Problem B:Longest Ordered Subsequence A numeric sequence of ai is ordered if a1 < a2 < ... &l

相 关 资 源

• Problem B:Longest Ordered Subsequence A numeric sequence of ai is ordered if a1 < a2 < ... &l免费下载

  资源简介：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. ...

  标签： Subsequence sequence Problem Longest

  上传时间： 2016-12-08

  上传用户：busterman

• This example shows how to perform a repeated sequence of conversions using // "repeat sequence-of-c免费下载

  资源简介：This example shows how to perform a repeated sequence of conversions using // "repeat sequence-of-channels" mode.

  标签： sequence-of-c conversions repeated sequence

  上传时间： 2016-09-04

  上传用户：gxmm

• A star Algorithm of AI免费下载

  资源简介：A star Algorithm of AI

  标签： Algorithm star AI of

  上传时间： 2016-03-20

  上传用户：stampede

• A new type of cloak is discussed: one that gives all cloaked objects the appearance of a flat conduc免费下载

  资源简介：A new type of cloak is discussed: one that gives all cloaked objects the appearance of a flat conducting sheet. It has the advantage that none of the parameters of the cloak is singular and can in fact be made isotropic. It makes broadban...

  标签： appearance discussed cloaked objects

  上传时间： 2017-03-30

  上传用户：rishian

• Presents a general picture of what is inside a Unix kernel and how Linux competes against other well免费下载

  资源简介：Presents a general picture of what is inside a Unix kernel and how Linux competes against other well-known Unix systems.

  标签： Presents competes general against

  上传时间： 2017-04-13

  上传用户：frank1234

• If you are a C++ programmer who desires a fuller understanding of what is going on "under the hood,"免费下载

  资源简介：If you are a C++ programmer who desires a fuller understanding of what is going on "under the hood," then Inside the C++ Object Model is for you! Inside the C++ Object Model focuses on the underlying mechanisms that support object-orient...

  标签： understanding programmer desires fuller

  上传时间： 2017-09-25

  上传用户：gtzj

• Problem Statement You are given a string input. You are to find the Longest substring of input su免费下载

  资源简介：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. De...

  标签： input Statement You are

  上传时间： 2015-09-21

  上传用户：sunjet

• LCS Algorithm, this is a c++ code for lcs(Longest Common Subsequence)免费下载

  资源简介：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必备习题免费下载

  资源简介：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

• 最长子序列,英文版,电子书 ACM文章1977+Algorithms+for+the+Longest+Common+Subsequence+Problem免费下载

  资源简介：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. 免费下载

  资源简介：Instead of finding the Longest common Subsequence, let us try to determine the length of the LCS. &#1048708 Then tracking back to find the LCS. &#1048708 Consider a1a2…am and b1b2…bn. &#1048708 Case 1: am=bn. The LCS must contain am,...

  标签： the Subsequence determine Instead

  上传时间： 2013-12-17

  上传用户：evil

• Generating next numbers in SQLServer should not be a Problem. But Problems arise when a customer ask免费下载

  资源简介：Generating next numbers in SQLServer should not be a Problem. But Problems arise when a customer asks for different types of next numbers that you cannot generate directly from SQL Server. This brief article describes how you would tackle t...

  标签： Generating SQLServer customer Problems

  上传时间： 2015-01-11

  上传用户：as275944189

• This an adaptive receiver for a direct-sequence spread spectrum (DS-SS) system over an AWGN channel.免费下载

  资源简介：This an adaptive receiver for a direct-sequence spread spectrum (DS-SS) system over an AWGN channel. The adaptive receiver block is modified from the LMS adaptive filter block in DSP Blockset. For DS-SS signal reception, the adaptive filter...

  标签： direct-sequence adaptive receiver spectrum

  上传时间： 2014-01-16

  上传用户：D&L37

• Evaluate a sequence of Bessel functions of the first and second kinds and their derivatives with i免费下载

  资源简介：Evaluate a sequence of Bessel functions of the first and second kinds and their derivatives with integer orders and real arguments

  标签： derivatives functions and Evaluate

  上传时间： 2015-05-27

  上传用户：685

• Fortran program: Evaluate a sequence of exponential integrals En(x)免费下载

  资源简介：Fortran program: Evaluate a sequence of exponential integrals En(x)

  标签： exponential integrals Evaluate sequence

  上传时间： 2015-05-27

  上传用户：sardinescn

• ％直接型到并联型的转换 ％ ％[C,B,A]=dir2par(b,a) ％C为当b的长度大于a时的多项式部分 ％B为包含各bk的K乘2维实系数矩阵 ％A为包含各ak的K乘3维实系数矩阵 ％免费下载

  资源简介：％直接型到并联型的转换 ％ ％[C,B,A]=dir2par(b,a) ％C为当b的长度大于a时的多项式部分 ％B为包含各bk的K乘2维实系数矩阵 ％A为包含各ak的K乘3维实系数矩阵 ％b为直接型分子多项式系数 ％a为直接型分母多项式系数 ％

  标签： dir par 系数 矩阵

  上传时间： 2014-01-20

  上传用户：lizhen9880

• ACM试题Problem K:Ones Description Given any integer 0 <= n <= 10000 not divisible by 2 or 5,免费下载

  资源简介：ACM试题Problem K:Ones Description Given any integer 0 <= n <= 10000 not divisible by 2 or 5, some multiple of n is a number which in decimal notation is a sequence of 1 s. How many digits are in the smallest such a multiple of n?

  标签： Description divisible Problem integer

  上传时间： 2015-08-23

  上传用户：zhenyushaw

• Distributed Median,Alice has an array A, and Bob has an array B. All elements in A and B are distinc免费下载

  资源简介：Distributed Median,Alice has an array A, and Bob has an array B. All elements in A and B are distinct. Alice and Bob are interested in finding the median element of their combined arrays.

  标签： array B. Distributed has

  上传时间： 2013-12-25

  上传用户：洛木卓

• Pai has been a good sequence array. A few are imported, as called for under the original law will in免费下载

  资源简介：Pai has been a good sequence array. A few are imported, as called for under the original law will insert its array

  标签： sequence imported original called

  上传时间： 2014-01-14

  上传用户：凤临西北

• // 带有列主元的高斯消元法 // 功能：　求解线性方程组 Ax = b // 参数：　A － 指向n*n系数矩阵的指针 // 　　　　b － 常数向量的指针 // 　　　　n － 方程组的维免费下载

  资源简介：// 带有列主元的高斯消元法 // 功能：　求解线性方程组 Ax = b // 参数：　A － 指向n*n系数矩阵的指针 // 　　　　b － 常数向量的指针 // 　　　　n － 方程组的维数 // 返回值：0 － 如果成功。线性方程组的解保存在 b 中 // 　　　　1 － 求解失败

  标签： 方程 指针 Ax 高斯

  上传时间： 2013-12-18

  上传用户：xcy122677

• Finds a (near) optimal solution to the Traveling Salesman Problem (TSP) by setting up a Genetic Algo免费下载

  资源简介：Finds a (near) optimal solution to the Traveling Salesman Problem (TSP) by setting up a Genetic Algorithm (GA) to search for the shortest path (least distance needed to travel to each city exactly once)

  标签： Traveling Salesman solution Problem

  上传时间： 2013-12-04

  上传用户：从此走出阴霾

• n皇后问题求解(8<=n<=1000) a) 皇后个数的设定 在指定文本框内输入皇后个数即可,注意: 皇后个数在8和1000 之间(包括8和1000) b) 求解 点击<免费下载

  资源简介：n皇后问题求解(8<=n<=1000) a) 皇后个数的设定 在指定文本框内输入皇后个数即可,注意: 皇后个数在8和1000 之间(包括8和1000) b) 求解 点击<Solve>按钮即可进行求解. c) 求解过程显示 在标有Total Collision的静态文本框中将输出当前棋盘上的皇后总冲...

  标签： 1000 lt 设定 输入

  上传时间： 2016-01-28

  上传用户：ztj182002

• 遗传算法和“货郎担” 问题: "The traveling salesman Problem, or TSP for short, is this: given a finite number of免费下载

  资源简介：遗传算法和“货郎担” 问题: "The traveling salesman Problem, or TSP for short, is this: given a finite number of cities along with the cost of travel between each pair of them, find the cheapest way of visiting all the cities and returning ...

  标签： traveling salesman Problem finite

  上传时间： 2013-12-24

  上传用户：watch100

• Description Calculate a+b Input Two integer a,b (0<=a,b<=101000) Output Outpu免费下载

  资源简介：Description Calculate a+b Input Two integer a,b (0<=a,b<=101000) Output Output a + b Sample Input 5 7 Sample Output 12

  标签： Description Calculate integer 101000

  上传时间： 2014-01-25

  上传用户：tonyshao

• RSA核心运算使用的乘模算法就是 M(A*B)。虽然M(A*B)并不是乘模所需要的真正结果免费下载

  资源简介：RSA核心运算使用的乘模算法就是 M(A*B)。虽然M(A*B)并不是乘模所需要的真正结果，但只要在幂模算法中进行相应的修改，就可以调用这个乘模算法进行计算了。本软件起初未使用Montgomery 乘模算法时，加密速度比使用Montgomery乘模算法慢，但速度相差不到一个数...

  标签： RSA 模 核心 运算

  上传时间： 2016-07-16

  上传用户：hullow

• 序列对齐 Compare a protein sequence to a protein sequence database or a DNA sequence to a DNA sequenc免费下载

  资源简介：序列对齐 Compare a protein sequence to a protein sequence database or a DNA sequence to a DNA sequence database using the Smith-Waterman algorithm.[.wat815.] \fCssearch3\fP is about 10-times slower than FASTA3, but is more sensitive ...

  标签： sequence protein DNA database

  上传时间： 2016-08-01

  上传用户：wangchong

• 计算满足a<=x<=b且是Fibonacci数的x个数。免费下载

  资源简介：计算满足a<=x<=b且是Fibonacci数的x个数。

  标签： Fibonacci lt 计算

  上传时间： 2013-12-11

  上传用户：Shaikh

• （一） 求a~b 之间各个数的约数个数之和。(其中包括a和b在内) ans = sigma(f(i)) , (a <= i <= b) , 其中f(i)表示i的约数的个数免费下载

  资源简介：（一） 求a~b 之间各个数的约数个数之和。(其中包括a和b在内) ans = sigma(f(i)) , (a <= i <= b) , 其中f(i)表示i的约数的个数

  标签： sigma lt ans

  上传时间： 2016-12-31

  上传用户：daoxiang126

• 数值分析高斯——列主元消去法主程序 说明如下： ％ a----input,matrix of coefficient ％ b----input,right vector ％ sol----o免费下载

  资源简介：数值分析高斯——列主元消去法主程序 说明如下： ％ a----input,matrix of coefficient ％ b----input,right vector ％ sol----output,returns the solution of linear equation

  标签： input coefficient matrix vector

  上传时间： 2017-01-01

  上传用户：dancnc

• This thesis presents a comprehensive overview of the Problem of facial recognition. A survey of avai免费下载

  资源简介：This thesis presents a comprehensive overview of the Problem of facial recognition. A survey of available facial detection algorithms as well as implementation and tests of di铿€erent feature extraction and dimensionality reduction methods...

  标签： comprehensive recognition presents overview

  上传时间： 2017-05-05

  上传用户：royzhangsz