Thinking in Java, 3rd ed. Revision 4.0 Preface Introduction 1: Introduction to Objects 2: Everything is an Object 3: Controlling Program Flow 4: Initialization & Cleanup 5: Hiding the Implementation 6: Reusing Classes 7: Polymorphism 8: Interfaces & Inner Classes 9: Error Handling with Exceptions 10: Detecting Types 11: Collections of Objects 12: The Java I/O System 13: Concurrency 14: Creating Windows & Applets 15: Discovering Problems 16: Analysis and Design A: Passing & Returning Objects B: Java Programming Guidelines C: Supplements D: Resources Index
标签： Introduction Thinking Revision Preface
上传时间： 2014-07-13
上传用户：netwolf
The "GEE! It s Simple" package illustrates Gaussian elimination with partial pivoting, which produces a factorization of P*A into the product L*U where P is a permutation matrix, and L and U are lower and upper triangular, respectively. The functions in this package are accurate, but they are far slower than their MATLAB equivalents (x=A\b, [L,U,p]=lu(A), and so on). They are presented here merely to illustrate and educate. "Real" production code should use backslash and lu, not this package.
标签： illustrates elimination Gaussian pivoting
上传时间： 2016-11-09
上传用户：wang5829
The "GEE! It s Simple" package illustrates Gaussian elimination with partial pivoting, which produces a factorization of P*A into the product L*U where P is a permutation matrix, and L and U are lower and upper triangular, respectively. The functions in this package are accurate, but they are far slower than their MATLAB equivalents (x=A\b, [L,U,p]=lu(A), and so on). They are presented here merely to illustrate and educate. "Real" production code should use backslash and lu, not this package.
标签： illustrates elimination Gaussian pivoting
上传时间： 2014-01-21
上传用户：lxm
一个小的telnet程序与大家共享，/* 本程序支持如一些参数： * --host IP地址 或者 -H IP地址 * --port 端口 或者 -P 端口 * --back 监听数量 或者 -B 监听数量 * --dir 服务默认目录 或者 -D 服务默认目录 * --log 日志存放路径 或者 -L 日志存放路径 * --daemon 使程序进入后台运行模式 */
上传时间： 2016-11-20
上传用户：515414293
成績顯示三個部份abc #include<stdio.h> #include<stdlib.h> int main(void) { float gread printf("請輸入分數\n") scanf("％f",&gread) if(gread>=80&&gread<=100) printf("成績為A\n") else if(gread>=60&&gread<=79) { printf("成績為B\n") } else if(gread>=0&&gread<60) { printf("成績為C\n") } else { printf("分數輸入錯誤\n") } system("pause") return 0 }
标签： include stdlib float gread
上传时间： 2014-01-15
上传用户：waizhang
AR6001 WLAN Driver for SDIO installation Read Me March 26,2007 (based on k14 fw1.1) Windows CE Embedded CE 6.0 driver installation. 1. Unzip the installation file onto your system (called installation directory below) 2. Create an OS design or open an existing OS design in Platform Builder 6.0. a. The OS must support the SD bus driver and have an SD Host Controller driver (add these from Catalog Items). b. Run image size should be set to allow greater than 32MB. 3. a. From the Project menu select Add Existing Subproject... b. select AR6K_DRV.pbxml c. select open This should create a subproject within your OS Design project for the AR6K_DRV driver. 4. Build the solution.
标签： installation Windows Driver March
上传时间： 2014-09-06
上传用户：yuzsu
一个基于GTK+的单词数值计算器，1、 按照规则计算单词的值，如果 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 26个字母（全部用大写）的值分别为 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26，如： WINJACK这个单词的值就为：W+I+N+J+A+C+K=23+9+14+1+3+11=71％ HARDWORK=H+A+R+D+W+O+R+D=8+1+18+4+23+15+18+11=98％ LOVE=L+O+V+E=12+15+22+5=54％ LUCK=L+U+C+K=12+21+3+11=47％ ATTITUDE= A+T+T+I+T+U+D+E=1+20+20+9+20+24+4+5=100％ 2、对程序的界面布局参考如下图所示，在第一个单行文本框输入一个单词，点击“计算”按钮，按照以上算法计算出该单词的值。 3、如果在最下面的单行文本框输入一个文件路径，此文件每行记录一个单词，那么经过程序计算出各个单词的值，并把结果输出到当前目录下result.txt文件中。如果文件不存在，应该提示错误。
上传时间： 2014-01-11
上传用户：康郎
专家点评: Y P7 `. @ {$ r％ pa.功能很强大，可以看出花了很多心血在算法上，非常好。算法上还有一点瑕疵，例如在删除一个员工的同时没有办法自动建立其他员工的上下级关系，必须删除全部下级员工，不是非常合理。此外，界面设计过于简单，应该加强. " W" R+ b* g$ a$ Sb.程序运用了自己的算法来提高Tree控件显示的速度和资源分配，这个非常值得肯定和鼓励。* C. c4 D0 e9 ` J$ w# U c.基本实现所有规定的功能，在所有参赛者中唯一熟 : O) l- F6 F9 f) S7 Q. l练使用面向对象设计方式开发程序的工程师，很不错！程序体现了作者非常扎实的数据结构功底，值得大家学习。工程管理也做得非常好，体现了作者在软件工程方面也有很深入的研究，该代码是很好的学习范例。 ％ G* H$ ~3 W1 ]. e! id.算法的创新是独特之处(hashtable算法建立),可见作者在数据结构方面的熟练掌握.此程序是很多专家会员学习典范.
上传时间： 2017-01-19
上传用户：奇奇奔奔
Instead of finding the longest common subsequence, let us try to determine the length of the LCS. Then tracking back to find the LCS. Consider a1a2…am and b1b2…bn. Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
标签： the subsequence determine Instead
上传时间： 2013-12-17
上传用户：evil
先用C-均值聚类算法程序，并用下列数据进行聚类分析。在确认编程正确后，采用蔡云龙书的附录B中表1的Iris数据进行聚类。然后使用近邻法的快速算法找出待分样本X（设X样本的4个分量x1=x2=x3=x4=6；子集数l=3）的最近邻节点和3-近邻节点及X与它们之间的距离。
上传时间： 2014-01-23
上传用户：frank1234
