rfidreader

集合的运算：交、并、补（难度系数：1.1） 全集：大写字母 ‘A’~’Z’ 要求实现以下功能： 1、集合的输入：自动去掉重复和非法的字符 2、集合的显示：输出集合的全部元素 3、输出一个给定集合的补集 4、输出两个给定集合的交集和并集 (jiao bing bu)

上传程序用于解析Novatel输出range 信息，解析结果已经验证。(The upload process is used to resolve the Novatel output range information, analytical results have been verified.)

基于Visual C++ 的win32控制台的程序，实现使用定时器的功能，(Visual C++ win32 console-based program, using the timer function,)

this is the library to compress and decompress any file. I found it on Net

求解符号三角形问题.给定一行符号(比如+和-),按照同号为+异号为-的原则计算第二行直至下面的行,最后的一行只有一个符号,要求统计各个符号的个数.(Problem solving triangle symbols. Given his symbols (such as+ And-), in accordance with the same number is+ Different number- the principle of calculation of the second line until the following line, the last line there is only one symbol, all symbols requested statistics number.)

to check the user has the administration rights is more complicated in windows 7, this is the sample to check it.

设X[ 0 : n - 1]和Y[ 0 : n – 1 ]为两个数组，每个数组中含有n个已排好序的数。找出X和Y的2n个数的中位数。  编程任务 利用分治策略试设计一个O (log n)时间的算法求出这2n个数的中位数。 数据输入 由文件input.txt提供输入数据。文件的第1行中有1个正整数n（n<=200），表示每个数组有n个数。接下来的两行分别是X，Y数组的元素。结果输出 程序运行结束时，将计算出的中位数输出到文件output.txt中(Let X [0: n- 1] and Y [0: n- 1] for the two arrays, each array containing the n number has been sorted. 2n X and Y to identify the number of digits.  programming tasks using the divide and conquer strategy try to design an O (log n) time algorithm to calculate this median number 2n. Data input by the input data provided input.txt file. The first line in the file has a positive integer n (n < = 200), that there are n numbers of each array. The next two lines are the X, Y array elements. The end result is output program runs, the calculated median output to file output.txt)

电位移控制台介绍和详细说明介绍了各种型号和实验(The electric displacement console and detailed description of the various models and experimental)

console application mineswiper

使用哈弗曼编码技术进行数据压缩和解压缩，缺点在于只能用于文本(Havermann coding techniques using data compression and decompression, disadvantage is that the text can only be used)