cPPReference

随意写的小程序，可在屏幕上显示星号表示的I LOVE YOU 有点小瑕疵 请多谅解啊 呵呵(Random write small programs can be displayed on the screen asterisk indicates the I LOVE YOU have a small flaw like him to understand ah Hehe)

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受阻力的球体运动轨迹程序，阻力与运动方向相反，满足阻力方程(By the resistance of the ball trajectory program, drag, and opposite to the direction of movement, and to meet the resistance equation)

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)

2410的WINCE蜂鸣驱动程序代码，提供给大家参考学习使用 (2410 WINCE beep driver code provided for your reference learning to use)

本源代码用三种不同的方法对三个数求最小公倍数，分别是循环穷举法，短除法和函数嵌套法。( Common multiple of three numbers Circular exhaustion Short division Function nesting)

proteus经典培训教程，值得期待，值得收藏(proteus classical training course, worth the wait, worth collecting)

带ID的lpc2368 CAN通信，包括初始化、中端、CAN通信和串口通信(With the ID lpc2368 CAN communications, including initialization, in the end, CAN communications and serial communication)

Chart2d 滚动 多轴 具有多重缩放功能，并能恢复至默认状态(Chart2d scrool multiaxis)

Flash+VC Flash+VC Flash+VC (Flash+ VC Flash+ VC Flash+ VC)