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Description: 最佳矩阵连乘 给定n个矩阵｛A1,A2,…An｝，其中Ai与A i+1是可乘的，i=1，2…，n-1。考察这n个矩阵的连乘积A1A2…An。矩阵A和B可乘的条件是矩阵A的列数等于矩阵B的行数。若A是一个p×q矩阵，B是一个q×r矩阵，则其乘积C=AB是一个p×r矩阵,需要pqr次数乘。
由于矩阵乘法满足结合律，故计算矩阵的连乘积可以有许多不同的计算次序。例如，设3个矩阵{A1,A2,A3}的维数分别为10×100，100×5，和5×50。若按加括号方式((A1A2)A3)计算，3个矩阵连乘积需要的数乘次数为10×100×5+10×5×50=7500。若按加括号方式(A1(A2A3))计算，3个矩阵连乘积总共需要10×5×50+10×100×50=75000次数乘。由此可见，在计算矩阵连乘积时，加括号方式，即计算次序对计算量有很大影响。
矩阵连乘积的最优计算次序问题，即对于给定的相继n个矩阵｛A1,A2,…An｝(其中矩阵Ai的维数为pi-1×p，i=1,2，…，n)，确定计算矩阵连乘积A1,A2,…An的计算次序，使得依此次序计算矩阵连乘积需要的数乘次数最少。
-best matrix continually multiply given n matrix (A1, A2, ... An), Ai and A is a mere i, i = 1, 2 ..., n-1. N explore the link matrix product ... An A1A2. Matrices A and B can either condition is out of the matrix A few matrix B is the number of rows. If A is a p q matrix B is a q-r-matrix, its product C = AB is a p r matrix, the number required by pqr. Because matrix multiplication meet the law of combination, it's even calculated matrix product can be calculated in many different priorities. For example, the matrix-based 3 (A1, A2, A3) dimension of 10 100, 100 5 5 and 50. If bracketed by the way ((A1A2) A3), even three product matrix multiplication in the number of 10 100 10 5 50 = 7,500. If bracketed by the way (A1 (A2A3)), three matrix product even need a total of 10 5 50
Platform: |
Size: 6148 |
Author: 肿事右 |
Hits:
Description: 1.能实现不同的个数的矩阵连乘.
2.最后矩阵大小是8X8.
3是最优的矩阵相乘.
描 述：给定n 个矩阵｛A1, A2,...,An｝，其中Ai与Ai+1是可乘的，i=1，2…，n-1。考察这n个矩阵的连乘积A1A2...An。矩阵A 和B 可乘的条件是矩阵A的列数等于矩阵B 的行数。若A 是一个p x q矩阵，B是一个q * r矩阵，则其乘积C=AB是一个p * r矩阵,需要pqr次数乘。-1. To achieve a number of different matrix continually multiply. 2. The final size of a 8x8 matrix. 3 is the best matrix multiplication. Description : given n matrix (A1, A2 ,..., An), and Ai Ai is a mere, i = 1,2 ..., n-1. N explore the link matrix product ... An A1A2. Matrices A and B can either condition is out of the matrix A few matrix B is the number of rows. If A is a p x q matrix B is a matrix q * r, its product C = AB is a p * r matrix, the number required by pqr.
Platform: |
Size: 2785 |
Author: 林小绵 |
Hits:
Description: This programme is to show the words like a air balloon around the control button.It ia about interface programme.-This program is to show the words like a ai r balloon around the control ia about button.It interface program.
Platform: |
Size: 49317 |
Author: linrui |
Hits:
Description: 最佳矩阵连乘 给定n个矩阵｛A1,A2,…An｝，其中Ai与A i+1是可乘的，i=1，2…，n-1。考察这n个矩阵的连乘积A1A2…An。矩阵A和B可乘的条件是矩阵A的列数等于矩阵B的行数。若A是一个p×q矩阵，B是一个q×r矩阵，则其乘积C=AB是一个p×r矩阵,需要pqr次数乘。
由于矩阵乘法满足结合律，故计算矩阵的连乘积可以有许多不同的计算次序。例如，设3个矩阵{A1,A2,A3}的维数分别为10×100，100×5，和5×50。若按加括号方式((A1A2)A3)计算，3个矩阵连乘积需要的数乘次数为10×100×5+10×5×50=7500。若按加括号方式(A1(A2A3))计算，3个矩阵连乘积总共需要10×5×50+10×100×50=75000次数乘。由此可见，在计算矩阵连乘积时，加括号方式，即计算次序对计算量有很大影响。
矩阵连乘积的最优计算次序问题，即对于给定的相继n个矩阵｛A1,A2,…An｝(其中矩阵Ai的维数为pi-1×p，i=1,2，…，n)，确定计算矩阵连乘积A1,A2,…An的计算次序，使得依此次序计算矩阵连乘积需要的数乘次数最少。
-best matrix continually multiply given n matrix (A1, A2, ... An), Ai and A is a mere i, i = 1, 2 ..., n-1. N explore the link matrix product ... An A1A2. Matrices A and B can either condition is out of the matrix A few matrix B is the number of rows. If A is a p q matrix B is a q-r-matrix, its product C = AB is a p r matrix, the number required by pqr. Because matrix multiplication meet the law of combination, it's even calculated matrix product can be calculated in many different priorities. For example, the matrix-based 3 (A1, A2, A3) dimension of 10 100, 100 5 5 and 50. If bracketed by the way ((A1A2) A3), even three product matrix multiplication in the number of 10 100 10 5 50 = 7,500. If bracketed by the way (A1 (A2A3)), three matrix product even need a total of 10 5 50
Platform: |
Size: 6144 |
Author: 肿事右 |
Hits:
Description: 1.能实现不同的个数的矩阵连乘.
2.最后矩阵大小是8X8.
3是最优的矩阵相乘.
描 述：给定n 个矩阵｛A1, A2,...,An｝，其中Ai与Ai+1是可乘的，i=1，2…，n-1。考察这n个矩阵的连乘积A1A2...An。矩阵A 和B 可乘的条件是矩阵A的列数等于矩阵B 的行数。若A 是一个p x q矩阵，B是一个q * r矩阵，则其乘积C=AB是一个p * r矩阵,需要pqr次数乘。-1. To achieve a number of different matrix continually multiply. 2. The final size of a 8x8 matrix. 3 is the best matrix multiplication. Description : given n matrix (A1, A2 ,..., An), and Ai Ai is a mere, i = 1,2 ..., n-1. N explore the link matrix product ... An A1A2. Matrices A and B can either condition is out of the matrix A few matrix B is the number of rows. If A is a p x q matrix B is a matrix q* r, its product C = AB is a p* r matrix, the number required by pqr.
Platform: |
Size: 2048 |
Author: 林小绵 |
Hits:
Description: This programme is to show the words like a air balloon around the control button.It ia about interface programme.-This program is to show the words like a ai r balloon around the control ia about button.It interface program.
Platform: |
Size: 2888704 |
Author: |
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Description: robocup中国科技大学2队的代码，守门员代码写得很好，总体写得也是相当不错的-RoboCup China Science and Technology University 2 team code, code written in a very good goalkeeper, on the whole is written in fairly good
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Size: 422912 |
Author: letmecc |
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Description: 问 题：矩阵连乘问题
描 述：给定n个矩阵｛A1,A2,...,An｝，其中Ai与Ai+1是可乘的，i=1，2…，n-1.考察这n个矩阵的连乘积A1A2...An。矩阵A 和B可乘的条件是矩阵A的列数等于矩阵B的行数。若A是一个p x q矩阵，B是一个q * r矩阵，则其乘积C=AB是一个p * r矩阵,需要pqr次数乘。
编程任务：对于给定的相继n个矩阵{A1,A2, ..., An }及其维数，编程计算矩阵连乘积A1A2...An 需要的最少数乘次数。
-Question: Even by the issue of Matrix Description: given n matrices (A1, A2 ,..., An), which Ai and Ai+ 1 is the multiplicative, i = 1,2 ..., n-1. Visit this n Even a matrix product A1A2 ... An. Matrix A and B take the condition that the matrix A the number of rows equal to the number of rows of matrix B. If A is a pxq matrix, B is a q* r matrix, then their product C = AB is a p* r matrix, the need to take the number of pqr. Programming tasks: one after another for a given n matrices (A1, A2, ..., An) and its dimension, programming computing matrix product A1A2 ... An even needed by the number of at least a few.
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Size: 14336 |
Author: Katherine |
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Description: ADT HuffmanTree{
数据对象：D={ai| ai∈CharSet,i=1,2,……,n, n≥0}
数据关系：R={< ai-1, ai > ai-1, ai∈D, ai-1基本操作P：
HuffmanTree() 构造函数
~ HuffmanTree() 析构函数
Initialization(int WeightNum)
操作结果：构造哈夫曼树。
Encoder()
初始条件：哈夫曼树已存在或者哈夫曼树已存到文件中。
操作结果：对字符串进行编码
Decoder()
初始条件：哈夫曼树已存在且已编码。
操作结果：对二进制串进行译码
Print()
初始条件：编码文件已存在。
操作结果：把已保存好的编码文件显示在屏幕
TreePrinting()
初始条件：哈夫曼树已存在。
操作结果：将已在内存中的哈夫曼树以直观的方式显示在终端上
-err
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Size: 4096 |
Author: 罗罗 |
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Description: 探讨了在 Mh T I AB环境中实现遗传算法仿真 的方法 ， 并 以一个 简单的求函数最值的问 题作为遗传算法的应用实铡， 说明遗传算法的全局寻优性及用 M AI I AB实现仿真的可行性。-A me f l ~dt o r e Aa z e g e me f i e t I 皿 i n MKI I AB i s d ~- u s s e d．A ha e t i o ~o p t h r f i z a f i o n p r o b l e m i s p r e s e n t e d
t o d l m： l o ml r a t et h e 龉 l y Ⅱ 面 me t h o d 翘 we l l翘 d e m咖曲越i t h e g l o b a l。 n 】 i 越d 衄 f i mc f ima l i t y g e n e t i c~ a g o- r i t h m
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Size: 97280 |
Author: 阿铁 |
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Description: 针对多目标优化问题，提出一种用于求解多目标优化问题的蚁群算法。-Ai m i n g a t mu l t i — o b j e c t i v e o p t i m iz a t i o n p r o b l e m， t h i s p a p e r p r o p o s e s a n An t C o l o n y A l g o r i t h m( AC A)f o r s o l v i n g Mu l t i — o b j e c t i v e Op t i mi z a t i o n P r o b l e m( MOP ACA) ．
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Size: 226304 |
Author: 薛大伟 |
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Description: 3PCF计算多线程实现
定义：
点集D、R。
定义D中的点为ai∈D，R中的点为bi∈R。
距离：r1、r2、r3、err
求：
满足以下条件的三元组（空间中三角形）的数目
<ai, bm, bn>，|ai-bm|=r1±err且|ai-bn|=r2±err且|bm-bn|=r3±err
原始解法：
对于D中每一点ai，在R中找到与之距离为r1的点集R’，找到与之距离为r2的点集R’’。在点集R’与R’’中，查找两点间距离为r3的点组数目。累加。 -3PCF Multithread computing definition: point sets D, R. The point of the definition of D ai ∈ D, R in the point of bi ∈ R. Distance: r1, r2, r3, err requirements: the triple meet the following conditions (space triangle) the number of <ai, bm, bn> , | Ai-bm | = r1 ± err and | ai-bn | = r2 ± err and | bm-bn | = r3 ± err original solution: For D, each point ai, found in R with the distance r1 point set R ' , find the point with a distance r2 set R' ' . In the point set R ' and R' ' , find the distance between two points in the point group for the number of r3. Accumulation.
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Size: 1024 |
Author: Junki Lee |
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Description: 导弹拦截飞机的源代码，通过算出两者间的距离，时间差，速度，利用拦截算法算出相关数值，从而实现-Aircraft missiles to intercept the source code, through the calculated distance between the two, the time difference, speed, calculated using the relevant numerical algorithm block in order to achieve
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Size: 4348928 |
Author: 哈哈哈 |
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Description: 游戏F.E.A.R（极度恐慌）里运用的基于目标驱动的AI决策框架。即G.O.A.P框架-Game FEAR (extreme fear) in the use of AI-based and goal-driven decision-making framework. Framework that GOAP
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Size: 2377728 |
Author: 刘志勇 |
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Description: C’est avec un grand plaisir que je remercie M Franç ois Costa, M Jean Luc Schanen et M
Faouzi Ben Ammar qui ont accepté de faire partie de mon jury de thèse.
Des remerciements aussi chaleureux vont à mes collègues du CEGELY avec qui j’ai partagé
ces années de travail, je pense à tout(e)s les doctorant(e)s ainsi qu’au personnel permanent (la
liste est longue et je suis sû r qu’ils vont se reconnaî tre facilement).
Je réserve la fin de mes remerciements à ma famille pour leur soutien quotidien.
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Size: 76800 |
Author: marouf araibi |
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Description: C++Builder中国象棋研发，里面含有人工智能象棋研发的基本教程，以及各种算法，搜索算法-C++ Builder development of Chinese chess, chess AI R & D which contains the basic tutorials, and a variety of algorithms, search algorithms
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Size: 14621696 |
Author: 吴鑫强 |
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Description: 目前市售的万年历千篇一律，不仅显示单一，而且在农历显示、星期显示、二十四节气显示都不符合我国的习惯，因此我制作了一款万年历，除克服了上述弊病外，还增加了数九三伏的显示、十二生肖显示、个性化语音正点半点报时、128个节日显示、节日来历自动、红外线遥控播出、家人生日纪念日字幕滚动显示、歌曲播放、电话来电显示等。
●家人生日祝福语提醒（可储存5人）
●重大节日显示（一百多个节日）
●节日来历解说（MP3音质）
●高精度环境温度显示（-50℃--127℃误差<0.1）
●真人正点，半点报时
●数伏，数九，二十四节气显示
●周六，周日突出变色显示
●内嵌SD卡播放器，内存多首世界名
●红外遥控操作
●电话来电显示（可储存20个号码）
●四组闹铃，三级亮度
●自动省电设置，（晚10点关大部分显示）
●软件无限升级，所有设置全部可以个性化
视频http://www.tudou.com/programs/view/jW5akQ0b-AI/
我的邮箱xiujiebaoguo@126.com
-Currently commercially available calendar fits, not only shows a single, and in calendar display, Day of week, twenty-four solar terms shown do not meet our habits, so I made a calendar, in addition to overcome these drawbacks, but also increased the number of Nine dog day of the show, zodiac show the slightest timekeeping personalized voice punctuality, 128 holiday displays, holiday origins automatic, infrared remote broadcast subtitles scroll family birthday celebration, song playback, telephone caller ID and so on.
● birthday greetings to remind family members (5 stores)
● major holiday display (more than one hundred holiday)
● explain the origins of holidays (MP3 audio)
● High ambient temperature display (-50 ℃- 127 ℃ error of <0.1)
● real hours, the half timekeeping
● the number of volts, beginning of winter frost, twenty-four solar terms Display
● Saturday, Sunday highlight color display
● embedded player, SD card, memory, the first world famous multi-
● IR r
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Size: 516096 |
Author: baoguo |
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Description: 组合问题
问题描述：找出从自然数1，2，…，n中任取r个数的所有组合。
采用回溯法找问题的解，将找到的组合以从小到大顺序存于a[0]，a[1]，…，a[r-1]中，组合的元素满足以下性质：
（1） a[i+1]>a，后一个数字比前一个大；
（2） a-i<=n-r+1。 -Combinatorial problems Problem Description: Find out from the natural numbers 1,2, ..., n r the number of either taking all the combinations. Problem using backtracking to find the solution, will find a combination of small to large order to save on a [0], a [1], ..., a [r-1], the combination of elements to meet the following properties: (1) a [i+1]> a, after a large number than the previous one (2) ai < = n-r+1.
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Size: 1024 |
Author: 明达 |
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Description: Book of IA with c++ Game programming with c-Book of IA with c++ Game programming with c++
Platform: |
Size: 3232768 |
Author: jorge lm |
Hits:
Description: 一
种基于概率潜在语义模型 的高分 辨率遥感影像分类方法-A P r obabi I i s t i c L at en t S eman t i c An aI y si s Bas ed CI a s s i f i cat i on f o r Hi g h Re s ol u t i on
Remot el y S en s ed I mag e r y
Platform: |
Size: 577536 |
Author: 卢卡斯 |
Hits: