daqiMTF

说明：  cegui使用说明,初学者可以看看,对api的介绍(cegui cegui cegui )

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本书详细介绍了平时编程过程中可能范的各种错误，介绍了高质量C编程的各种知识以及各种技巧。(This book details the normal process of programming a variety of errors that might Fan introduced a variety of high-quality C programming knowledge and a variety of skills.)

使用与matlab中的小波熵程序，易于实现，注释清晰容易理解(With the the wavelet entropy program in matlab, easy to implement easy to understand, clear notes)

学习wimax，尤其针对ip技术，适合初学者(Learning wimax)

遗传算法的源程序，基于matlab，对于毕业设计会起到很大的作用；(Genetic algorithm source code, based on matlab, will play a big role for a graduation project )

VC++与Matlab混合编程实现图像处理(VC and Matlab Image Processing Mixed Programming)

某市有一码头，每次仅容一辆船停泊装卸货，由于经常有船等候进港，部分人提出要扩建码头。经过调查历史资料发现，码头平均每月停船24艘，每艘船的停泊时间为24±20小时，相邻两艘船的到达时间间隔为30±15小时，如果一艘船因有船在港而等候1小时，其消耗成本为1000元。经预算，扩建码头大约需要1350万元，故市长决策如下：如果未来五年内停泊船只因等候的成本消耗总和超过扩建码头花费则扩建码头，否则，不予扩建。因此，希望你能够帮助市长做出决策。此问题已知到达的大概时间和大概停泊时间，对于此问题用概率统计的方法来做比较复杂，可用程序随机产生到达时间和停泊时间来模拟未来五年内船的停泊，多次模拟预测停泊情况，以做出决策(。-Problem in a city with a terminal expansion dock, each only allow a boat moored loading and unloading, as often there are waiting boat into port, some people proposed to expand pier. Terminal stopping 24 per month, per vessel berthing time of 24 ± 20 hours, the arrival of two adjacent vessels interval 30 ± 15 hours, if a ship due to boat in the harbor while waiting for one hour, and its consumption cost of 1,000 yuan. The budget of about 13.5 million yuan expansion dock, so the mayor decision as follows: If a ship is moored next five years, the cost of waiting more than the sum of consumption spending is expanding marina pier extension, otherwise, not expanded.)

一个简单的matlab通讯的小程序，希望对大家有帮助了(A simple sort of small procedures in the hope we have helped )

这是关于matcom和C++联合编程的资料，可以实现脱离matlab环境来进行编程(This is joint programming on matcom and C++, the data can be achieved out of the matlab environment for programming)