demod_2fsk

CSR与NFC集合工具，NFC是只能适合复旦微的芯片，CSR适合ROM版本和FALSE版本(CSR and NFC collection tools, NFC is only suitable for fudan micro chips, CSR for ROM version and FALSE version)

说明：  C+陷阱与缺陷,适合于各个阶段的学习者学习!(C+ traps and defects, suitable for all phases of learners!)

This is qpsk ber curve simulation.

清楚,MATLAB命令大全 管理命令和函数 help 在线帮助文件 doc 装入超文本说明 what M、MAT、MEX文件的目录列表 (clear)

一个可以用来分析程序复杂度的程序，可以评价程序的复杂度(A program can be used to analyze the complexity of the process, the complexity of procedures can be evaluated)

C/C++中调用MATLAB引擎的数值仿真的实现， 微分方程文件“fun.m”放在matlab的work目录下(C/C++ in the MATLAB engine called the realization of the numerical simulation, differential equations documents " fun.m" on the work directory matlab)

可以解开QR二维码,VISUAL c++(it can read the QR)

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.)

这个是霍夫曼编码的问题，要求输入只含有ABCD的四句话，且每句8个字符的长度，然后求出四个字符每个的概率，然后当成权值计算霍夫曼编码，我的这个程序已经实现了输入、概率统计、计算出霍夫曼编码存在了bian[]这个数组中(hoffuman decoding)

matlab教程ppt,适用于初学者,有程序(matalb )