sokudo 算法

Freescale MC13211的无线收发程序(Freescale MC13211 wireless transceiver)

回声消除本质上就是把输出信号和它产生的回声信号之间建立一个回声数学模型，利用开始的数据训练这个模型的参数。在远端有说话，但近端没有说话的时候，录音应该是静音，即回声完全消除。所以算法就朝着这个方向努力，一旦回声为0，则滤波器收敛。自适应滤波器算法多种多样，但是目前流行的还是最经典的LMS和NLMS，NLMS是LMS的优化 。 判断标准：收敛速度快，运算复杂度低，稳定性好，失调误差小

ST2.0库的FOC程序，现在已经精简了，用起来更加方便和容易理解。(ST2.0 library FOC procedures, streamline now, it is more convenient to use and easy to understand)

This a ddk filter sample for keyboard

ad7812采集程序源码10位的，运用c语言言开发的，环境是KEIL 51。 可直接使用。 (The ad7812 acquisition program source 10, developed using c language words, the environment is the KEIL 51. Can be used directly.)

C# 串口调试 C# 串口调试 C# 串口调试 实现收发(C# serial debugging C# serial debugging C# serial debugging transceiver)

数据库课程设计（完整版）报刊订阅系统 付：代码，文档，数据库(Database Course Design (full version) press Subscribe Systems to pay: code, documentation, database)

DEM数据读取显示 能读取各种DEM数据(DEM data can be read read various shows DEM data)

自建词典的开发界面和底层数据库操作整个包上载(The development interface of the self-built dictionary and the operation of the underlying database for the whole package upload)

说明：  C++源代码，递归实现。原问题描述：鼓上蚤时迁到一个山洞里去盗宝，里面有n件宝贝，价值分别是a1,a2......an，并且满足ai大于a1+a2+....+a(i-1)的和。他身上有一个背包，能装的东西最大价值是K，他只能进洞一次，请问他最多能带出多少？ 要求：输入K，n，a1，a2......an 输出：他能最多带出的宝物价值 (C++ Source code Recursive achieve. The original problem description: drum when flea盗宝going to move in a cave, there are n pieces of treasure, valued are a1, a2 ...... an, and to meet ai than a1+ A22B !....+ a (i-1) and. Him to have a backpack, things can hold the greatest value is K, he could only hole once, to ask him to bring out the maximum number of? Requirements: Enter K, n, a1, a2 ...... an output: he can bring out the maximum value of the treasures)