white-answer

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)

用sph（光滑流体粒子动力学）方法模拟经典流动泊肃叶流，适合sph方法初学者练习使用！(With sph (smooth fluid particle dynamics) simulations classic Poiseuille flow stream, suitable for beginners to practice using the method sph!)

socket头文件，VC++6.0头文件库中不存在，下载后即可方便使用。(socket header files, VC++6.0 library header file does not exist, you can download easy to use.)

你将学到为何利用其他开发人员的经验与智慧解决问题，我们会看看设计模式的用途与优点，并通过实例来了解模式是如何运作。(You will learn why to use the experience and wisdom of other developers to solve the problem, we will look at the uses and benefits of design patterns, and an example to understand the model is how it works.)

基于VC++的,多个串口的,带通讯协议的应用程序,两个通讯口,可分别接受,发送协议数据包.()

RAS api的MFC类封装(MFC class encapulation of RAS api)

一个用gdi实现了编辑的矢量绘图程序(a gdi achieved with the vector graphics editing procedures)

具有播放媒体文件，全屏控制，音量控制，定不显示等功能(Play media files, full control, volume control, given does not display and other functions)

嵌入式处理器MCU的选型及开发注意事项（包括软硬件使用注意）(MCU embedded processor selection and development of Attention )

获取计算机IP地址。可以试试看。源码用C++实现(C++)