自行开发的j2me小软件,用于管理个人日常开支收入,midp1.0,开发环境为eclipse 3.21,j2me toolkit 2.5 jsdk1.6
上传时间: 2014-01-07
上传用户:851197153
The code, images and designs for this book are released under a Creative Commons Attribution-NonCommercial-ShareAlike 2.5 License. http://creativecommons.org/licenses/by-nc-sa/2.5/ You are free: * to copy, distribute, display, and perform the work * to make derivative works Under the following conditions: *Attribution. You must attribute the work in the manner specified by the author or licensor. *Noncommercial. You may not use this work for commercial purposes. *Share Alike. If you alter, transform, or build upon this work, you may distribute the resulting work only under a license identical to this one. *For any reuse or distribution, you must make clear to others the license terms of this work. *Any of these conditions can be waived if you get permission from the copyright holder. CONTACT ME Please address any questions to info@andybudd.com.
标签: Attribution-NonComm Creative released Commons
上传时间: 2014-01-19
上传用户:chfanjiang
写一个程序,列出在0和1之间的所有分母不大于N的最简分数,下面是N=5时的情况: 0/1 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5 1/1 总共有11个分数! 还需要进行排序。
标签: 程序
上传时间: 2013-12-11
上传用户:chenbhdt
Hard-decision decoding scheme Codeword length (n) : 31 symbols. Message length (k) : 19 symbols. Error correction capability (t) : 6 symbols One symbol represents 5 bit. Uses GF(2^5) with primitive polynomial p(x) = X^5 X^2 + 1 Generator polynomial, g(x) = a^15 a^21*X + a^6*X^2 + a^15*X^3 + a^25*X^4 + a^17*X^5 + a^18*X^6 + a^30*X^7 + a^20*X^8 + a^23*X^9 + a^27*X^10 + a^24*X^11 + X^12. Note: a = alpha, primitive element in GF(2^5) and a^i is root of g(x) for i = 19, 20, ..., 30. Uses Verilog description with synthesizable RTL modelling. Consists of 5 main blocks: SC (Syndrome Computation), KES (Key Equation Solver), CSEE (Chien Search and Error Evaluator), Controller and FIFO Register.
标签: symbols length Hard-decision Codeword
上传时间: 2014-07-08
上传用户:曹云鹏
遗传算法源程序,求解一个简单优化问题f(x)=x1^2+x2^2,-5<=x1<=5,-5<=x2<=5
上传时间: 2015-09-09
上传用户:xiaoxiang
请将您需要的文件或目录拷贝到硬盘中。当\Source目录下的源文件被拷贝到硬盘上时,源文件的属性可能变为只读的,在编译之前应该将它们的属性改为可读写的。 \Source目录中的vcip.dsw包含了全书的所有项目,在Visual C++中打开它就能控制全部的代码。 光盘中各目录的内容如下所示: \Bin 本书所有示例的可执行文件,可直接在光盘中运行 \Images 用于图像处理测试的图像文件 \Source\ShowDIB 第2章2.5节示例的源程序 \Source\ViewDIB 第3章3.4节示例的源程序 \Source\EffectShow 第4章4.10节示例的源程序 \Source\TransformShow 第5章5.7节示例的源程序 \Source\ColorProcess 第6章6.10节示例的源程序 \Source\ImageProcess 第7章7.14节示例的源程序 \Source\Huffman 第8章8.1节示例的源程序 \Source\LZW 第8章8.1节示例的源程序 \Source\RLE 第8章8.1节示例的源程序 \Source\ViewImage 第8章8.7节示例的源程序 \Source\ImageLib 第9章9.1节图像工具库ImageLib的源程序 \Source\ImageBoard 第9章9.2节图像处理工具ImageBoard的源程序
上传时间: 2013-12-22
上传用户:jiahao131
用VHDL编写的一个出租车计费器,起步6元计2公里,此后每半公里计0.8元,停车等待每2.5分计0.8元。通过仿真,但未下载到CPLD测试
上传时间: 2013-12-24
上传用户:caixiaoxu26
Input The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. The first line of each case contains N, and the second line contains N integers giving the time for each people to cross the river. Each case is preceded by a blank line. There won t be more than 1000 people and nobody takes more than 100 seconds to cross. Output For each test case, print a line containing the total number of seconds required for all the N people to cross the river. Sample Input 1 4 1 2 5 10 Sample Output 17
标签: the contains integer number
上传时间: 2015-10-27
上传用户:plsee
关于C++的一个PPT教程C++语言概述,包括介绍 2.2 基本数据类型和表达式 2.3 数据的输入与输出 2.4 算法的基本控制结构 2.5 自定义数据类型
上传时间: 2014-10-30
上传用户:zuozuo1215
高斯列主元消去法,计算方法实现,已编译通过。使用例子 3x1+2x2+2x3+3x4=2.5 5x1+2x2+3x3+4x4=2.5 2x1+2x2+x3+2x4=2 3x1+x2+3x3+2x4=1.5 输入N=4,A={3 2 2 3 5 2 3 4 2 2 1 2 3 1 3 2},B={2.5 2.5 2 1.5}
标签: 高斯
上传时间: 2015-12-23
上传用户:yan2267246