Unit1

说明：  酒店管理，下载解压就可以，不是源码，是已经做好的成品压缩文件(Hotel management, download and decompress, not source code, but finished product compressed file)

用STL实现了贪吃蛇的基本功能，算法比较精练，有参考价值-Using STL to achieve the basic functions of the Snake, concise algorithm, have reference value

宠物管理系统,就使用简单的java面向对象编程!(Pet management system)

SPI connection of STM32f407

介绍如何用Labview与视觉助手还有VDM中的函数进行图像处理程序的开发(How to use Labview, visual assistant and VDM functions to develop image processing procedures)

说明：  有关copula函数的代码，里面有常用的阿基米德copula函数和高斯，t copula(Code about the copula function, which has the commonly used Archimedes copula function and Gauss, t copula)

矩阵导致小程序，对输入的矩阵求倒置后的矩阵-Matrix led to a small program, the input of the matrix after the matrix inversion

说明：  修改ET199锁的客户号和ATR值，适用于模拟的ET199锁。(Modify the customer number and ATR of ET199 lock)

很好用的声源定位资料，用labview开发，教你怎么入门，怎么使用labview，以及定位的方法(Good use of sound localization information, with labview development, teach you how to get started, how to use labview, and positioning method)

算法实现题 最长公共子序列问题 给定两个序列 X = { x1 , x2 , ... , xm } Y = { y1 , y2 , ... , yn } 求X和Y的一个最长公共子序列 举例 X = { a , b , c , b , d , a , b } Y = { b , d , c , a , b , a } 最长公共子序列为 LSC = { b , c , b , a }-Algorithm to achieve the longest common sub-title sequence of the problem given two sequences X = (x1, x2, ..., xm) Y = (y1, y2, ..., yn) for X and Y, a maximum common substructure sequence, for example X = (a, b, c, b, d, a, b) Y = (b, d, c, a, b, a) the longest common sub-sequence for LSC = (b, c, b, a )