ACCESS常用窗体查询，例举多种窗体查询的方法

数据库基础知识的笔记题集,对初学者进行复习很有用-database

发送短信模块源代码！PCON &= 0x7F; //波特率不倍速 SCON = 0x50; //8位数据,可变波特率 AUXR &= 0xBF; //定时器1时钟为Fosc/12,即12T AUXR &= 0xFE; //串口1选择定时器1为波特率发生器 TMOD &= 0x0F; //清除定时器1模式位 TMOD |= 0x20; //设定定时器1为8位自动重装方式 TL1 = 0xFD;  //设定定时初值 TH1 = 0xFD;  //设定定时器重装值 ET1 = 0;    //禁止定时器1中断 TR1 = 1;    //启动定时器1   ES=1;  //开串口中断

C/C++语言指针使用指南,超过80%的死机现象是由于指针操作不当引起的。因而学会正确使用指针是C/C++语言使用者非常必要的一项小技能-C/C language pointer using the guidelines, more than 80% of Vista is due indicator of improper operation. So learn how to correctly use is the target C/C language users very necessary skills of a small

几种常用排序算法的效率比较,通过测试可得出这几种排序算法的优劣.-several commonly used algorithm efficiency comparison, the test may come this algorithm several merits.

VHDL实现布斯算法-VHDL Booth algorithm

vfp的界面，系统登陆时的表单。（压缩文件）-Probe interface, the system of landing form. (Compressed files)

图的建立、遍历、最小生成树、最短路径 图的建立、遍历、最小生成树、最短路径-plans the establishment, traverse, the minimum spanning tree and the shortest road map for the establishment of traversal, minimum spanning tree, shortest path

db2日常维护汇总-db2 routine maintenance Summary

pb 源码 进度条程序-progress of the process

迷宫算法（maze） /* Maze * Starting point is m[0][0], need to find a path go to m[9][9]. 0 means OK, * 1 means cannot go there, boundary is 0 and 9, cannot go beyond boundary. * Each step can be made horizontally or vertically for one more grid (diagonal * jump is not allowed). * Your program should print a series of grid coordinates that start from m[0][0] * and go to m[9][9] * Hint: No need to find the shortest path, only need to find one path that gets * you to desitination. */ -/* Maze * Starting point is m[0][0], need to find a path go to m[9][9]. 0 means OK, * 1 means cannot go there, boundary is 0 and 9, cannot go beyond boundary. * Each step can be made horizontally or vertically for one more grid (diagonal * jump is not allowed). * Your program should print a series of grid coordinates that start from m[0][0] * and go to m[9][9] * Hint: No need to find the shortest path, only need to find one path that gets * you to desitination.