NCRE all titles, including the title, testing tools, exercises, on topics such a...

labview树形控件的使用，树形控件内容添加和显示-LabVIEW controls the use of tree, tree controls to add and display content

一个可以锁定系统三组热键 ctrl+esc alt + F4 ctrl + alt + del的代码-can lock a system hotkey three groups esc ctrl alt F4 ctrl alt del code

VC++创建定时器-VC create timer

具有基本的加、减、乘、除算术运算能力，可带有计算函数的扩充功能-Have a basic add, subtract, multiply, with the exception of arithmetic ability to function can be calculated with the expansion of functions

适合作为Struts框架入门练习的案例。数据库表包含最为常见的员工表和用户表。方便代码的阅读和理解。基本的增删改查操作均通过了测试，基本功能已经完善。

本源码实现了树这个数据结构。可以直接拿来用-Realize the source of the data structure of the tree. Can be directly used by

一个WIN32命令行分析程序的源代码.-a command-line analysis of a program"s source code.

汉诺塔问题用c++解决，这是一个非常经典的问题，如何实现递归。-HANOR c solve the problem with this is a very classic problem, how to achieve recursive.

i2c通信模拟器。模拟一个内存芯片到i2c的eeprom估计，该芯片的实现是可能的，并写入eeprom

编写一个求解Josephus问题的函数。用整数序列1, 2, 3, ……, n表示顺序围坐在圆桌周围的人。然后使用n = 9, s = 1, m = 5，以及n = 9, s = 1, m = 0，或者n = 9, s = 1, m = 10作为输入数据，检查你的程序的正 确性和健壮性。最后分析所完成算法的时间复杂度。定义JosephusCircle类，其中含完成初始化、报数出圈成 员函数、输出显示等方法。-Josephus write a problem solving function. With the sequence of integers 1, 2, 3, ... ..., n the order of sitting around the round table, said the people around. And then use n = 9, s = 1, m = 5, and n = 9, s = 1, m = 0, or n = 9, s = 1, m = 10 as input data, check the correctness of your program and robustness. Completed the final analysis, the algorithm" s time complexity. Definition JosephusCircle class, which included the completion of initialization, reported that the number of member functions of a circle, the output display and other methods.