Problem Statement You are given a string input. You are to find the Longest substring of input such that the reversal of the substring is also a substring of input. In case of a tie, return the string that occurs earliest in input. Definition Class: ReverseSubstring Method: findReversed Parameters: string Returns: string Method signature: string findReversed(string input) (be sure your method is public) Notes The substring and its reversal may overlap partially or completely. The entire original string is itself a valid substring (see example 4). Constraints input will contain between 1 and 50 characters, inclusive. Each character of input will be an uppercase letter ( A - Z ). Examples 0) "XBCDEFYWFEDCBZ" Returns: "BCDEF" We see that the reverse of BCDEF is FEDCB, which appears later in the string. 1)
上传时间： 2015-09-21
上传用户：sunjet
LCS Algorithm, this is a c++ code for lcs(Longest Common Subsequence)
标签： Subsequence Algorithm Longest Common
上传时间： 2013-12-25
上传用户：李梦晗
Longest Ordered Subsequence,acm必备习题
标签： Subsequence Longest Ordered acm
上传时间： 2014-10-27
上传用户：1159797854
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
最长子序列,英文版,电子书 ACM文章1977+Algorithms+for+the+Longest+Common+Subsequence+Problem
标签： Subsequence Algorithms Longest Problem
上传时间： 2014-01-27
上传用户：zhuimenghuadie
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.
标签： the subsequence determine Instead
上传时间： 2013-12-17
上传用户：evil
This scheme is initiated by Ziv and Lempel [1]. A slightly modified version is described by Storer and Szymanski [2]. An implementation using a binary tree is proposed by Bell [3]. The algorithm is quite simple: Keep a ring buffer, which initially contains "space" characters only. Read several letters from the file to the buffer. Then search the buffer for the Longest string that matches the letters just read, and send its length and position in the buffer.
标签： initiated described modified slightly
上传时间： 2014-01-09
上传用户：sk5201314
参考算法导论写的LCS算法，仿照STL的泛型风格，适用于多种STL容器中的各种类型数据构成的序列的最大公共子序列（Longest Common Subsequence）问题求解。
上传时间： 2014-11-22
上传用户：stvnash
这是动态规划中，求最长公共子序列（Longest common string）的源代码。自己编写执行。程序简单，有注释。
标签： 动态规划
上传时间： 2013-12-29
上传用户：xuanjie
This program sets up EVA Timer 1, EVA Timer 2, EVB Timer 3 and EVB Timer 4 to fire an interrupt on a period overflow. A count is kept each time each interrupt passes through the interrupt service routine. EVA Timer 1 has the shortest period while EVB Timer4 has the Longest period.
上传时间： 2013-12-25
上传用户：康郎
