesh meshc meshz surf 来绘制 z=x*exp(-x^2-y^2) 在 -5=<x,y<=5 内的空间曲面图形,说明他们的区别,其中要求在用surf绘制的窗口内加入位置为[1,0.5,2]的光源,加入颜色标尺,采用spring色系 -1. 在No.1图形窗口中绘制 y=sin(x)在[0,2*pi]内的曲线。要求曲线的颜色为绿色,线型为 点划线,用*标示坐标点,在x轴的附近用 黑体 标注 ‘x轴’字样,在图形的上方加上标题 ‘正弦函数’,严格控制x,y轴分度相等,并开启网格。 2. 在No.2图形窗口中创建四个子窗口,在第一、二子窗口中用不同的方法同时绘制 y=x^2,y=-x^2,y=x^2*sin(x) 在[0,2*pi]内的曲线,并要给出标注 在第三个子窗口中绘制 三维曲线 3. 把No.3图形窗口分成五个子窗口,分别用plot3 mesh meshc meshz surf 来绘制 z=x*exp(-x^2-y^2) 在-5=<x,y<=5 内的空间曲面图形,说明他们的区别,其中要求在用surf绘制的窗口内加入位置为[1,0.5,2]的光源,加入颜色标尺,采用spring色系
Platform: | Size: 1024 | Author: 李子木 | Hits:

[Windows Developencryption

Description: The program takes 3 inputs: 1. A letter of the alphabet that will remain unencoded (e.g. "J") 2. A 5-letter keyword (e.g. "BREAK") 3. A message to be encrypted (e.g. "COMPUTERSCIENCE"). You may assume that it does not contain any spaces or punctuation. To form the code, first create a 5x5 matrix in which the first row is your keyword and the other elements are the remaining letters of the alphabet (minus the unencoded letter): B R E A K C D F G H I L M N O P Q S T U V W X Y Z -Encryption code
Platform: | Size: 1024 | Author: 王梓 | Hits:

[JSP/Javagenetic_algorithm

Description: 基于求函数f(x,y,z)=xyz*sin(xyz)最大值问题的演示程序 解压后在命令行输入:java -jar genetic_algorithm.jar-Based on the demand function f (x, y, z) = xyz* sin (xyz) Maximum problem after decompression of the demo program at the command line, type: java-jar genetic_algorithm.jar
Platform: | Size: 7460864 | Author: wrgdggfgfg | Hits:

[Data structsHaffmancode

Description: 课程设计: 1.求出在一个n×n的棋盘上,放置n个不能互相捕捉的国际象棋“皇后”的所有布局。 2.设计一个利用哈夫曼算法的编码和译码系统,重复地显示并处理以下项目,直到选择退出为止。 【基本要求】 1) 将权值数据存放在数据文件(文件名为data.txt,位于执行程序的当前目录中) 2) 分别采用动态和静态存储结构 3) 初始化:键盘输入字符集大小n、n个字符和n个权值,建立哈夫曼树; 4) 编码:利用建好的哈夫曼树生成哈夫曼编码; 5) 输出编码; 6) 设字符集及频度如下表: 字符 空格 A B C D E F G H I J K L M 频度 186 64 13 22 32 103 21 15 47 57 1 5 32 20 字符 N O P Q R S T U V W X Y Z 频度 57 63 15 1 48 51 80 23 8 18 1 16 1 -Curriculum design: 1. Obtained in an n × n chessboard, the place to catch each other should not n个chess "Queen" of all the layout. 2. The design of a use of Huffman coding and decoding algorithms systems, and deal with duplicate to show the following items until the exit date selection. The basic requirements 【】 1) will be the right value data stored in data files (file named data.txt, located in the implementation of procedures in the current directory) 2), respectively, dynamic and static storage structure 3) Initialization: keyboard input character set size of n, n and n characters of the right value, set up Huffman tree 4) Coding: Using the built Huffman tree generated Huffman coding 5) output coding 6) The character set and the frequency of the following table: Space characters A B C D E F G H I J K L M Frequency of 186 64 13 22 32 103 21 15 47 57 1 5 32 20 Character N O P Q R S T U V W X Y Z Frequency 57 63 15 1 48 51 80 23 8 18 1 16 1
Platform: | Size: 550912 | Author: 赵刚 | Hits:

[Algorithmmath

Description: 同一坐标系统下坐标有多种不同的表现形 式,一种形式实际上就是一种坐标系。如空间直 角坐标系( , , )、大地坐标系( , )、 平面直角坐标( , )等。通过坐标统的转换我 们得到了BJ54坐标系统下的空间直角坐标,我们 还须在BJ54坐标系统下再进行各种坐标系的转 换,直至得到工程所需的坐标。 1)将空间直角坐标系转换成大地坐标系,得-Sub btnCal_OnClick dim SpaceForm dim X, Y, Z, B, L, H, Elli, i, A_earth, E_earth set SpaceForm=document.Space for i = 0 to 2 if SpaceForm.optElli(i).checked Then Elli = SpaceForm.optElli(i).value
Platform: | Size: 16384 | Author: 天使 | Hits:
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