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编写程序：基类base，有一个私有数据成员n，派生类Derived公有继承基类base，Derived中新增加一个数据成员m(Programming: the base class base, there is a private data member n, the derived class Derived inherits the base class public base, Derived add a new data member m)

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.)

一种VC++6.0与MATLAB混合编程的方法。有多个pdf文件。可以供需要者参考(One kind of VC++6.0 and MATLAB mixed programming method. There are several pdf files. Reference for those who need it can be)

自己编写的迭代小程序，小程序，没问题额的啊(I have written a small program iteration, a small program, the amount of, ah no problem)

这是关于信号处理方面的知识，对matlab的卡尔曼滤波十分重要(This is knowledge about signal processing and it is very importance of kalman filter for matlab )

Fuzzy Extractor Sketch

内含五个程序 1.zoeppritz Aki K I 和Richards P G的纵波近似程序 2.地震记录合成程序 3.大地电磁一维正演程序 4.雷克子波合成程序 5.地震反射系数生成程序(1.zoeppritz Aki KI and Richards PG longitudinal wave approximation procedures 2. Seismogram synthesis program 3. Magnetotelluric one-dimensional forward modeling program 4. Ricker wave synthesis procedure 5. Seismic reflection coefficient generator)

bom provide the orginal details before calculation

在二维空间里有两个点A=(X1，Y1)和B=(X2，Y2)，如果X1>X2并且Y1>Y2 ，我们认为A优于B，A是这个集合的最大极值点，在给定的二维空间里的N个点的集合S中，找出最大极值点。(In two-dimensional space, there are two points = A = (X1, Y1) and B (X2, Y2), if the X1 and Y1 > X2 > Y2, we think that A is superior to B, A is the biggest extreme value point of the set, in A given set of N points in the two-dimensional space S, find out the most extreme value point.)

说明：  BP 算法的实现，分别包含了C++，Matlab，Forturn等几种语言的实现方法