Instead of finding the longest common
subsequence, let us try to determine the
length of the LCS.
Then tracking back to find the LCS.
Consider a1a2…am and b1b2…bn.
Case 1: am=bn. The LCS must contain am,
we have to find the LCS of a1a2…am-1 and
b1b2…bn-1.
Case 2: am≠bn. Wehave to find the LCS of
a1a2…am-1 and b1b2…bn, and a1a2…am and
b b b
b1b2…bn-1
Let A = a1 a2 … am and B = b1 b2 … bn
Let Li j denote the length of the longest i,g g
common subsequence of a1 a2 … ai and b1 b2
… bj.
Li,j = Li-1,j-1 + 1 if ai=bj
max{ L L } a≠b i-1,j, i,j-1 if ai≠j
L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
復接入，B/W雙用戶使用直接擴頻序列
％ >>>multiple access b/w 2 users using DS CDMA
％ >>>format is : cdmamodem(user1,user2,snr_in_dbs)
％ >>>user1 and user2 are vectors and they should be of equal length
％ >>>e.g. user1=[1 0 1 0 1 0 1] , user2=[1 1 0 0 0 1 1],snr_in_dbs=-50
％ >>>or snr_in_dbs=50 just any number wud do
％ Waqas Mansoor
％ NUST , Pakistan
This article presents GISCoordinate.java - a class that allows you to represent a GIS coordinate in your JAVA code in decimal degrees (38.4443, e.g. 122.33433) , minute degrees (33 44 22E, 122 33 44N), or radian degrees. Also, you can use this class to manipulate the coordinate, moving it around the globe by giving it distances in feet and direction of travel. You can then extract the new coordinate that is calculated after the travel.
【問題描述】
設計一個利用哈夫曼算法的編碼和譯碼系統，重復地顯示并處理以下項目，直到選擇退出為止。
【基本要求】
(1)初始化：鍵盤輸入字符集大小n、n個字符和n個權值，建立哈夫曼樹；
(2)編碼：利用建好的哈夫曼樹生成哈夫曼編碼；
(3)輸出編碼；
(4)設字符集及頻度如下表：
字符：A B C D E F
頻度：4 9 23 2 17 15
字符：G H I J K
頻度：1 2 3 3 4
LBreakout a breakout-style arcade game for Linux
Requirements:
X11, SDL[1.0 or better]
Installation:
1. type ./configure
options:
--disable-sound [no sound]
--disable-install [no installation play from source directory]
[e.g. ./configure --disable-sound --disable-install ]
2. type make
3. become root
4. type make install [only if install is not disabled]
5. type lbreakout to play
