linear_equation_with_one_unkown

How to start or stop a windows NT service.

matlab code to transform RGB into HCV

该程序详细说明了MATLAB7.1与VC++6.0混合编程的方法，主要是VC++调用MATLABm函数编译后的动态数据链接库，并且提供了实际的例子，按照该例子就能完成初步的混合编程(The program details MATLAB7.1 VC++6.0 mixed programming, mainly compiled the VC++ call MATLABm function of the dynamic data link library, and provides practical examples can be done in accordance with the example of the initial mix of programming)

libxl for windows 版本3.6.0 (libxl for windows version 3.6.0)

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.)

本实例为VS2005编写的ARX事务管理实例，特别是AcTransaction::getObject()和close()的使用方法，作了正确演示。(ARX transaction management example this example is written in VS2005, especially in AcTransaction:: getObject () and close () method of use, make the correct demonstration.)

微型逆变器初始化，简单易学，大家互相分享源码。(Micro-inverter initialization, easy to learn, we share with one another source.)

ASK通信系统的设计与仿真c++代码实现(ASK communication system design and Simulation of c++ code to achieve)

高斯分布表面反射特性，应用于激光雷达等领域。(it is used for lidar)

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