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src
说明: Petri网中XML根结点为net,其属性有id、type,其子结点有四个,分别为token,place,transition,arc。根据传入矩阵生成Petri网(In Petri nets, the root node of XML is net, and its attributes are ID and type. There are four sub-nodes: token, place, transition and arc. Generating Petri Nets from Input Matrix)
- 2019-03-15 15:46:29下载
- 积分:1
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bgkms6
说明: 这是一个按键精灵脚本的插件程序,实现后台鼠标键盘操作(This a anjian jinglinng de program,This is a test program.)
- 2020-03-29 21:17:12下载
- 积分:1
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dataprocessing
该程序实现了基于LabView的板卡数据输入输出,及频谱分析、谐波分析等数据处理。(Realize the program)
- 2008-07-14 10:34:54下载
- 积分:1
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houosr-rent-accessPcSharp
c#+SQL开发的房屋租赁系统,很适合做毕业设计使用,(c#+ SQL development for the housing lease system, it is suitable for use graduation,)
- 2011-05-05 20:57:09下载
- 积分:1
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新建文件夹.part01
维尼通编程例程,简单易学,容易上手,希望对你有所帮助(Veneton Programming Routine)
- 2019-02-21 13:55:12下载
- 积分:1
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这是动画设计的一些资料,希望对有兴趣的朋友们参考参考
这是动画设计的一些资料,希望对有兴趣的朋友们参考参考-This is the animation of some of the information, and I hope to interested friends information.
- 2022-03-21 06:44:10下载
- 积分:1
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VC++转换COLORREF值为RGB分量
VC++转换COLORREF值为RGB分量,举个例子,在本例中,是将COLORREF值:0x000080FF转换成:R分量:0xFF、G分量:0x80、B分量:0x00,注意,并不是分离。请看定义:
//R分量
BYTE RED = GetRValue(cr);
//G分量
BYTE GREEN = GetGValue(cr);
//B分量
BYTE BLUE = GetBValue(cr);
CString strText = _T("");
strText.Format(_T("COLORREF值:0x%08X"), cr);
pDC->TextOut(100, 50, strText);
strText.Format(_T("R分量:0x%02X"), RED);
pDC->TextOut(100, 100, strText);
strText.Format(_T("G分量:0x%02X"), GREEN);
pDC->TextOut(100, 150, strText);
strText.Format(_T("B分量:0x%02X"), BLUE);
pDC->TextOut(100, 200, strText);
- 2022-06-13 02:13:45下载
- 积分:1
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mimo
说明: 基于MINO的非正交多址接入系统的功率分配算法
用于多天线(Power allocation algorithm for MINO-based non-orthogonal multiple access system)
- 2020-12-11 10:49:17下载
- 积分:1
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PMA
模拟101、104、modbus主站从站功能(Analog 101, 104, MODBUS main station slave function)
- 2017-09-20 10:28:38下载
- 积分:1
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FloodingSimulation
说明: 在具有至少十五个节点的网络中模拟洪泛路由。每个数据包应包含一个计数器(n = 10),该计数器在每一跳上递减。(Simulate flood routing in a network with at least fifteen nodes. Each packet should contain a counter (n=10) that is decremented on each hop. When the counter gets to zero, the packet is discarded. Time is discrete, and each
link can handle only one packet per time interval (i.e. only one packet in total may traverse
the link in either direction). Nodes should include a buffer to queue any additional packets.
Present your results in terms of the number of duplicate packets produced, and the
congestion at each node, and suggest a possible solution for reducing this problem.)
- 2020-06-24 09:40:07下载
- 积分:1