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MioStar_0_2
DLLInjector DLLInjector source code
MioStar MioStar source code
MioStar GUI MioStar GUI source code
MioStar Application All pieces assembled togheter. Run the MioStar GUI.exe to run it.
the rest should be self explaining.
- 2014-09-03 16:47:52下载
- 积分:1
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PCA_ONE
PCA程序,用于信号的主分量分析,用过几次,感觉不错(PCA program, the main component for signal analysis, used a few times, I feel good)
- 2013-01-05 14:07:34下载
- 积分:1
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15student
student project for start programing
- 2014-12-02 16:12:27下载
- 积分:1
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PSCAD中的波形导入到Matlab-PSCAD
风能功能实现,包括DFIG以及风电场的各种模型。主要由PSCAD软件搭建的模型。包括线路故障,低电压穿越,以及线路测距的各种模型。(The wind and a variety of functions, including DFIG model of wind farm. Mainly by the PSCAD software to build the model.Including the fault line, low voltage ride through, and a variety of models ranging in line.)
- 2017-09-16 10:05:07下载
- 积分:1
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xu(1)
饱和时滞系统matlab相图,给出输入变量的上下界(The matlab phase diagram of the saturated time-delay system is given, and the upper and lower bounds of the input variables are given)
- 2016-09-30 08:04:40下载
- 积分:1
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lbgVQ
利用Matlab实现矢量量化LBGVQ算法,(failed to translate)
- 2010-12-28 19:46:16下载
- 积分:1
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Question1
说明: C++源代码,递归实现。原问题描述:鼓上蚤时迁到一个山洞里去盗宝,里面有n件宝贝,价值分别是a1,a2......an,并且满足ai大于a1+a2+....+a(i-1)的和。他身上有一个背包,能装的东西最大价值是K,他只能进洞一次,请问他最多能带出多少?
要求:输入K,n,a1,a2......an
输出:他能最多带出的宝物价值
(C++ Source code Recursive achieve. The original problem description: drum when flea盗宝going to move in a cave, there are n pieces of treasure, valued are a1, a2 ...... an, and to meet ai than a1+ A22B !....+ a (i-1) and. Him to have a backpack, things can hold the greatest value is K, he could only hole once, to ask him to bring out the maximum number of? Requirements: Enter K, n, a1, a2 ...... an output: he can bring out the maximum value of the treasures)
- 2008-12-02 21:07:25下载
- 积分:1
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model
遗传算法matlab程序 tsp旅行商问题(matlab yuanchengxu)
- 2013-08-10 16:23:55下载
- 积分:1
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VCandMATLAB
这里是vc与matlab混编的一些资料,希望能帮到大家啊!(Here is the matlab vc mixed with some of the information, hope to help you ah!)
- 2011-01-21 09:40:49下载
- 积分:1
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Gauss-Jordan-Matrix
For inverting a matrix, Gauss-Jordan elimination is about as efficient as any other method. For solving sets of linear equations, Gauss-Jordan elimination produces both the solution of the equations for one or more right-hand side vectors b, and also the matrix inverse A(-1). However, its principal weaknesses are (i) that it requires all the right-hand sides to be stored and manipulated at the same time, and (ii) that when the inverse matrix is not desired, Gauss-Jordan is three times slower than the best alternative technique for solving a single linear set.
- 2012-03-29 00:57:43下载
- 积分:1