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COMTestCode
vc环境下COM组件编程小例子，一个文件生成COM组件，另一个文件调用该COM组件。(COM component programming small example, a file generated COM component vc environment, another file to call the COM component.)
- 2013-12-19 16:21:03下载
- 积分：1
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You-are-no-longer-afraid-pointer
要了解指针,多多少少会出现一些比较复杂的类型,所以我先介绍一下如何完全理解一个复杂类型,要理解复杂类型其实很简单,一个类型里会出现很多运算符,他们也像普通的表达式一样,有优先级,其优先级和运算优先级一样(To understand pointers, more or less there will be some of the more complex types, so I' ll explain how to fully understand a complex type, to understand the complex type is actually very simple, there will be a lot of one type of operator, they, like the general expression style, have a priority, the priority and operational priorities as)
- 2014-05-19 10:43:47下载
- 积分：1
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zoj1094
zoj094
Matrix multiplication problem is a typical example of dynamical programming.
Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices. Since matrix multiplication is associative, the order in which multiplications are performed is arbitrary. However, the number of elementary multiplications needed strongly depends on the evaluation order you choose.
For example, let A be a 50*10 matrix, B a 10*20 matrix and C a 20*5 matrix.
There are two different strategies to compute A*B*C, namely (A*B)*C and A*(B*C).
The first one takes 15000 elementary multiplications, but the second one only 3500.
Your job is to write a program that determines the number of elementary multiplications needed for a given evaluation strategy.
(Matrix multiplication problem is a typical example of dynamical programming.
Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices. Since matrix multiplication is associative, the order in which multiplications are performed is arbitrary. However, the number of elementary multiplications needed strongly depends on the evaluation order you choose.
For example, let A be a 50*10 matrix, B a 10*20 matrix and C a 20*5 matrix.
There are two different strategies to compute A*B*C, namely (A*B)*C and A*(B*C).
The first one takes 15000 elementary multiplications, but the second one only 3500.
Your job is to write a program that determines the number of elementary multiplications needed for a given evaluation strategy.
)
- 2012-06-10 10:38:53下载
- 积分：1
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live555H.264
VC下的RTSP源代码，live555+H264(RTSP source code under VC, live555+ H264)
- 2010-09-25 11:43:50下载
- 积分：1
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aawmutual
说明： 此为windows风格的单文档窗体,比较通用欢迎在此基础上,加一改进,创造出更好的界面来(This is a single document form with windows style, which is generally welcome to add an improvement to create a better interface.)
- 2018-12-24 03:36:53下载
- 积分：1
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WavDest
directshow 的 示范例子。完成WAV 流的转换。(directshow demonstration examples. WAV flow completed the conversion.)
- 2007-06-03 03:28:21下载
- 积分：1
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CompactDataBase
mfc，才有用ado操作access数据库，解决数据库删除内容后无法自动压缩大小的问题(mfc, only access the database using ado action to resolve the contents of the database are not automatically remove the compressed size of the problem)
- 2011-06-25 10:47:24下载
- 积分：1
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traffic
[基本要求]
（1）提供对城市信息进行编辑（如：添加功删除）的功能
（2）城市之间有两面种交通工具：火车和飞机。提供对列车时间表和飞机航班进行编辑（增设或删除）的功能。
（3）提供两面种最优决策：最快到达或最省钱到达。全程只考虑一种交通工具。
（4）旅途中耗费的总时间应该包括中转站的等候时间。
（5）咨询以用户和计算机的对话方式进行。由用户输入起始站、终点站、最优决策原则和交通工具，输出信息：最快需要多长时间才能到达或者最少需要多少旅费才能到达，并详细说明依次于何时坐哪一趟列车哪一次班机到何地。
[实现提示]
（1）对全国城市交通图和列车时刻表及飞机航班表的编辑，应该提供文件形式输入和键盘输入两种方式。飞机航班表的信息包括：起始站台票出发时间、终点站的到达时间和票价；列车时刻表则根据交通图给出各个路段的详细信息。
（2）以邻接表作交通图的存储结构，表示边的结点内除了含有邻接点的信息外，还应包括交通工具、路程中消耗的时间和花费以及出发和到达的时间等多项属性。
()
- 2008-07-07 00:17:13下载
- 积分：1
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sniffer1
C语言实现的监听网络流量的程序，并对截取的报文进行分析。(C language to monitor the flow of network procedures, and analysis of the interception of the message.)
- 2017-01-04 15:17:53下载
- 积分：1
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MAX6675
MAX6675 的说明文件，用于温度采集方面很好用，电路简单，温度准确。(MAX6675 documentation, for the temperature acquisition with good, simple circuit, accurate temperature.)
- 2014-12-19 16:09:42下载
- 积分：1
