一些原来的代码，对于学习和提高很有帮助的帮助

三连棋： 人机对弈，三子连在一起就算赢。 人人对弈，规则相同。-Third, even the game: man-machine chess, three sons together even win. All game, the same rules.

此程序用到了,ADO连接ACCESS数据库,切分窗口,以及不同视图之间进行通信.List Control和Tree Control,用户UI线程等技术.-use of this procedure, ADO connection ACCESS database segmentation window different Views of communication between. List Control and Tree Control, users UI thread technology.

实现卷积，DFT FFT 算法的源代码，希望对一些朋友会有帮助 -Realize convolution, DFT FFT algorithm source code, in the hope that some of my friends will be helpful

#include < iostream.h > #define类int公共voidvoidvoid sortLogic() ； void bubsort:: getdata() { cout arr [i] ； void bubsort:: showdata() { cout

C语言经典程序的源码对初学者很有用,内容涉及面广-C language source of the classic program useful for beginners, covering a broad area

1.使用串口同两块west4170(PID控制温度仪表). 2.可以显示当前温度,并可设定参数(主要是设定温度)-1. The use of serial port with two west4170 (PID Control Temperature Instrumentation) .2. Can display the current temperature, and set the parameters (mainly temperature)

数根 数也有根吗？有！数根就是把一个数的各个数位上的值相加，直到得到一个一位数为止。请你编写一个程序来计算任意数N（0≤N≤1000000000000）的根。 输入： 输入有多个案例。每一个案例为一行，每行只有一个整数N。当输入的N是-1时，输入结束。 输出： 每一个案例输出一行。每行只有一个数字，就是N对应的数根。 Sample input 39 -1 Sample output 3-Number of root number also has roots do? There is! The number of roots is to a number of different values on the digital sum until they have been a number of up to one. Please write a program to calculate an arbitrary number N (0 ≤ N ≤ 1000000000000) roots. Input: Enter a number of cases. Each case for the line, each line is only an integer N. When the input of N is-1, the input end. Output: the output line of each case. There is only one number per line, that is, N corresponds to the number of roots. Sample input 39-1 Sample output 3

路径最短算法(C 语言开发完成)，希望对大家有所帮助，-min topo path

10025，一款使用软件-1005

dds->9954的pcb供大家参考,这个扳子我调通了的可以放心开办-dds-