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tname
说明： 大家好，本程序功能是VC环境下Matlab组件后绑定的实现，希望对大家有所帮助(Hello everyone, this program features is the VC environment Matlab after the binding components to achieve, I hope all of you to help)
- 2008-10-19 09:54:44下载
- 积分：1
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huatu
VC调用MATLAB的简单事例，没有复杂的功能，只适合初学者(A simple example of calling MATLAB VC)
- 2011-08-09 15:15:53下载
- 积分：1
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R
说明： R语言地理统计及空间分析的函数包，可以做空间插值，相关分析等。(Function package R language geostatistical and spatial analysis, spatial interpolation, correlation analysis can be done.)
- 2015-04-02 08:57:52下载
- 积分：1
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HsT2Hlp_Demo
HsT2Hlp_Demo,中信证券api简单的api示例(HsT2Hlp_Demo, CITIC Securities api api simple example)
- 2014-02-13 10:49:59下载
- 积分：1
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MyDEA1.0
做DEA用的,功能很强大,大家可以试试看,真的很好用(DEA do use, very powerful, we can try, really well)
- 2014-02-05 17:32:37下载
- 积分：1
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SOFM
一个关于神经网络的 SOFM算法的程序 改程序用C语言编写(A SOFM neural network algorithm on the procedure to change procedures for using C language)
- 2009-12-27 12:59:56下载
- 积分：1
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3zuoye
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.)
- 2015-04-17 18:59:46下载
- 积分：1
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ms-bcs-spl-1.2-2
matlab code for image compressive sensing.
block compressive sensing is used.
- 2013-04-22 23:05:20下载
- 积分：1
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2DMaxima
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.)
- 2013-12-24 21:02:56下载
- 积分：1
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Quantisation
image quantisation technique
- 2010-01-31 03:51:51下载
- 积分：1
