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EasyDraw是一个简单的绘图软件，对综合运用的例子…
EasyDraw是一个简单的绘图软件，这个实例综合运用了Visual C++ 2005的基础知识。EasyDraw拥有以下功能：
1.直线、矩形、椭圆的绘制。
2.可以更改绘制线条的类型、宽度、颜色。-EasyDraw is a simple mapping software, the examples of the integrated use of the Visual C++ 2005 the basic knowledge. EasyDraw have the following features: 1. Linear, rectangular, oval drawing. 2. You can change the type of drawing lines, width, color.
- 2022-10-23 13:45:04下载
- 积分：1
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基于CUDA的心脏四腔室跳动模拟程序，数据来源www.ices.utexas.edu/~jessica/medical_data/heart/Heart_Val...
基于CUDA的心脏四腔室跳动模拟程序，数据来源www.ices.utexas.edu/~jessica/medical_data/heart/Heart_Valve_new.htm-CUDA-based four-chamber heart beating simulation program, data sources www.ices.utexas.edu/ ~ jessica/medical_data/heart/Heart_Valve_new.htm
- 2022-04-11 20:51:25下载
- 积分：1
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[1]
考勤时间=2005
[1]
考勤时间=2005-10-27
员工编号=1
状态=迟到
上班时间(早)=16:59:40
备注=无
[2]
考勤时间=2005-10-27
员工编号=2
状态=迟到
上班时间(早)=16:59:40
备注=无
[3]
考勤时间=2005-10-27
员工编号=3
状态=迟到
上班时间(早)=16:59:40
备注=无
[6]
考勤时间=2005-10-27
员工编号=6
状态=早退
上班时间(早)=上午 18:52:45
备注=无
下班时间(晚)=上午 18:52:45
[5]
考勤时间=2005-10-27
员工编号=5
状态=早退
上班时间(早)=上午 19:01:40
备注=无
下班时间(晚)=上午 19:01:40
[4]
考勤时间=2005-10-27
员工编号=4
状态=早退
上班时间(早)=上午 19:04:14
备注=无
下班时间(晚)=上午 19:04:14
-[1] time = 2005-10-27 employees attendance number = 1 state = late working hours (early) = 16:59:40 Note = No [2] time = 2005-10-27 employees attendance number = 2 status = late working hours (early) = 16:59:40 Note = No [3] time = 2005-10-27 employees attendance number = 3 state = late working hours (early) = 16:59:40 Note = No [6] Attendance time = 2005-10-27 employee ID = 6 state = leave early to go to work hours (early) = 18:52:45 AM Remarks = no working hours (late) = 18:52:45 AM [5] attendance period = 2005-10-
- 2022-08-15 07:26:19下载
- 积分：1
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There is a infinite series, whose general term is expressed as: a [n] = 10* 10*....
有一个无穷数列，其通项表示为：
a[n]=10*10*... n=0,1,2,3.....
构成了1,10,100,1000...
把它连起来，就成了数串。
判断数串的第i位到底是0还是1.性能要求1s
-There is a infinite series, whose general term is expressed as: a [n] = 10* 10*... n = 0,1,2,3 ..... constitute 1,10100,1000 it even ... together, it becomes the number of strings. Determine the number of strings in the end of the first i-bit is 0 or 1. Performance requirements 1s
- 2022-03-22 03:32:03下载
- 积分：1
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文件SQL入库实例
文件SQL入库实例-document examples of SQL storage
- 2022-03-29 14:40:28下载
- 积分：1
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一个论坛的原创代码，是不是太多说了自己看啊
一个论坛的原代码，~多的就不说了自己看吧，~-a forum of the original code, was not much said his Behold, O ~
- 2022-05-10 22:12:28下载
- 积分：1
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VC++ Serial MSComm
VC++串行命令
- 2022-01-26 17:42:01下载
- 积分：1
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WINSOCKET LAN
基于WINSOCKET局域网聊天程序，使用Visual C++编写，帮助大家熟悉了解套接字。-WINSOCKET LAN-based chat program, using Visual C++ development, to help everyone to become familiar with socket.
- 2022-03-25 20:39:30下载
- 积分：1
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约瑟夫环（Joseph）问题的一种描述是：编号为1，2，…，n的n个人按顺时针方向围坐一圈，每人持有一个密码（正整数）。一开始人选一个正整数作为报数上限值m，从...
约瑟夫环（Joseph）问题的一种描述是：编号为1，2，…，n的n个人按顺时针方向围坐一圈，每人持有一个密码（正整数）。一开始人选一个正整数作为报数上限值m，从第一个人开始按顺时针方向自1开始顺序报数，报到m时停止报数。报m的人出列，将他的密码作为新的m值，从它在顺时针方向的下一个人开始重新从1报数，如此下去，直至所有人全部出列为止。试设计一个程序求出出列顺序-Joseph Central (Joseph) a description of the problem is: numbered 1,2, ..., n of n individuals according to a clockwise direction around a circle, each holding a password (positive integer). The beginning of a positive integer as a candidate at a number of upper limit of m, from the first individual to embark on the clockwise direction since the beginning of the order of 1 is reported the number of stops required to report back at a few m. Those who reported m out of his password as the new value of m, in a clockwise direction from the next person to start off again from the 1, and so on, until all that out until all. Try to design a procedure to derive the column order
- 2022-02-10 01:58:28下载
- 积分：1
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jforum 论坛 速度非常优秀的开源论坛,虽然功能还不足.利用了很多JSP新技术，并且支持HSQLDB数据库，是学习研究的好东东哦....
jforum 论坛 速度非常优秀的开源论坛,虽然功能还不足.利用了很多JSP新技术，并且支持HSQLDB数据库，是学习研究的好东东哦.-Jforum source code, a very fast forum source code, using the newest technology of JSP, support HSQLDB database.
- 2022-03-02 16:47:40下载
- 积分：1
