I have written a small program, can you specify the file under the name of all f...

这是一个利用Delphi提供的系统服务类制作的后台服务程序，演示了定时备份文件的功能。注册服务的方法，在Delphi的Run parameters对话框中输入“/install”，然后运行代码。卸载输入“/uninstall”，然后运行代码。-This is a Delphi system services provided by the class production services for the background, demonstrated a regular backup files. Registration services, in Delphi"s Run dialog box input parameters "/ install," and then execute code. Unloading the importation of "/ uninstall", and then run code.

VC++环境下，用于制作一个多功能的状态栏-VC++ Environment used to create a multi-purpose status bar

捆绑机代码 捆绑机代码 捆绑机代码-bundling machine code bundling machine code bundling machine code bundling machine code bundle tied machine code bundled machine code

利用DELPHI不需要任何代码自动生成图章工具．-DELPHI using automatically generated code does not require any stamp tool.

视频 监控 程序, 值得参考, 调试通过!-Video surveillance program, it is also useful, debugging through!

非子窗口的滚动条定制-Feizi rolling window of Custom

在本次设计中，通过对新乡市自学考试和考生信息管理方式的调查，利用ASP、Dreamweaver、Fireworks等网站开发软件创建了新乡市自学考试网站，实现了网上报考、成绩和座号查询的功能。 新乡市自学考试网站开通的网上报考是自学考试报名方式的一项重大改革，也是新乡市自学考试网站的关键技术和难点。网上报考系统使考生省去了来回奔波、到现场排队填写报名卡等手续，它充分发挥网络的优势，使考生的报考时间更充裕，网上报考比社会报名点更早开始，更迟截止。无论考生身在何处都可以轻松报考，完全没有时空的限制。当考生在报考期间内想修改在线报考的结果， 随时重新登陆系统，即可以重新选择报考的科目，提高了信息采集的及时性、准确性。自学考试网站方便了考生的报考、查询，了解自学考试政策、自学考试办公室最新通知，便于自学考试政策的执行。 自学考试网站的创建使新乡市的自学考试管理方式又上升了一个层次，适应了社会发展的需求。-in the design, Xinxiang City of self-examination and information management candidates to the survey, the use of ASP, Dreamweaver, Fireworks such as web site development and software to create a self-examination of Xinxiang City website, sit through the Internet, achievements and seat inquiry functions. Self-examination of Xinxiang City website launched on the Internet is self-sit the examination form of a major reform, as well as self-examination Xinxiang City website the key technical and difficult. Online system allows candidates to sit eliminating the back and forth, to the scene to fi

这是一个关于属性页的源代码不知道能否对大家有点用处-This is a page on the attributes of the source code do not know whether we are a bit useless right

求解迷宫问题，输入一个迷宫矩阵，矩阵只有1和0，1表示通行，0表示不行。是同过递归求解-Maze solving the problem, enter a maze matrix, matrix, said the passage of only one, and 0,1, 0 not do that. Been solved with the recursive

这是本人做的一个计算器，可以计算+-*/cos sin tan asin atan，还可以识别括号，大家快来看看哈-This the one I do calculators, calculation-*/cos sin tan asin atan can also identify brackets, we hurry to see Kazakhstan