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fleurydaima
这个是一个旅行商的问题，在多个路段中选择合适的路径是解决问题的关键，这个算法就解决了这个问题(This is a traveling salesman problem, in many sections to choose the right path is the key to solve the problem, this algorithm has solved this problem)
- 2010-09-03 23:21:13下载
- 积分：1
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BLS-GSM_Denoising
基于小波域隐马尔可夫模型的图像降噪，性能最好的图像降噪程序。
(Based on wavelet-domain Hidden Markov Model of image noise, the performance of the best image noise reduction procedures.)
- 2009-04-12 18:06:56下载
- 积分：1
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yuyinshouji
语音信号的实时采集，利用电脑自带的麦克进行语音信号的实时采集(failed to translate)
- 2013-05-07 17:02:01下载
- 积分：1
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Robotics
Keywords: colahi, virtual reality (Simulate 3-bar robotic arms with Virtual Reality toolbox)
- 2009-05-14 04:51:45下载
- 积分：1
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ROVsim
水下机器人控制matlab仿真，对水下机器人控制的仿真有很大帮助，内含好多源代码(Underwater robot control matlab simulation of underwater robot control simulation , containing a lot of source code
)
- 2021-03-17 15:19:20下载
- 积分：1
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Recursive_integer_division
递归法求解整数划分。
整数划分，是指把一个正整数n写成如下形式：
n=m1+m2+…+mi （其中mi为正整数，并且1 <= mi <= n），则{m1,m2,...,mi}为n的一个划分。
如果{m1,m2,...,mi}中的最大值不超过m，即max(m1,m2,...,mi)<=m，则称它属于n的一个m划分。这里我们记n的m划分的个数为f(n,m) (Recursive method integer division. Integer division, refers to a positive integer n written as follows: n = m1+m2+ ...+mi (where mi is a positive integer, and 1 < = mi < = n), then {m1, m2, ..., mi} is a division of n. If {m1, m2, ..., mi} does not exceed the maximum value of m, i.e., max (m1, m2, ..., mi) < = m, m n is said that it belongs to a division. Where m is the number of division n our mind is f (n, m) )
- 2014-02-06 14:21:38下载
- 积分：1
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Matlab
matlab课程设计大作业，内含十道题，是对matlab应用的一个综合(matlab)
- 2010-03-04 18:27:35下载
- 积分：1
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Fast-Automatic-T-emplate-Matching--
Fast Automatic T emplate Matching for Spike Sorting
- 2014-02-20 10:11:15下载
- 积分：1
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modulation&demodulation
支持多种星座符号调制和解调，包括m-PSK和m-QAM(Supports a variety of constellation symbol modulation and demodulation, including m-PSK and m-QAM)
- 2018-11-30 11:03:14下载
- 积分：1
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Newton--chazhi
说明： 能够很好的解决牛顿插值问题，对于构建差商表的程序给出了解释和说明！！！(Newton interpolation can be a good solution to the problem, divided difference table for building the program and instructions are given to explain! ! !)
- 2011-04-07 14:29:15下载
- 积分：1
