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[Console] PROGRAMMIFA DL : 0: 幂法求矩阵的特征值和特征向量算法9.1 1．输入矩阵A，初始向量X，误差限ε最大迭代次数N。 2．置k=1，u=0。 3．求xr=>λ，|xr|=max|xr|,i=1,….,n。 4．计算 Y=X/λ，X=AY 5．若|λ-u|〈ε，输出λ，X，停机；否则，转步骤6。 6. 若k<N，k+1=>k，λ=>u,转步骤3；否则，输出错误信息，停机
Date : 2008-10-13 Size : 18.66kb User : Gellss
[Console] PROGRAMMIFA DL : 0: 幂法求矩阵的特征值和特征向量算法9.1 1．输入矩阵A，初始向量X，误差限ε最大迭代次数N。 2．置k=1，u=0。 3．求xr=>λ，|xr|=max|xr|,i=1,….,n。 4．计算 Y=X/λ，X=AY 5．若|λ-u|〈ε，输出λ，X，停机；否则，转步骤6。 6. 若k<N，k+1=>k，λ=>u,转步骤3；否则，输出错误信息，停机 -Power-law matrix eigenvalue and eigenvector algorithm 9.11. Input matrix A, the initial vector X, the error ε maximum iteration limit N. 2. Home k = 1, u = 0. 3. For xr => λ, | xr | = max | xr |, i = 1, ...., N. 4. Calculation Y = X/λ, X = AY5. If | λ-u | <ε, the output λ, X, downtime Otherwise, switch to step 6.6. If k <N,k+1=> k, λ => u, to Step 3 Otherwise, output an error message , parking
Date : 2025-12-28 Size : 18kb User : Gellss
[Console] Matching_on_Weighted_Bipartite_Graph DL : 0: 加权二分图匹配，对于给定权重的二分图求出最大匹配-find the max matching of a given weighted bipartite graph
Date : 2025-12-28 Size : 1kb User : LS
[Console] Proiect DL : 0: A graphics application, made in Visual C++ and using the glut library. Models created in 3ds Max and saved in .obj format.
Date : 2025-12-28 Size : 14.76mb User : fpganoob
[Console] prio DL : 0: 　　优先队列是0个或多个元素的集合,每个元素都有一个优先权或值,对优先队列执行的操作有1) 查找 2) 插入一个新元素 3) 删除.在最小优先队列(min priorityq u e u e)中,查找操作用来搜索优先权最小的元素,删除操作用来删除该元素对于最大优先队列(max priority queue),查找操作用来搜索优先权最大的元素,删除操作用来删除该元素.优先权队列中的元素可以有相同的优先权,查找与删除操作可根据任意优先权进行. -Priority queue is 0 or more elements of the set, each element has a priority or value, on the implementation of priority queue operations are 1) search 2) insert a new element 3) Delete. At a minimum priority queue ( min priorityq ueue), the search operation to search for the smallest element of priority, the delete operation to remove the element to the maximum priority queue (max priority queue), the largest search operation to search the priority elements of the delete operation to Remove the element. priority queue elements can have the same priority, search and delete operation can be carried out according to any priority.
Date : 2025-12-28 Size : 1.03mb User : bob
[Console] ParsePcap DL : 0: 使用VS2008进行开发，对WireShack抓包中的发包时间进行分析，得到min/max/avg/mdev值以判断网络或者系统的负载情况，pcap包需要用Wireshack导出成txt格式才能进行分析。-Using VS2008 to develop contract WireShack capture, analyze, and get min/max/avg/mdev value to determine the network or system load, the pcap packet using Wireshack export txt format can be analyzed.
Date : 2025-12-28 Size : 1.47mb User : 李一文
[Console] max-min DL : 0: max-min聚类算法实现，机器学习方面的经典算法，值得一看-Realization of max-min clustering algorithm, classical machine learning algorithms, worth a visit
Date : 2025-12-28 Size : 3.56mb User : 何凯
[Console] Seek-the-max-and-min DL : 0: 用递归求一个数组中的最大数和最小值，算法思想是从数组的最后一个值向前比较-Recursive number of maximum and minimum values in an array, the algorithm thinking forward from the last value of the array
Date : 2025-12-28 Size : 81kb User : 辛文飞
[Console] 20112219_04_02 DL : 0: 数据结构的实验，C语言编写的一个52张牌，洗牌后发给四家，每一家手里的牌可以按照斗地主方式排序（3最小2最大），或者按照花色和牌面大小排序的小程序。-Experimental data structures, C language, a 52 cards, shuffle issued after four, each one can follow his cards Landlords sort (3 min 2 max), or in accordance with the color and size of the sign face sort of applet.
Date : 2025-12-28 Size : 1kb User : 龙常
[Console] maxflow-v3.01 DL : 0: 最新版本的max flow 算法，功能很强大很有作用-max—flow algorithm
Date : 2025-12-28 Size : 14kb User : dongdong chen
[Console] FindingPixelAverage-(2) DL : 0: Compute min max values of image pixels in image processing
Date : 2025-12-28 Size : 1kb User : choopan
[Console] MAXInt DL : 0: max int , simple project to demo
Date : 2025-12-28 Size : 1.33mb User : New Mem
[Console] Eight-queen DL : 0: 八皇后问题，是一个古老而著名的问题，是回溯算法的典型案例。该问题是国际西洋棋棋手马克斯·贝瑟尔于1848年提出：在8X8格的国际象棋上摆放八个皇后，使其不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上，问有多少种摆法-Eight queens problem is an old and famous, is a typical case of backtracking algorithm The problem is the international chess player Max bethel in 1848: in the 8 by 8 lattice chess put eight queen, make its can t attack each other, that any two queens are not in the same line on the same column or the same slash and asked how many kinds of pendulum method
Date : 2025-12-28 Size : 3.82mb User : zhuobengang
[Console] MaksMin DL : 0: Find the max and min number in an array
Date : 2025-12-28 Size : 11kb User : Mrt
[Console] Non-linear-filters DL : 0: Apply non-linear filters(Median, max, min filters).
Date : 2025-12-28 Size : 5.54mb User : Himali weerasinghe
[Console] FindStudent DL : 0: 把长度为l1,l2…ln 的n个程序放在磁带T1和T2上，并且希望按照使用最大检索时间取得最小值的方式存储，即如果存放在T1和T2上的程序集合分别为A和B，则希望所选择的A和B使得max{∑li 1,∑li2}(i1属于A，i2属于B)取得最小。使用回溯法实现-The length of L1, L2... Ln n programs put in the tape on T1 and T2, and hope that in accordance with the maximum search time to obtain the minimum value is stored, that is, if stored in T1 and T2 procedures set respectively a and B, hope that the choice of a and B the max {sigma Li 1, sigma li2} (I1 and belong to a, I2 b) has the smallest. Realization of using backtracking
Date : 2025-12-28 Size : 4.24mb User : 321
[Console] TSPSearch DL : 1: c++使用禁忌算法解决MAX-CUT问题-c++ use tabu search algorithm to solve the problem of MAX-CUT
Date : 2025-12-28 Size : 1kb User : 郭俊虎
[Console] divmnu.c DL : 0: Computes the n-m+1-halfword quotient and m-halfword remainder of n halfwords divided by m halfwords, unsigned. Max line length is 57, to fit in hacker.book.
Date : 2025-12-28 Size : 3kb User : vocesper
[Console] Max-Multiply DL : 0: i.最大K乘积问题问题描述设I是一个n位十进制整数。如果将I划分为k段，则可得到k个整数。这k个整数的乘积称为I的一个k乘积。试设计一个算法，对于给定的I和k，求出I的最大k乘积。例如十进制整数 1234 划分为 3 段可有如下情形： 1 × 2 × 34 68 1 × 23 × 4 92 12 × 3 × 4 144 编程任务对于给定的I 和k，编程计算I 的最大k 乘积。数据输入输入的第1 行中有2个正整数n和k。正整数n是序列的长度；正整数k是分割的段数。接下来的一行中是一个n位十进制整数。（n< 10）结果输出计算出的最大k乘积。输入文件示例输出文件示例 input.txt output.txt 3 2 312 62 实现提示参考矩阵连乘问题。 -That is a difficult problem which i used dynamic ways to solve it.
Date : 2025-12-28 Size : 1kb User : 武磊