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lsyqc
说明: 现代谱估计的matlab的程序。用的是最小二乘法。(Modern Spectral Estimation matlab procedures. Using a least squares method.)
- 2008-11-25 19:38:14下载
- 积分:1
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d009_dyn
说明: 标准9节点psat动态模型,包含模型描述以及simulink下搭建的模型(9 nodes psat standard dynamic model, including model descriptions and model structures under simulink)
- 2010-04-28 10:32:22下载
- 积分:1
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570486690TDIDF_Demo
TFIDF计算方法详细解释,代码解释很多,也有帮助文档(TFIDF calculation method explained in detail, explain a lot of code, but also help documentation)
- 2015-04-03 10:18:35下载
- 积分:1
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libev-4.15.tar
libev-4.15源码
libev是Marc Lehmann用C写的高性能事件循环库。通过libev,可以灵活地把各种事件组织管理起来,如:时钟、io、信号等。libev在业界内也是广受好评,不少项目都采用它来做底层的事件循环。node.js也是其中之一(libev-4.15 source libev Marc Lehmann is written in C, a high-performance event loop library. By libev, the flexibility to manage the organization of various events, such as: clock, io, signals. libev is also widely acclaimed in the industry, many projects have adopted it as the underlying event loop. node.js is one of them)
- 2014-01-27 01:37:10下载
- 积分:1
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3
说明: lms algorethem update
- 2010-05-25 22:01:49下载
- 积分:1
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PSOBPNNM
自己编写的PSO算法优化BPM文件,用于时滞系统的辨识,辨识效果还不错。(This is a M file of BPNN I have written for the identification of delay systems identification.The simulation result shows that the method is valid.)
- 2013-04-29 21:32:16下载
- 积分:1
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jiyuMATLABdefangzhenyanjiu
基于matlab的直流伺服电机驱动进给系统的仿真研究 ,(Matlab-based DC servo motor drive feed system simulation,)
- 2013-05-14 10:40:35下载
- 积分:1
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PCA
使用Matlab实现的关于主分量分析PCA的方法的案例,可先实现仿真(Matlab achieved using principal component analysis PCA on the case method, the simulation can be achieved)
- 2013-08-26 10:33:46下载
- 积分:1
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floyd
floyd算法主要用于最小生成树,最短路径,把散点输进去,就可以生成最短路径(loyd algorithm is mainly used for minimum spanning tree, shortest path, the scatter lost in, you can generate the shortest path)
- 2013-11-26 18:30:38下载
- 积分:1
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2
说明: 有一个长度为整数L(1<=L<=10000)的马路,可以想象成数轴上长度为L的一个线段,起点是坐标原点,在每个整数坐标点有一棵树,即在0,1,2,...,L共L+1个位置上有L+1棵树。
现在要移走一些树,移走的树的区间用一对数字表示,如 100 200表示移走从100到200之间(包括端点)所有的树。
可能有M(1<=M<=100)个区间,区间之间可能有重叠。现在要求移走所有区间的树之后剩下的树的个数。
输入:
两个整数L(1<=L<=10000)和M(1<=M<=100)。
接下来有M组整数,每组有一对数字。
输出:
可能有多组输入数据,对于每组输入数据,输出一个数,表示移走所有区间的树之后剩下的树的个数。(Has a length of integer L (1 < = L < = 10000) of the road, one can imagine a number line segment of length L, the starting point is the coordinate origin, each integer coordinate points in a tree, i.e. 0,1 , 2, ..., L total has L+1 L+1 positions on the tree. Now to remove some trees, tree removed by a range of digitally represented, such as 100 200 represents removed (inclusive) between 100 and 200 all the trees. There may be M (1 < = M < = 100) intervals, there may be overlap between the intervals. Now, after the number of trees require the removal of all remaining sections of the tree. Input: two integer L (1 < = L < = 10000) and M (1 < = M < = 100). Then there is an integer M groups, each with a pair of numbers. Output: there may be multiple sets of input data, for each set of input data, the output of a number indicating the number of the removal of all remaining sections of the tree after tree.)
- 2015-03-22 23:35:47下载
- 积分:1