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[2D GraphicARC

Description: 逐点圆弧插补编程 用于数控中的 可实现在某一象限的圆弧插补 -Point by point circular interpolation for CNC programming can be achieved in a certain quadrant of circular interpolation
Platform: | Size: 1024 | Author: 孙来东 | Hits:

[androidQuadrant.apk

Description: Android手机推出的性能测试软件《象限》(Quadrant)是Google应用商店中评价高达4.5星的性能测试软件,针对CPU、内存、I/O输入输出、2D及3D图像的性能,提供了一键式的完整测试,或是根据需要选择其中某些测试项目单独测试。 -android benchmark tool
Platform: | Size: 852992 | Author: 李雷 | Hits:

[AlgorithmCreateMagicMatrix

Description: 《洛书》增强版算法解密 包括偶数阶、双偶数阶、单偶数阶等各类魔方(幻方)的生成算法源码。 其中单偶数阶最难: 单偶数魔方,n为偶数,且不能被4整除(n=6,10,14,18,22……)(n=4k+2,k=1,2,3,4,5……)。这是奇数阶魔方,双偶数阶魔方,单偶数阶魔方三种魔方里面最复杂的魔方。 以n=10为例,这时k=2。 (1)把方阵从左到右,从上到下分成A,B,C,D四个象限,这样每个象限肯定是奇数阶,用罗伯法,依次在A,D,B,C象限按奇数阶魔方的填法填数。 (2)在A象限的中间行中间列开始,按自左向右的方向标出k列,A象限的其他行则标出最左边的k列,将这些标出的数和C象限相同位置上的数互换位置。 (3)在B象限任意一行的中间列,自左向右标出k-1列,将这些标出的数和D象限相同位置上的数互换位置。魔方完成。-" Luo Shu" decryption algorithm enhancements include even order, double even order, even order, and other single cube (magic square) generation algorithm source code. Wherein the single most difficult to even order: single even Cube, n is an even number, and can not be divisible by 4 (n = 6,10,14,18,22 ......) (n = 4k+2, k = 1,2,3,4, 5 ......). This is the odd-order cube, double even order cube, Rubik three kinds of even order single most complex Rubik cube inside. Case with n = 10, then k = 2. (1) the square left to right, top to bottom is divided into A, B, C, D four quadrants, and each quadrant is certainly odd order, with Rob law, followed by the A, D, B, C quadrant press Rubik' s odd order number fill fill method. (2) A quadrant in the middle of the middle row of columns, beginning with the left-right directions marked k columns, other rows A quadrant is marked k leftmost column, the same number and location of these marked C quadrant Number of swap positions. (3) in
Platform: | Size: 2048 | Author: zngsai | Hits:

[Otherzhixian

Description: 四象限逐点比较法直线插补matlab仿真程序源代码。输入起点、终点、给进步长等即可实现仿真。-Quadrant point by point comparison Linear Interpolation matlab simulation program source code. Enter the start, end, and so long to progress to achieve the simulation.
Platform: | Size: 1024 | Author: 梁生 | Hits:

[Otheryuanhu

Description: 四象限逐点比较法圆弧插补matlab仿真程序源代码。输入起点、终点、给进步长、顺时针或逆时针等即可实现仿真。-Quadrant point by point comparison circular interpolation matlab simulation program source code. Enter the start and end points, to progress long, clockwise or counterclockwise, etc. can be realized simulation.
Platform: | Size: 1024 | Author: 梁生 | Hits:

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