二维码生成和读取

资源描述判断矩阵是否规律，如m x n。判断两个矩阵是否为同阶矩阵，是否为同型矩阵。计算矩阵的转置矩阵。判断两个矩阵是否一模一样。计算两个同型矩阵的乘法、加法、减法。计算数与矩阵的乘法。

数值计算的程序-numerical calculation procedures

The cart with an inverted pendulum, shown below, is "bumped" with an impulse force, F. Determine the dynamic equations of motion for the system, and lin earize about the pendulum s angle, theta = Pi (in other words, assume that p endulum does not move more than a few degrees away from the vertical, chosen to be at an angle of Pi). Find a controller to satisfy all of the design re quirements given below. -The cart with an inverted pendulum, shown below, is "bumped" with an impulse force, F. Determine the dynamic equations of motion for the system, and lin earize about the pendulum s angle, theta = Pi (in other words, assume that p endulum does not move more than a few degrees away from the vertical, chosen to be at an angle of Pi). Find a controller to satisfy all of the design re quirements given below.

c/c++爱好者学习的一个很好的例子。对于提高初学者的水平，开拓思维有很大的帮助-c/c++ enthusiasts to learn a good example. For beginners improve the level of development thinking has a great help

卷积码的编译码，信道为AWGN信道，BPSK调制，码率为1/2，译码为viterbi硬判决，并画出了BER的曲线

Dijkstra的推广之一： 最短路径的子路径是最短路径。证明:(算法导论P358) 次短路径的子路径是最短路或次短路.(这个是次短路径的最优子结构). 证明： 请点击左侧文件开始预览 ！预览只提供20%的代码片段，完整代码需下载后查看 加载中 侵权举报

样条曲线拟合,以简单的有理样条曲线的拟合方式，最后根据算法原理，最终需要实现NURBS曲线的生成及反算

牛顿-拉格朗日插值法C++源代码，最新修改的通码，网友可以-Newton-Lagrange interpolation C source code, the latest revision of immigration code, net surfers can s

这是代码来实现多径 dsr.this 使用 DSR protocol.there 是使用 multipath.we 使用 3 多径基于的纸张的路由协议实现的方法。

该代码实现了am，f-msk，qpsk，mdpsk-the code achieved am, f-msk, qpsk, mdpsk