* "Copyright (c) 2006 Robert B. Reese ("AUTHOR")" * All rights reserved. * (R. Reese, reese@ece.msstate.edu, Mississippi State University) * IN NO EVENT SHALL THE "AUTHOR" BE LIABLE TO ANY PARTY FOR * DIRECT, INDIRECT, SPECIAL, INCIDENTAL, OR CONSEQUENTIAL DAMAGES ARISING OUT * OF THE USE OF THIS SOFTWARE AND ITS DOCUMENTATION, EVEN IF THE "AUTHOR" * HAS BEEN ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
上传时间： 2015-09-24
上传用户：mpquest
design LP,HP,B S digital Butterworth and Chebyshev filter. All array has been specified internally,so user only need to input f1,f2,f3,f4,fs(in hz), alpha1,alpha2(in db) and iband (to specify the type of to design). This program output hk(z)=bk(z)/ak(z),k=1,2,..., ksection and the freq.
标签： Butterworth internally Chebyshev specified
上传时间： 2015-11-08
上传用户：253189838
Floyd-Warshall算法描述 1)适用范围： a)APSP(All Pairs Shortest Paths) b)稠密图效果最佳 c)边权可正可负 2)算法描述： a)初始化：dis[u,v]=w[u,v] b)For k:=1 to n For i:=1 to n For j:=1 to n If dis[i,j]>dis[i,k]+dis[k,j] Then Dis[I,j]:=dis[I,k]+dis[k,j] c)算法结束：dis即为所有点对的最短路径矩阵 3)算法小结：此算法简单有效，由于三重循环结构紧凑，对于稠密图，效率要高于执行|V|次Dijkstra算法。时间复杂度O(n^3)。 考虑下列变形：如(I,j)∈E则dis[I,j]初始为1,else初始为0，这样的Floyd算法最后的最短路径矩阵即成为一个判断I,j是否有通路的矩阵。更简单的，我们可以把dis设成boolean类型，则每次可以用“dis[I,j]:=dis[I,j]or(dis[I,k]and dis[k,j])”来代替算法描述中的蓝色部分，可以更直观地得到I,j的连通情况。
标签： Floyd-Warshall Shortest Pairs Paths
上传时间： 2013-12-01
上传用户：dyctj
该程序用以查找任意两个整数之间的所有素数。 Prime number finder can find all primes between a and b and will write the results to the file PRIMES.TXT
上传时间： 2014-01-14
上传用户：cccole0605
I wrote this code early this year using ColdFire MCF5213 in codewarrior IDE. The LCD is STN B/W 320x240 dot matrix LCD. The code include 3 different fonts, and basic LCD driver. All original!
标签： this codewarrior ColdFire wrote
上传时间： 2013-12-20
上传用户：皇族传媒
Program main BIOS image | | /B - Program Boot Block | | /N - Program NVRAM | | /C - Destroy CMOS checksum | | /E - Program Embedded Controller Block | | /K - Program all non-critical blocks | | /Kn - Program n th non-critical block only(n=0-7) | | /Q - Silent execution | | /REBOOT - Reboot after programming | | /X - Don t Check ROM ID | | /S - Display current system s ROMID | | /Ln - Load CMOS defaults
标签： Program Destroy Block NVRAM
上传时间： 2016-07-26
上传用户：wfl_yy
Problem B:Longest Ordered Subsequence A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).
标签： Subsequence sequence Problem Longest
上传时间： 2016-12-08
上传用户：busterman
21世纪大学新型参考教材系列 集成电路B 荒井
上传时间： 2013-04-15
上传用户：eeworm
家电维修(最基础的教程B)1-20.Torrent
上传时间： 2013-06-10
上传用户：eeworm
jk-b交通信号控制机原理图
上传时间： 2013-07-13
上传用户：eeworm
