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[Mathimatics-Numerical algorithms] 零件切割问题 DL : 0: 给定一块宽度为W的矩形板，矩形板的高度不受限制。现需要从板上分别切割出n个高度为hi，宽度为wi的矩形零件。切割的规则是零件的高度方向与矩形板的高度方向保持一致。要求求出一种切割法使得所使用的矩形板的高度h最小．用递归及分治法解此问题-given a width of the rectangular plate W, the height of rectangular plate unrestricted. Board is required from the respective cut out n height of the hi, wi width of the rectangular parts. Cutting parts to the rules of the height of the rectangular plate with a high degree of consistent direction. Asked obtained a law made by cutting the use of the rectangular plate height h minimum. Using recursive and partition Solving this problem
Date : 2008-10-13 Size : 1.34kb User : 叶黎
[Mathimatics-Numerical algorithms] my0-1knapsack DL : 0: 给定n 个物品, 物品i重为wi 并且价值为 vi ，背包所能承载的最大容量为 W. 0-1 背包问题即是选择含有着最大总价值的物品的子集且它的容量 ≤W ．用动态规划实现-given n goods, items i weight of wi and value of vi, the backpack can carry a maximum capacity of W. 0-1 knapsack problem that is a choice with a maximum total value of the goods but a subset of the W capacity. Dynamic Programming
Date : 2008-10-13 Size : 1002byte User : 叶黎
[Mathimatics-Numerical algorithms] codeing000 DL : 0: 从空格(ASCII码32)到~(ASCII码126)。表内的第一行与表头相同，下面的每一行都与上一行的内容相同，只是字符相左移动了一个位置。这样，下一行的最后一个字符与上一行的第一个字符相同。为了进行文本编码，可以任意选择一个字符串，称之为编码字符串，也就是常说的密钥。为解释编码方法，我们假设密钥是Walrus，待编码的文本（即常说的明文）是： Meet me in St. Louis 我们在待编码的文本之上重复书写上述密钥，使得其长度与待编码文本相同： WalrusWalrusWalrusWa Meet me in St. Louis 从上述两行文本中按列对应方式依次提取一个字符，可得到多个字符对：WM、ae、le等，这些字符对可用作上表的索引。这样，依次以这些字符对作为索引可从上表查到一系列字符，这些字符就构成了文本编码，即常说的密文。例如，第W行第M列队应得字符是%，因此编码的第一个字符就是%；第a行第e列对应的字符是G；第l行第e列对应的是R。依次进行上述查找操作，可以得到完整的密文 %Grgua=aVauGLol?eiAU 进行相反的操作就可对该文本解码。编写编码/解码程序，可以对文本文件或键盘输入的字符串进行编码/解码，在选择编码解码后，需要提示用户输入密钥。 -from space (ASCII 32) to ~ (ASCII 126). The schedule of the first line with the same head table, each of the rows below the previous line with the same content. Character movement is a curator position. So, the next line of characters with a final party on the first character of the same. To text encoding, can choose an arbitrary string, the string as coding, which is often said that the key. To explain the encoding method, we assume that key is Walrus when coding the text (often said that an express) : Meet me in St.. Louis coding in question above the text written above key repeat, making its length and text encoding the same question : WalrusWalrusWalrusWa Meet me in St.. Louis from the above two OK text counterpart way out by the extraction followed a character, a number of characters
Date : 2008-10-13 Size : 1.83kb User : 梁清华
[Mathimatics-Numerical algorithms] ElasticNet DL : 0: 使用到的参数跟谈到弹性网络的那一章里头所讲的是一样的， ke 则是终止条件。如果 step 被打勾，则程式在每一步之间会暂停 100毫秒（或其他使用者输入的数值）。如果 Random 被打勾，则程式会以系统时间作为乱数产生器的种子数，否则，就以使用者输入的数（ Random 右边那一格）为种子数。你可以利用 load 来载入推销员问题档与其最佳解，如此便可比较弹性网络所找出来的解与最佳解差了多少。 Central, Radius, and Error 这三个参数的前两个，只影响弹性网络的起使位置和大小，对求解没有影响。第三个参数代表城市与网络点之间的容忍距离，也就是说，如果某城市与某网络点之间的距离，小于容忍距离，那就把这个城市当成是被该网络点所拜访。按下小 w按钮会将目前的结果与参数值写到“en.out”这个档案。使得我们可以很方便地来比较不同参数的效果。 -use to turn the parameters of the network flexibility chapter Erlitou have said are the same, ke is the termination conditions. If the step was ticking, every step of the program will be suspended between one hundred milliseconds (or other user input value). If Random was ticking, the program will be the system time as the Random Number Generator the seeds, otherwise, Take the user input (which Random right field) for seeds. You can use load to stall contained salesman problem with the optimal solution, so can be more flexible network to find the best solution and the deficit with a number of questions. Central, Radius, and Error these three parameters of the former two, flexible network affects only the starting position and size so that the solution will not be affected. The third parame
Date : 2008-10-13 Size : 38.76kb User : 林尚义
[Mathimatics-Numerical algorithms] 零件切割问题 DL : 0: 给定一块宽度为W的矩形板，矩形板的高度不受限制。现需要从板上分别切割出n个高度为hi，宽度为wi的矩形零件。切割的规则是零件的高度方向与矩形板的高度方向保持一致。要求求出一种切割法使得所使用的矩形板的高度h最小．用递归及分治法解此问题-given a width of the rectangular plate W, the height of rectangular plate unrestricted. Board is required from the respective cut out n height of the hi, wi width of the rectangular parts. Cutting parts to the rules of the height of the rectangular plate with a high degree of consistent direction. Asked obtained a law made by cutting the use of the rectangular plate height h minimum. Using recursive and partition Solving this problem
Date : 2025-12-29 Size : 1kb User : 叶黎
[Mathimatics-Numerical algorithms] my0-1knapsack DL : 0: 给定n 个物品, 物品i重为wi 并且价值为 vi ，背包所能承载的最大容量为 W. 0-1 背包问题即是选择含有着最大总价值的物品的子集且它的容量 ≤W ．用动态规划实现-given n goods, items i weight of wi and value of vi, the backpack can carry a maximum capacity of W. 0-1 knapsack problem that is a choice with a maximum total value of the goods but a subset of the W capacity. Dynamic Programming
Date : 2025-12-29 Size : 1kb User : 叶黎
[Mathimatics-Numerical algorithms] codeing000 DL : 0: 从空格(ASCII码32)到~(ASCII码126)。表内的第一行与表头相同，下面的每一行都与上一行的内容相同，只是字符相左移动了一个位置。这样，下一行的最后一个字符与上一行的第一个字符相同。为了进行文本编码，可以任意选择一个字符串，称之为编码字符串，也就是常说的密钥。为解释编码方法，我们假设密钥是Walrus，待编码的文本（即常说的明文）是： Meet me in St. Louis 我们在待编码的文本之上重复书写上述密钥，使得其长度与待编码文本相同： WalrusWalrusWalrusWa Meet me in St. Louis 从上述两行文本中按列对应方式依次提取一个字符，可得到多个字符对：WM、ae、le等，这些字符对可用作上表的索引。这样，依次以这些字符对作为索引可从上表查到一系列字符，这些字符就构成了文本编码，即常说的密文。例如，第W行第M列队应得字符是%，因此编码的第一个字符就是%；第a行第e列对应的字符是G；第l行第e列对应的是R。依次进行上述查找操作，可以得到完整的密文 %Grgua=aVauGLol?eiAU 进行相反的操作就可对该文本解码。编写编码/解码程序，可以对文本文件或键盘输入的字符串进行编码/解码，在选择编码解码后，需要提示用户输入密钥。 -from space (ASCII 32) to ~ (ASCII 126). The schedule of the first line with the same head table, each of the rows below the previous line with the same content. Character movement is a curator position. So, the next line of characters with a final party on the first character of the same. To text encoding, can choose an arbitrary string, the string as coding, which is often said that the key. To explain the encoding method, we assume that key is Walrus when coding the text (often said that an express) : Meet me in St.. Louis coding in question above the text written above key repeat, making its length and text encoding the same question : WalrusWalrusWalrusWa Meet me in St.. Louis from the above two OK text counterpart way out by the extraction followed a character, a number of characters
Date : 2025-12-29 Size : 2kb User : 梁清华
[Mathimatics-Numerical algorithms] ElasticNet DL : 0: 使用到的参数跟谈到弹性网络的那一章里头所讲的是一样的， ke 则是终止条件。如果 step 被打勾，则程式在每一步之间会暂停 100毫秒（或其他使用者输入的数值）。如果 Random 被打勾，则程式会以系统时间作为乱数产生器的种子数，否则，就以使用者输入的数（ Random 右边那一格）为种子数。你可以利用 load 来载入推销员问题档与其最佳解，如此便可比较弹性网络所找出来的解与最佳解差了多少。 Central, Radius, and Error 这三个参数的前两个，只影响弹性网络的起使位置和大小，对求解没有影响。第三个参数代表城市与网络点之间的容忍距离，也就是说，如果某城市与某网络点之间的距离，小于容忍距离，那就把这个城市当成是被该网络点所拜访。按下小 w按钮会将目前的结果与参数值写到“en.out”这个档案。使得我们可以很方便地来比较不同参数的效果。 -use to turn the parameters of the network flexibility chapter Erlitou have said are the same, ke is the termination conditions. If the step was ticking, every step of the program will be suspended between one hundred milliseconds (or other user input value). If Random was ticking, the program will be the system time as the Random Number Generator the seeds, otherwise, Take the user input (which Random right field) for seeds. You can use load to stall contained salesman problem with the optimal solution, so can be more flexible network to find the best solution and the deficit with a number of questions. Central, Radius, and Error these three parameters of the former two, flexible network affects only the starting position and size so that the solution will not be affected. The third parame
Date : 2025-12-29 Size : 38kb User : 林尚义
[Mathimatics-Numerical algorithms] apsignalproc DL : 0: The Window Design Method The basic idea behind the design of linear-phase FIR filters using the window method is to choose a proper ideal frequency-selective filter [which always has a noncausal, infinite duration impulse response] and then truncate its impulse response hd[n] to obtain a linear-phase and causal FIR filter h[n]. To truncate the impulse response of the ideal filter a time window w[n] is used. Available windows in Matlab are rectangular [or boxcar in Matlab], bartlett, hamming, hanning
Date : 2025-12-29 Size : 857kb User : Rizwan
[Mathimatics-Numerical algorithms] w DL : 0: 单原点最短路径问题算法，在这个问题中，给出有向图G，它的每条边都有一个非负的长度（耗费） a [i ][ j ]，路径的长度即为此路径所经过的边的长度之和。对于给定的源顶点s，需找出从它到图中其他任意顶点（称为目的）的最短路径。-Single-origin issue of the shortest path algorithm, in this issue, given directed graph G, each of which has a non-negative edge length (cost) a [i] [j], that is, the length of the path for this path by After the length of the edge of and. For a given source vertex s, find the need to map from its other arbitrary vertex (known as the purpose of) the shortest path.
Date : 2025-12-29 Size : 25kb User : 韩鑫
[Mathimatics-Numerical algorithms] abc DL : 0: 单循环赛中选手胜负序列求解问题名称：单循环赛中选手胜负序列求解问题内容：有n个选手 P 1 ,P 2 ,P 3 ,… ,P n 参加了的单循环赛，每对选手之间非胜即负。现要求求出一个选手序列 P 1 ,P 2 ,P 3 ,… ,P n , 使其满足 P i 胜 P i+ 1 (i=1,… ,n-1) 。 -Single round robin tournament players winning sequence to solve the problem name: the outcome of single round-robin tournament players to solve the problem sequence elements: There are n-player P 1, P 2, P 3, ..., P n participated in a single round-robin between each pair of players Non-win or negative. Players are now required to find a sequence of P 1 ' , P 2' , P 3 ' , ..., P n' , to meet the P i ' W P i+ 1' (i = 1, ..., n-1).
Date : 2025-12-29 Size : 142kb User : yang
[Mathimatics-Numerical algorithms] fastKICA DL : 0: 盲源分离FastICA、matalab程序，-FAST KERNEL ICA | |--------------------| Version 1.0- February 2007 Copyright 2007 Stefanie Jegelka, Hao Shen, Arthur Gretton This package contains a Matlab implementation of the Fast Kernel ICA algorithm as described in [1]. Kernel ICA is based on minimizing a kernel measure of statistical independence, namely the Hilbert-Schmidt norm of the covariance operator in feature space (see [3]: this is called HSIC). Given an (n x m) matrix W of n samples from m mixed sources, the goal is to find a demixing matrix X such that the dependence between the estimated unmixed sources X *W is minimal. FastKICA uses an approximate Newton method to perfom this optimization. For more information on the algorithm, read [1], and for more information on HSIC, refer to [3]. The functions chol_gauss and amariD are taken from and based on, respectively, code from Francis Bach (available at http://cmm.ensmp.fr/~bach/kernel-ica/index.htm). The derivative is com
Date : 2025-12-29 Size : 95kb User : wanghqiang
[Mathimatics-Numerical algorithms] tiaoshi DL : 0: 需要求3个长方柱的体积，请编写一个基于对象的程序。数据成员包括length（长）、 width（宽）、height（高）。要求用成员函数实现以下功能： 1、由键盘分别输入3个长方柱的长、宽、高； 2、计算长方柱的体积； 3、输出3个长方柱体积；请编写程序，上机调试并运行。-Demand for the volume of 3 rectangular column, please write a object-based programs. Data members, including length (length), width (W), height (H). Require member functions to achieve the following functions: 1, 3 are input from the keyboard long rectangular column, width, height 2, calculate the volume of rectangular column 3, the output of three rectangular column volumes write the program, the Machine commissioning and operation.
Date : 2025-12-29 Size : 1kb User : 王帝
[Mathimatics-Numerical algorithms] SVD-TLS-LS DL : 0: SVD-TLS和LS方法估计AR模型功率谱,假定仿真的观测数据由产生，其中w(n)是一高斯白噪声，其均值为0，方差为1，并取n=1,…..,128，这里分别用LS方法和SVD-TLS方法估计观测数据的AR模型参数。-SVD-TLS and the LS method to estimate AR model for power spectrum, assuming that the simulation produced by the observed data, where w (n) is a Gaussian white noise with mean 0 and variance 1, and take n = 1, ... .. , 128, LS methods were used here and SVD-TLS method to estimate the AR model parameters observed data.
Date : 2025-12-29 Size : 1kb User : 云卷云舒
[Mathimatics-Numerical algorithms] modified-stfd_esprit DL : 0: 提出了基于修正空间时频分布( S TF D) 矩阵的 ES P RI T算法以实现对宽带线性调频信号的到达角估计-Th e a l g or i t h m f o r di r e c t i o n- o f- a r r i va l o f t he wi d e ba n d c hi r p s i gna l s ba s e d 0 1 3 ．ESPRI T u s i n g t he mo di f i e d s p a t i a l t i me- f r e q ue n c y ma t r i x i S pr e s e nt e d．The mod i f i e d STFD ma t r i x whi c h h as t he s i mi l ar ma t he ma t i c a l c on st r uc t i o n wi t h t he c o va r i a nc e ma t r i x c a n be ob t a i ne d wi t h t h e c r o s s W i gn e r- Vi l l e di s t r i but i o ns o f t he o ut pu t s of t he ar r a y．Unde r t he c o nd i t i on of un i f o r m l i n e a r r a y，t he mo di f i e d STFD ma t r i x c a n b e t r a n s f or me d i nt o t he ma t r i x wh i c h h a s t he pr op e r t y of r ot a t i on a l i nva r i a nc e ． The n t he ESPRI T c a n be a p pl i e d t o D O A e s t i ma t i on
Date : 2025-12-29 Size : 142kb User : fjp119
[Mathimatics-Numerical algorithms] Statistical-letters-number DL : 0: Description: 给定一段文章，请输出每个字母出现的次数。 Input: 只有一组输入数据，该数据大小<10KB。在文章中除最后一个字符外，只有小写字母、空格和换行符，没有另外的标点、数字和大写字母等。该文章以’#’结尾。 Output: 输出格式为“C A”，C为’a’..’z’中的字母，A为出现次数，C和A之间空一格。 Sample Input: here is the input this is the article# Sample Output: a 1 b 0 c 1 d 0 e 5 f 0 g 0 h 4 i 5 j 0 k 0 l 1 m 0 n 1 o 0 p 1 q 0 r 2 s 3 t 5 u 1 v 0 w 0 x 0 y 0 z 0 -Description: for a given section of the article, please output the number of occurrences of each letter. Input: only one set of input data, the data size & lt10KB. In the article, except the last character, only lowercase letters, spaces and line breaks, no other punctuation, numbers and uppercase letters. The article in ' #' at the end. Output: Output format is " CA" , C as ' a' .. ' z' in the letters, A is the number of occurrences, C and A space between. Sample Input: here is the input this is the article# Sample Output: a 1 b 0 c 1 d 0 e 5 f 0 g 0 h 4 i 5 j 0 k 0 l 1 m 0 n 1 o 0 p 1 q 0 r 2 s 3 t 5 u 1 v 0 w 0 x 0 y 0 z 0
Date : 2025-12-29 Size : 9kb User : 沙魄郎君
[Mathimatics-Numerical algorithms] free-pie DL : 0: 游戏在一个舞台上进行。舞台的宽度为W格，天幕的高度为H格，游戏者占一格。开始时游戏者站在舞台的正中央，手里拿着一个托盘。下图为天幕的高度为4格时某一个时刻游戏者接馅饼的情景。-On a stage in the game. The width W cell stage, the canopy height H cells, accounted for a game grid. The beginning of the game who stood center stage, holding a tray. Below the canopy height of 4 cells at a time when the player received a pie scene.
Date : 2025-12-29 Size : 1kb User : tsxlyh
[Mathimatics-Numerical algorithms] 8数码 DL : 0: 需要说明的是：本文示例图中的目标状态在计算上是最快捷的，首先取数很方便，一般地查看目标在第i个位置上的值，则需要访问数组goal[i]，而这里goal[i]==i，故而无需访问数组；第二，要想知道数码n的目标位置，则需要找到goal[i]==n，然后row=i/3,col=i%3. 但是这里的话，row=n/3,col=n%3. 当我们用8位无符号整型来表示各个数码值(0~8)时，n/3和n%3操作是非常快的，比访问数组还快。程序还提供一种玩游戏模式，即用户自己通过W,S,A,D四个键分别控制空格往上、下、左、右四个方向移动一步，功能很简单，界面部分是用OpenCV做的。界面部分和搜索部分是完全分离的，demo文件中main函数部分显示了如何将它们组合使用。(Need explanation is: This paper sample figure in the target state is the most efficient in calculation, the first number is very convenient, generally view the targets in the I position on the value, requires access to an array of goal[i], where goal[i]==i, therefore no need to access the array; second, to know the digital n the position of the target, you need to find goal[i]==n, row=i/3 and col=i%3., but here, row=n/3, col=n%3. when we use 8 bit unsigned integer to the digital value (0~8), n/3 and n%3 operations are very fast, faster than accessing array. Program also provides a play game mode, that is, users themselves through W, S, A, D, four keys, respectively, control space, up, down, left, right four directions move a step, the function is very simple, the interface part is done with OpenCV. The interface section is completely separate from the search section, and the main function section in the demo file shows how to combine them.)
Date : 2025-12-29 Size : 456kb User : llii
[Mathimatics-Numerical algorithms] 07 极限学习机（Extreme Learning Machine, ELM） DL : 0: ELM算法指出，其实隐层的权值矩阵W和偏置b其实是没有必要调整的，在学习算法开始时，任意随机给定W和b的值，利用其计算出H（隐层节点的输出），并令其保持不变，需要确定的参数就只有β了。这是一个比较重要的理论基础。(The ELM algorithm is pointed out, in fact, hidden layer weights matrix W and B is not necessary to adjust the bias of the learning algorithm, in the beginning, randomly given W and b values, the calculated H (output nodes), and remained unchanged, only need to determine the parameters of the beta. This is a very important theoretical basis.)
Date : 2025-12-29 Size : 690kb User : ZJN27
[Mathimatics-Numerical algorithms] idocdown_33lc.com DL : 0: 帮助学习分水岭变换是一种常用有效的图像分割方法。针对分水岭的过分割问题,对分水岭分割后的图像,(Study aid A ccordin g tO the analysis of eigenvalue and relativity it w as fcIund that ban ds of T M 3 T M 4 and T M 5 had sm all relativity and contain am ple inform ation)
Date : 2025-12-29 Size : 4.85mb User : 发光管3