-
momh_effectiveness_mopgp
MOEAs对于度欧表的有 花纹那(MOEAs for the degree of European tables have patterns that)
- 2008-05-31 16:26:10下载
- 积分：1
-
TABLEAU-DE-BORD-V1
Overview: In the application, often the need for custom report format, such as the printing contract, cargo manifests,
- 2015-04-22 23:14:26下载
- 积分：1
-
PatternsofEnterpriseapplicationrchitecture
Patterns of Enterprise application architecture
- 2007-12-11 12:52:16下载
- 积分：1
-
Flow-Calculation
电力系统及其自动化专业，电力系统分析潮流计算编程，采用N-R法(Systems and automation, power system analysis power flow calculation program，Using the N-R method)
- 2014-11-12 10:31:54下载
- 积分：1
-
Wavelet_denoising
小波去噪matlab程序，有demo，本人已运行，能用，大家可以试一试。(Wavelet denoising matlab program, demo, I have to run, can be used, we can try.)
- 2013-10-25 21:10:04下载
- 积分：1
-
MSChart
MSCart使用方法。MSChart各个属性介绍。数据如何传送到MSChart等。(MSCart use. MSChart introduced various attributes. How data is sent to the MSChart etc..)
- 2013-04-10 09:45:03下载
- 积分：1
-
ceshi
Kalman滤波器的MATLAB实现，一个Kalman滤波的应用实例(Lt RTI ID 0.0 & gt u5 & lt /RTI & gt & lt /RTI & gt & lt RTI ID 0.0 & gt u5 & lt /RTI & gt )
- 2017-04-24 15:47:40下载
- 积分：1
-
houzizhaitao
猴子摘桃海滩上有一堆桃子，五只猴子来分。第一只猴子把这堆桃子凭据分为五份，多了一个，这只
猴子把多的一个扔入海中，拿走了一份。第二只猴子把剩下的桃子又平均分成五份，又多了
一个，它同样把多的一个扔入海中，拿走了一份，第三、第四、第五只猴子都是这样做的，
问海滩上原来最少有多少个桃子？(Peach Monkeys on the beach there is a pile of peaches, five monkeys for points. The first monkey peach credentials this heap is divided into five copies, more than one, the monkey to more than one thrown into the sea, took a. The second monkey the rest of the peaches and what the average is divided into five, and there are one, and it is likewise more than one thrown into the sea, took a third, fourth, fifth monkeys are actually doing that , the question on the beach for at least the number of the original peach?)
- 2009-12-24 15:33:40下载
- 积分：1
-
cma-8psk
说明： 关于8psk的用星座图检验结果的matlab仿真程序(Constellation on the 8psk of test results with matlab simulation program)
- 2011-04-07 16:00:41下载
- 积分：1
-
surshrHEDA
基于直方图的分布估计算法matlab源程序，测试函数选用的是30维函数(for h=1:m
for j=1:((x2-x1)/binswidth)
if (x(h)>=x1+(j-1)*binswidth)&(x(h)<x1+j*binswidth)
H(j)=H(j)+1
end
end
end
[wid,len]=size(H)
H=H/(wid*len)
objvalue=sort(objvalue, descend )
for j=1:((x2-x1)/binswidth))
- 2013-03-15 10:44:15下载
- 积分：1
|
联系客服（QQ：3324461878）
