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 |
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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 |
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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: 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: C++primer第三版中文版，c++经典书籍，Stanley B Lippman著，潘爱民译-C++ Primer third edition of the Chinese version, c++ Classic books, Stanley B Lippman, and PAN Ai-min translation Platform: |
Size: 3993600 |
Author:pj |
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Description: C++Primer中文版 第三版
深入系列
Primer 第三版
著
中中文文版版潘爱民张丽译
Addison-Wesley 中国电力出版社
www.infopower.com.cn
Stanley B Lippman
J o s é e L a j o i e
-C++ Primer Chinese version of the third edition of Primer-depth series in the third edition of the Chinese version translated version of PAN Ai-min Zhang Addison-Wesley Publishing House of China Platform: |
Size: 10724352 |
Author:天佑 |
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Description: 适合C++初学者的入门书籍。深入C++系列，中文版；Stanley B Lippman著；潘爱民 张丽译；中国电力出版社；内容分五篇，共20章，很全面。-For C++ Introductory books for beginners. Depth C++ Series, the Chinese version Stanley B Lippman forward PAN Ai-min Zhang Li translation China Electric Power Publishing House the contents of sub-5, a total of 20 chapters, very comprehensive. Platform: |
Size: 3996672 |
Author:lixiaoyang |
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Description: 人工智能教材中的圆盘问题：【设有3个大小不等的圆盘A,B,C套在一根轴上，每个圆盘上都标有数字1、2、3、4，并且每个圆盘都可以独立的绕轴做逆时针转动，每次转动90度，初始状态S0和目标状态Sg，用深度优先搜索和广度优先搜索求从S0到Sg的路径。】。包括Python代码和C++代码，相关题目可参考百度文库（http://wenku.baidu.com/link?url=s3iHFwJ2zOumzqGPJ_BoaW0cYwOSBUhyAM3_KmhQ_BoZNWSj3tLiVFM8j5ZWmokau5UiE0hs8QSpKxYAirSJEmhg7TieL-0IfYE8eHZGnpW）-AI textbooks disc problem: [features three sizes of discs A, B, C set on a shaft, each disc is marked with figures 1, 2, and Each disc can be independently rotated counter-clockwise around the shaft, each rotated 90 degrees, the initial state S0 and the target state Sg, with a depth-first search and breadth-first search request S0 to Sg path. ]. Including Python code and C++ code, related topics may refer Baidu library (http://wenku.baidu.com/link?url=s3iHFwJ2zOumzqGPJ_BoaW0cYwOSBUhyAM3_KmhQ_BoZNWSj3tLiVFM8j5ZWmokau5UiE0hs8QSpKxYAirSJEmhg7TieL-0IfYE8eHZGnpW) Platform: |
Size: 3072 |
Author:wangkai |
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Description: 人工智能及其应用蔡自兴第五版计算智能部分中遗传算法最优计算的自己编写的求解例题-
39/5000
Réngōng zhìnéng jí qí yìngyòng càizìxìng dì wǔ bǎn jìsuàn zhìnéng bùfèn zhōng yíchuán suànfǎ zuì yōu jìsuàn de zìjǐ biānxiě de qiújiě lìtí
Artificial Intelligence and Its Application Cai Zixing s Fifth Edition Computational Intelligence Part of the Genetic Algorithm Optimization Calculation of the preparation of their own example Platform: |
Size: 10240 |
Author:liudy |
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Description: //使用方法：
//首先创建m_pRobot = new CRobot()对象
//叫分：
//1.初始化手牌到m_pRobot->m_szHandCards数组，手牌按牌面从小到大排序,例:200,300,101,201,102,403,205,311,210,310....,注意数组大小，不要有空值或者0值
//2.调用解析函数m_pRobot->Resolution();
//3.获取叫分m_pRobot->getCallF(),返回0不叫，1一倍 2两倍 3三倍
//出牌：
//1.初始化手牌到m_pRobot->m_szHandCards数组
//2.调用解析函数m_pRobot->Resolution();
//3.设置牌池中的出牌数据：
//a.m_pRobot->m_nLastPutCardIndex最后那个出牌人的座位号[0,1,2]
//b.m_pRobot->m_nLastPutCardNum牌池内牌型有多少张
//c.m_pRobot->m_szLastPutCard牌池内牌型，按牌面从小到大排序(// Method of use:
// First create m_pRobot = new CRobot () object
/ / points:
Initialize the hand to the m_pRobot-> m_szHandCards array. Hands are sorted from small to large, for example: 200, 300, 101, 201, 102, 403, 205, 311, 210, 310.... Pay attention to the size of the array. Do not have empty or zero values.
//2. Call the analytic function m_pRobot - > Resolution ();
//3. Get m_pRobot - > getCallF (), return 0 no call, double 2 times 3 times
/ / playing cards:
//1. Initialize the hand to the m_pRobot-> m_szHandCards array
//2. Call the analytic function m_pRobot - > Resolution ();
//3. Set the license data in the pool:
//a.m_pRobot-> m_nLastPutCardIndex Last Player's Seat Number [0,1,2]
//b.m_pRobot-> m_nLastPutCardNum pool number of cards
//c.m_pRobot-> m_szLastPutCard pool type, sorted from small to large) Platform: |
Size: 12288 |
Author:Los瘦身 |
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