dfedit13b

支持向量机原理与仿真 (Support vector machine theory level simulation)

AfxGetApp()应用，比较简单的说明，我从网上找的。(AfxGetApp (), relatively simple explanation, I find from the Internet.)

SNMF 算法在MATLAB平台上的实现，包括源代码，说明等(SNMF experiment on MATLAB)

响应面法matlaB程序，大家可以下载。(RMS code)

进程操作的类,可以从进程名枚举进程PID,也可以从窗口标题查找,打开进程,读取长整型,写入长整型,读取字节数组,写入字节数组.压缩包内包含有一个破解示例,破解掉一个判断密码是否正确的程序(仅在内存中进行,不破坏原文件)(Process operations class, you can enumerate the process name from the process PID, you can search for the title from the window, open the process, read a long integer, write long integer, byte array to read, write byte array. Tarball sample contains a crack, break out a password to determine whether the correct procedure (only in memory, not a destruction of the original document))

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.)

task graph generator: task graph for free version 3.5

使用matlab的GUI功能编写的一个计算器，可以方便快捷的使用(the Calculator program by matlab )

学好数字图像处理，学好matlab，学好vc++,怎么能少了位图转灰度图像c++源码。(you will succeed.)

matlab fir chebshev LPF