Location:
Search - X Y Z
Search list
Description: 重力感应，三个方向，x，y，z
不同方向不同值-Gravity sensing, three directions, x, y, z different values in different directions
Platform: |
Size: 133120 |
Author: 周飞 |
Hits:
Description: 输入三个整数x,y,z，请把这三个数由小到大输出。
程序分析：我们想办法把最小的数放到x上，先将x与y进行比较，如果x>y则将x与y的值进行交换，
然后再用x与z进行比较，如果x>z则将x与z的值进行交换，这样能使x最小。-Enter three integers x, y, z, the number of small to large output of these three please. Program analysis: we find a way to put the smallest number x, the first to compare x and y, if x> y x and y values will be exchanged, and then compare the x and z, if x> z the value of x and z will be exchanged, so make x minimum.
Platform: |
Size: 2048 |
Author: liumeihong |
Hits:
Description: 单表密码
定义：
1、 明表：
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
0 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
2、 明文
they will arrive tomorrow
3、 密钥
K=Monday
4、 密码实现
1） 将明文与密钥转为数字串：
K=(12,14,13,3,0,24)
M=(19,7,4,24,22,8,11,11,0,17,17,8,21,4,19,14,12,14,17,17,14,22,)
2）将明文数字串依密钥长度分段，并逐一与密钥数字相加（模26），得到密文数字串。
19 7 4 24 22 8…
12 14 13 3 0 24…
5 21 17 1 22 6
C=(5，21，17，1，22，6，23，25，13，20 。。。)
3） 密文数字串转换为字母串
c=（FVRBWG XZNURG HSGRMM DFBZ）
5、 解密：模26减运算
1） C=(5，21，17，1，22，6，23，25，13，20 。。。)
2）模26减运算
5 21 17 1 22 6
— 12 14 13 3 0 24…
-7 7 4 -2 22 -18
+ 26 26 26 26 26 26
M=（19 7 4 24 22 8…）
3）将M转换为字符串，得到明文。
they will arrive tomorrow-Single table code
Platform: |
Size: 4096 |
Author: lss |
Hits:
Description: 输入三个整数x,y,z，请把这三个数由小到大输出。
1.程序分析：我们想办法把最小的数放到x上，先将x与y进行比较，如果x>y则将
x与y的值进行交换， 然后再用x与z进行比较，如果x>z则将x与z的
值进行交换，这样能使x最小。-Enter three integers x, y, z, the number of small to large output of these three please. 1 program analysis: we find a way to put the smallest number x, the first to compare x and y, if x> y x and y values will be exchanged, and then compare the x and z, if x > z x and z values will be exchanged, so make the x minimum.
Platform: |
Size: 1024 |
Author: Joanna99 |
Hits:
Description: 在已经知道X,Y,Z轴坐标，画出该些点立体图。-Already know that X, Y, Z axis, draw some point in the three-dimensional map.
Platform: |
Size: 1024 |
Author: 邱旭东 |
Hits:
Description: 超声波空间定位程序 设定，X,Y,Z坐标用NAKIO5110屏实时显示 物体的坐标-Ultrasonic spatial orientation program set, X, Y, Z coordinates of the object with NAKIO5110 real-time display screen coordinates
Platform: |
Size: 55296 |
Author: 孔斌 |
Hits:
Description: the tranformation from X,Y Z to geodetic latitude and longitude
Platform: |
Size: 2048 |
Author: dingzbxx |
Hits:
Description: 可通过拖拽鼠标旋转坐标轴，有x,y,z三个坐标轴，并且能显示坐标轴的名字-By dragging the mouse to rotate the axes, there are x, y, z three axes, and can display the name of the axis
Platform: |
Size: 2033664 |
Author: 林顶顶 |
Hits:
Description: 曲面位场处理。1、坐标单位为m。
2、观测面上的x,y,z坐标以及重力异常存放在gravity.dat中。第一列为x坐标（向东），第二列为y坐标（向北），第三列为z坐标（铅垂向下），第四列为重力异常（g.u.）。
3、计算面坐标存放在xyz.dat中。第一列为x坐标（向东），第二列为y坐标（向北），第三列为z坐标（铅垂向下）。
4、实验要求：利用空间域等效源法（点质量）得到计算面的重力异常。
-curved surface process
Platform: |
Size: 20480 |
Author: 叶青 |
Hits:
Description: 345加速度传感器，x,y,z三轴数据准确读取，并显示在1602液晶上-345 acceleration sensors, x, y, z three axis data accurate read, and displayed in 1602 LCD
Platform: |
Size: 32768 |
Author: 陈泽宴 |
Hits:
Description: Writes a 3D (x,y,z,u,v,w) vector field to a binary VTK file for Paraview.
Platform: |
Size: 2048 |
Author: lsx |
Hits:
Description: Assignment 4:
1. Analyze and simulate the following code lists (code1 and code 2) with the same input signals shown below by presenting POW and OL. If the data type of “a, b, c, d, u, v, w, x, y, z” is declared as std_logic, what will the simulation outputs be changed?
Platform: |
Size: 172032 |
Author: 魏攸 |
Hits:
Description: 输入三个整数x,y,z，请把这三个数由小到大输出。
1.程序分析：我们想办法把最小的数放到x上，先将x与y进行比较，如果x>y则将x与y的值进行交换，
然后再用x与z进行比较，如果x>z则将x与z的值进行交换，这样能使x最小。-Enter three integers x, y, z, please these three the number of small to large output.
A program analysis: we want to approach the smallest number on the x and compare the first x and y, if x> y then x and y values are exchanged,
Then compare the x and z, for x> z then x and z values to be exchanged, so make x the smallest.
Platform: |
Size: 1024 |
Author: 小神经 |
Hits:
Description: java实现10个数的全排列的应用
给定等式 A B C D E 其中每个字母代表一个数字，且不同数字对应不
D F G 同字母。编程求出这些数字并且打出这个数字的
+ D F G 算术计算竖式。 ─────── 语句； X Y Z D E-The number 10 Java realize the application of the arrangement
Platform: |
Size: 29696 |
Author: 宋艳超 |
Hits:
Description: 生成100个点的坐标（x,y,z），x和y在10-10000之内，z在0-10之内，用TXT输出x，y，z以及平均高程h。-Generate 100 points of coordinates (x, y, z), x and y in 10-10000, z 0-10, TXT output x, y, and z and the average elevation h.
Platform: |
Size: 236544 |
Author: t |
Hits:
Description: 个人原创代码：Point2D和Point3D作图。编写Java Application程序，分别编写两个类Point2D，Point3D来表示二维空间和三维空间的点，使之满足下列要求：
(1) Point2D有两个整型成员变量x, y (分别为二维空间的X,Y方向坐标)，Point2D的构造方法要实现对其成员变量x, y的初始化。
(2)Point2D有一个void型成员方法offset(int a, int b)，它可以实现Point2D的平移。
(3)Point3D是Point2D的直接子类，它有有三个整型成员变量x,y,z (分别为三维空间的X,Y,Z方向坐标)，Point3D有两个构造方法：Point3D(int x,int y,int z)和Point3D(Point2D p,int z)，两者均可实现对Point3D的成员变量x, y,z的初始化。
(4)Point3D有一个void型成员方法offset(int a, int b,int c)，该方法可以实现Point3D的平移。
(5)在Point3D中的主函数main()中实例化两个Point2D的对象p2d1，p2d2，打印出它们之间的距离，再实例化两个Point2D的对象p3d1，p3d2，打印出他们之间的距离。-Personal original code: Point2D and Point3D mapping. Write a java application program, namely the preparation of two classes Point2D, the Point3D to the point of the two-dimensional space and three-dimensional space, to meet the following requirements:
(1) Point2D two integer member variables x and y (respectively for the two-dimensional space X, the Y-direction coordinate), the construction method in Point2D to achieve its member variables x, y initialization.
(2) Point2D a void member offset (int a, int b), it can Point2D translational.
(3) the Point3D is a direct subclass of Point2D, it has three integer member variables x, y, and z (respectively, three-dimensional space X, Y, Z direction coordinate), Point3D has two constructors: the Point3D (int x int y, int z) and the Point3D (Point2D p-int z), both of which can Point3D member variable x, y, and z initialization.
(4) the Point3D a void member offset (int a, int b, int c), the method can achieve the Point3D the pan.
(5) the P
Platform: |
Size: 1024 |
Author: 李瑞 |
Hits:
Description: 个人原创代码：Java异常处理程序。编写Java Application程序，自定义类Sanj，其中有成员 x,y,z,作为三边长，构造方法Sanj(a,b,c)分别给x,y,z赋值,方法求面积getArea和显示三角形信息(三个边长)showInfo，这2个方法中当三条边不能构成一个三角形时要抛出自定义异常NotSanjiaoException，否则显示正确信息。在另外一个类中的主方法中构造一个Sanj对象(三边为命令行输入的三个整数)，显示三角形信息和面积，要求捕获异常。-Personal original code: Java exception handling procedures.Write a Java application program, Sanj custom classes, including members of the x, y, and z, as long, triangular and assignment constructor Sanj (a, b, and c), respectively, to x, y, z, and the area the getArea and display a triangle(three side length) showInfo these two methods when the three sides can not form a triangle to throw self-defined exceptions NotSanjiaoException, or display the correct information. The Sanj object (triangular command line input three integers) a method to construct a display triangle and the area required to capture the exception.
Platform: |
Size: 2048 |
Author: 李瑞 |
Hits:
Description: 是实现一个漫游功能，当按下left键后向左转，right向右转，up前进，down后退，page up向上仰。g_Angle,rad_xz的初始值均为0。g_eye[0],g_eye[1],g_eye[2],分别代表眼睛的x,y,z g_look是参考点-Is to implement a roaming, when you press the left button, left turn, right turn to the right, up forward, down back page up up the belief. g_Angle rad_xz the initial value is 0. g_eye [0], g_eye [1], g_eye [2], representing the eye x, y, and z is the reference point g_look
撤消修改
新! 点击上方的字词即可查看其他翻译。 删除
字典
Platform: |
Size: 1781760 |
Author: 韩婷 |
Hits:
Description: function [] = circleagain(a,b,c,r)
pixel = 0.1
theta1 = 0
theta2 = 360*pi/180
pix = pixel/r
theta = theta1:pix:theta2
global x y z
x = a + r*cos(theta)
y = b + r*sin(theta)
z = ones(1,length(x))*c
x=round(x*10)/10
y=round(y*10)/10
z=round(z*10)/10
plot3(x,y,z, c )
Platform: |
Size: 378880 |
Author: boom |
Hits:
Description: 从笛卡尔坐标系到地理坐标系的转换，输入X，Y, Z，输出phi,lambda 和 h-conversion of Cartesian coordinates (X,Y,Z) to geographical coordinates (phi, lambda, h)
Platform: |
Size: 1024 |
Author: Angela Zhang |
Hits:
«
1
2
...
5
6
7
8
9
1011
12
13
14
15
...
37
»