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kszd
- LISP快速展点程序 LISP展点程序,主要地形测量中展绘碎部点等。 展1000点:在HP(AMD Athlon64 3000+ 256MB)电胶上仅耗时0.142秒; 在金利(Geleron(R) CPU 2.40GHz 256MB)电胶上耗时0.882秒 数据文件格式为:每一点的数据(点号、X、Y、H)为一行,用逗号或空格作为分隔符,即 点号1 X1 Y1 H1 或者 点号1, X1, Y1, H1 点号2 X2 Y2 H2 或者 点号2, X2, Y2
HH2
- Algorithm (H2):the algorithme H2 is a heuristic for csp problem.
somKohonen
- Herustic H2 Algorithm:the algorithme H2 is a algorithm for csp problems.
CHANNEL
- 假设信号产生和传输信道模型为: 而抽头维纳滤波器为: 假设 的方差为0.27, 的方差为0.1,均值都为零。并且: , 并假设权向量初始值为 ,分别使用步长0.015、0.025和0.05进行 LMS算法仿真。 分析:d(n)是子系统H1受到v1(n)激励产生的信号,而H2与加性噪声构成了加性噪声传输信道。将u(n)作为维纳滤波器的输入,且滤波器的期望响应为d(n)。问题就是如何求出滤波器的权系数使得估计误差e(n)在MMSE意义下最小。 -Estim
html
- HTML Headings HTML headings are defined with the <h1> to <h6> tags. Example <h1>This a heading</h1> <h2>This is a heading</h2> <h3>This is a heading</h3> -HTML Headings HTML headings
33673-1081694-editeur-tag-mp3
- Editeur tag mp3--- --- --- Url : http://codes-sources.commentcamarche.net/source/33673-editeur-tag-mp3Auteur : tof2503Date : 13/08/2013 Licence : ========= Ce document intitulé « Editeur tag mp3 » issu de CommentCaMarche (codes-s
One-yuan-polynomial-multiplication
- 题目说明: 要求采用链表形式,求两个一元多项式的乘积:h3 = h1*h2。函数原型为:void multiplication( NODE * h1, NODE * h2, NODE * h3 )。 输入: 输入数据为两行,分别表示两个一元多项式。每个一元多项式以指数递增的顺序输入多项式各项的系数(整数)、指数(整数)。 例如:1+2x+x2表示为:<1,0>,<2,1>,<1,2>, 输出: 以指数递
robustcontrol
- for robust control and h2 control
multiplication
- 要求采用链表形式,求两个一元多项式的乘积:h3 = h1*h2。函数原型为:void multiplication( NODE * h1, NODE * h2, NODE * h3 )。-Require the use of a linked list, find the product of two one yuan polynomial: h3 = h1* h2. Function prototype: void multiplication (NODE* h1, NODE* h2, NOD
abcds
- WAP to check whether a given system H1(z)and H2(z)in cascade is stable
debug-leds
- Many OMAP development platforms reuse the same debug board these platforms include H2, H3, H4, and Perseus2. -Many OMAP development platforms reuse the same debug board these platforms include H2, H3, H4, and Perseus2.
NEE_ospf
- 工具:Dynamips, VPCS结点:R1: 10.0.0.0/8, 90.0.0.0/8R2: 20.0.0.0/8, 90.0.0.0/8R3: 30.0.0.0/8, 90.0.0.0/8H1: 10.0.0.2H2: 20.0.0.2H3: 30.0.0.2要求:R1,R2,R3之间直接用串口(serial) 相连,采用X.25封装。分配给串口的X.25地址不作限制。目的:使得H1,H2,H3可以在IP层互通-err
debug-leds
- Many OMAP development platforms reuse the same debug board these platforms include H2, H3, H4, and Perseus2.-Many OMAP development platforms reuse the same debug board these platforms include H2, H3, H4, and Perseus2.
Dichotomy
- Dichotomy Methods Successive dichotomy uses a totally different approach to the problem of indexing. The constants A to F in the equation Q = A h2 + B k2 + C l2 + D kl + E hl + F hk are assigned minimum and maximum values so that for a gi
results
- 这是一个为E马当先团队制作的物流信息查询结果的前端设计(实际系统中只要将h1,h2……中的代码换成dataset中的数据),使用的是html和css编写的。-This is a team for the production of E horse at the head of the front-end design of logistics information query results using html and css written.
Hoo
- Hoo/H2控制方法进行卫星姿态容错控制,包括卫星模型,控制器设计和参数文档-Hoo/H2 control method for satellite attitude fault-tolerant control, including satellite model, controller design and parameter document
Illustration_Norms
- or a thorough descr iption of the norms, see the Chapter 4 of Robust and Optimal Control, Zhou, Glover, Doyle 1996. Here the point is to give some lines of code illustrating what are the H2 and Hinf norms of LTI systems. In that sense, it i
rapid-calculation24CPP
- 速算24 要求: a. 一副牌54张牌,黑桃(SA,SK,SQ,SJ,S10,??,S2),红桃(HA,HK,HQ,HJ,H10,??,H2),方块(DA,DK,DQ,DJ,D10,??,D2),草花(CA,CK,CQ,CJ,C10,??,C2)以及大鬼Q1和小鬼Q2。其中,A,K,Q,J及Q1,Q2的点值分别为:1,13,12,11,1,1。其余点值就是牌值。 b. 由计算机随机出四张牌。 c. 用户输入能算出24的表达式(只能用加、减、乘、除及括号组成的四则运算)。
h2-2014-08-06-src
- 一种用Java实现的内存数据据库的完整源码-An in-memory data implemented in Java, according to the complete source code library
3_cubic_puzzle
- 立方数码问题(8数码问题的三维拓展) 使用C/C++实现4个算法,即,使用启发函数h1(n)的A*算法:Astar_h1.cpp,使用启发函数h2(n)的A*算法:Astar_h2.cpp,使用启发函数h1(n)的IDA*算法:IDAstar_h1.cpp,使用启发函数h2(n)的IDA*算法:IDAstar_h2.cpp。输出文件名与算法对应。以h1为启发函数的A*算法输出到output_Ah1.txt;以h2为启发函数的A*算法输出到output_Ah2.txt;以h1为启发函数的IDA