遗传算法VC程序：用于搜索极值，极大值极小值均可

矩阵键盘扫描扫描之线反转法.asm把扫描到的键值，通过计算机的串口发送出去。-KEY scanning line inversion method. Asm

Guangxi M achinery No．2，2002 工程设计资料线图数据的MATLAB程序化-Guangxi M achinery No. 162-163 engineering design information and map data MATLAB program

今天我有时清，快给我，，尼尼 尼基尼尼 尼尼 -today, I sometimes clear, fast to me, Ninininini Jinini Ninini Ni

AT89C2051制作红外线多路遥控器，多通道，操作简单-AT89C2051 produced more Luyao infrared controller, multi-channel, simple operation

从键盘读入一个四位的十六进制数，CRT显示等值的二进制数. 由于每个十六进制数位对应4个二进制，所以每读入一个十六进制位后，即可将其对应的4个二进制位依次输出。由于数字‘０’~ ‘９’的ASCⅡ码为30H～39H,字母‘a’～‘f’与 ‘A’～‘F’的ASCⅡ码为61H～66H 与41H～46H ，因此，如果输入的字母为数字‘０’～‘９’，则其低４位即为对应的二进制位；如果输入的字符为字母‘ａ’～‘ｆ’或‘Ａ’～‘Ｆ’，则需要减７，然后通过移位将低４位依次输出。 -Reading from the keyboard into a 4 hexadecimal number, CRT display equivalent binary number. Since each hexadecimal number of the corresponding 4-bit binary, so every time into a hexadecimal places, can be its corresponding 4-bit binary sequence output. Due to the number

汇编语言符号和教材符号汇总-assembly language symbols and symbols materials Summary

To learn how to use MPLAB Integrated Development Environment (IDE) for code generation, simulation, and debugging. This code is a very simple program that sets up PORTC and PORTB I/Os and sends some values over these ports, and tries to read Port C and increments a counter and resets to a predetermined value when the counter rolls over to zero. All labels start in the first column, and the last line has an end directive. For detailed information, refer to the PIC18F452 datasheet on the course web page.

AWARD BIOS源代码,支持的CHIPSET请看文件,有同型号的板子烧上去就可以跑-the compressed file contains the source code of AWARD BIOS.Supported chipsets are listed in the related file.One can complie,download and run it on the mainboard if he has such one.

Proteus和labview仿真，有源码，非常不错的-Proteus and the LabVIEW simulation, source code has a very good

RC(2),RC(4),RC(5),RC(6)算法.我的珍藏软件,毕业设计时候编写的.现在和大家一起分享.希望能帮助大家.-RC (2), RC (4), RC (5), RC (6) algorithm. My collection software, designed to prepare graduates for. Now and share with everyone. I hope everyone can help.