DL : 9
设∑={α1, α2…… αn }是n个互不相同的符号组成的符号集。 Lk={β1β2…βk | βi ∑,1≤i≤k}是∑中字符组成的长度为k 的全体字符串。 S是Lk的子集,S是Lk的无分隔符字典是指对任意的S中元素a1a2…ak, b1b2…bk. {a2a3…akb1, a3a4…akb1b2, ……, akb1b2… bk-1 }∩S=Φ。该算法算法,对于给定的正整数n 和k,计算 Lk的最大无分隔符字典。
Update : 2009-02-20
Size : 1.26kb
Publisher : wild_lily
DL : 0
对于矩阵连乘积的最优计算次序问题,设计算Ai…j ,1≤i≤j≤n,所需的最少数乘次数为m[i,j],原问题的最优值为m[1,n]。
当i=j时,Ai…j=Ai为单一矩阵,无需计算,因此m[i,i]=0,i=1,2,…,n ;
当i<j时,可利用最优子结构性质来计算m[i,j]。事实上,若计算Ai…j的最优次序在Ak和Ak+1之间断开,i≤k<j,则:m[i,j]=m[i,k]+m[k+1,j]+rirk+1rj+1 。
-Even for the matrix product of the calculation of the optimal order problem, the design of operator Ai ... j, 1 ≤ i ≤ j ≤ n, the required number of at least a few by m [i, j], the original problem of optimal values for m [1 , n]. When i = j when, Ai ... j = Ai for a single matrix, without basis, it is m [i, i] = 0, i = 1,2, ..., n when i <j, to be used to sub-optimal structure the nature of computing m [i, j]. In fact, if the calculation of Ai ... j the optimal order in Ak and Ak+ 1 disconnect between, i ≤ k <j, then: m [i, j] = m [i, k]+ M [k+ 1 , j]+ rirk+ 1rj+ 1.
Update : 2025-02-17
Size : 1kb
Publisher : uhy
DL : 0
the famous tool used to detect the memory leak-the famous tool used to detect the memory le ak
Update : 2025-02-17
Size : 10.75mb
Publisher : 谢国
DL : 0
VC6.0从入门到精通的电子书找了很就才找的拿来分-VC6.0 proficient from the portal to find the e-book is only to find on the used-
Update : 2025-02-17
Size : 3.15mb
Publisher : 涂世昌
DL : 0
AK BOT network testing
Update : 2025-02-17
Size : 88kb
Publisher : darrenhzx
DL : 0
Singh-AK-2015-PhD-Thesis=Decentralized Estimation and Control
Update : 2025-02-17
Size : 3.93mb
Publisher : nmz2009
DL : 0
ak-47 Paladin for zombie plague the game cso or CS: Online found in zombie-plague.ru-ak-47 Paladin for zombie plague the game cso or CS: Online found in zombie-plague.ru
Update : 2025-02-17
Size : 705kb
Publisher : Smarty
DL : 0
AK5384规格书,英文编写,非常正确有需要请下载!(Ak 5383 specification, written in English, very correct and need to download!)
Update : 2025-02-17
Size : 219kb
Publisher : allen52
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