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[Other] warshall DL : 0: a、要求已知一个关系矩阵R，利用warshall算法求它的闭包R+ b、有良好的用户界面 c、有一个用于演示warshall算法的demo关系矩阵 d、能够求出用户输入的任何关系矩阵的闭包 e、能够将用户要求的结果清晰的显示在终端上 -a, a requirement known matrix R, warshall algorithm for the use of its closure R b, a good user interface c, an algorithm for demonstrating warshall demo matrix d, user input can be obtained in any relationship matrix Closure e, can be user requirements the results clearly show that the terminal
Date : 2008-10-13 Size : 7.46kb User : 张大鹏
[Other] 256色液晶取模软件 DL : 0: 256色液晶取模软件适用于S3C44b0X A R M处理器的 L C D 256色字模提取程序,适用于新手使用.也适合于其他支持 256色 l c d,色位为 R(3) G(2) B(2)的 Lcd 控制器,最大支持640*480的液晶屏.-256 color LCD module from the application software in S3C44b0X R A L M processor 256 C D Color Abstraction procedures applicable to the use of newcomers. Also suitable for other support 256-color l c, d, color-R (3) G (2) B (2) Lcd controller supports 640 x 480 LCD screen.
Date : 2008-10-13 Size : 588.45kb User : 郑家福
[Other] 抓交通肇事犯 DL : 0: 一辆卡车违反交通规则，撞人后逃跑。现场有三人目击事件，但都没有记住车号，只记下车号的一些特征。甲说：牌照的前两位数字是相同的；乙说：牌照的后两位数字是相同的，但与前两位不同；丙是数学家，他说：四位的车号刚好是一个整数的平方。请根据以上线索求出车号。 *问题分析与算法设计按照题目的要求造出一个前两位数相同、后两位数相同且相互间又不同的整数，然后判断该整数是否是另一个整数的平方。 *-a truck traffic violation, the young man escaped. Scene three witnesses to the incident, but did not remember running, only to alight in mind, a number of characteristics. A : The first two digits of the license is the same; B : license after two figures is the same, but with two different; Group C was a mathematician, he said : the license plate number four is an integral exactly the square. Please get more clues kept. * Analysis and design algorithm with the subject of a request to create a double-digit same before and after double-digit same but also between different integers, and then judge whether the integer is another integral to the square. *
Date : 2008-10-13 Size : 1011byte User : 姚紫欣
[Other] 平分七筐鱼 DL : 0: 甲、乙、丙三位鱼夫出海打鱼，他们随船带了21只箩筐。当晚返航时，他们发现有七筐装满了鱼，还有七筐装了半筐鱼，另外七筐则是空的，由于他们没有秤，只好通过目测认为七个满筐鱼的重量是相等的，7个半筐鱼的重量是相等的。在不将鱼倒出来的前提下，怎样将鱼和筐平分为三份？ -A, B, C 3 fisherman fishing, they brought on board 21 baskets. That night back, they found seven baskets full of fish, with seven baskets in a half-loaded fish baskets, and seven basket is empty, because they did not weigh, that only through a visual seven baskets full of fish is equivalent to the weight of seven semi-fish basket weight of the same. The fish will not reversing the premise, how to fish and divided into three baskets?
Date : 2008-10-13 Size : 1.84kb User : 姚紫欣
[Other] 哪个大夫哪天值班 DL : 0: 医院有A、B、C、D、E、F、G七位大夫，在一星期内(星期一至星期天)每人要轮流值班一天。现在已知： A大夫比C大夫晚一天值班； D大夫比E大夫晚二天值班； B大夫比G大夫早三天值班； F大夫的值班日在B和C大夫的中间，且是星期四；请确定每天究竟是哪位大夫值班 -Hospital A, B, C, D, E, F, G 7 doctor within a week (Monday-Sunday) to shifts each day. Now known : A doctor later than C day the doctor on duty; D E doctor than two days later the doctor on duty; Doctor G B than the doctor on duty three days earlier; F doctor on duty at the B and C doctor in the middle, which is Thursday; Please identify any day is the doctor on duty
Date : 2008-10-13 Size : 1.59kb User : 姚紫欣
[Other] c 语言开发b-tree数据文件索引 DL : 0: c 语言开发b-tree数据文件索引 .zip -b-tree data files Index c language b-tree index data files. Z IPS language b-tree index data files. zipc language b-tree data files Index . zip
Date : 2008-10-13 Size : 15.53kb User : 唐颖
[Other] 0132 DL : 0: 数字计时器程序 Clock.asm -- Analogue Clock Demo based on Charles Petzold s DIGCLOCK.C Translated into assembly 20/9/99 by Ron Thomas Ron_Thom@Compuserve.com -digital timer procedures Clock.asm -- Analogue Clock Demo b ased on Charles Petzold s DIGCLOCK.C Translate d into assembly by Ron Thomas 20/9/99 @ Ron_Thom compuserve.com
Date : 2008-10-13 Size : 13.04kb User : 刘明
[Other] 汇编编写报时小闹钟 DL : 1: 本程序是一个用汇编编的精致的图形时钟，运行时双击clock图标即可，钟表显示的时间为本机系统的时间。　　按b键可扩大画面；按s键可缩小画面；按c键可改变颜色；按e键可听音乐；按q键退出本程序.-this program is compiled using a series of intricate graphics clock, double-click the clock running icon, the clocks show the time-based system of the machine. By b Bond picture can be expanded; According s keys can narrow picture; By c Bond can change color; By e keys to listen to music; q keys on the withdrawal procedures.
Date : 2008-10-13 Size : 9.6kb User : 成朝夕
[Other] 矩阵相加的算法 DL : 0: 当稀疏矩阵A和B均以三元组表作为存储结构时，试写出矩阵相加的算法，其结果存放在三元组表C中。解：这个算法有点繁，要考虑到两个稀疏矩阵的非零元素不是一一对应的，在建立新的三元组表C时，为了使三元组元素仍按行优先排列，所以每次插入的三元组不一定是A的，按照矩阵元素的行列去找A中的三元组，若有，则加入C，同时，这个元素如果在B中也有，则加上B的这个元素值，否则这个值就不变如果A中没有，则找B，有则插入C，无则查找下一个矩阵元素。 -sparse matrix A and B were 3 groups, as a storage structure and try to write together the matrix algorithm, results stored in ternary Group C table. Solution : This is a bit complicated algorithm, taking into account the two nonzero sparse matrix element is not one-to-one. the establishment of a new ternary Group C table, in order to enable groups of elements remaining three yuan prioritize OK, So inserted three yuan each group is not necessarily A, in accordance with the matrix elements ranks to find a group of three yuan, and if so, C is added, and if this element is B, plus the B value of the element, Otherwise, the value on the same if not A, then B to find, then insert C, no one will find under the matrix elements.
Date : 2008-10-13 Size : 2.49kb User : snow
[Other] watchxianshi DL : 0: 报时小闹钟本程序是一个用汇编编的精致的图形时钟，运行时双击clock图标即可，钟表显示的时间为本机系统的时间。　　按b键可扩大画面；按s键可缩小画面；按c键可改变颜色；按e键可听音乐；按q键退出本程序. -timekeeping small alarm clock this procedure is a compilation series with exquisite graphics clock, double-click the clock running icon, watches showed the time-based system of the machine. By b Bond picture can be expanded; According s keys can narrow picture; By c Bond can change color; By e keys to listen to music; Q keys on the withdrawal procedures.
Date : 2008-10-13 Size : 13.69kb User : 张定
[Other] simpleversion DL : 1: 一个用C编写的汉字行编辑程序,功能虽然不强,但包括了绝大部分的编辑命令,类似于原来的文字编辑程序.详见<<软件工程导论>>附录B-C,,,.lt;lt;gt;gt;B
Date : 2008-10-13 Size : 15.37kb User : 张阳
[Other] 1773 DL : 0: 求最长公共子系列的长度问题若给定序列X={x1,x2,…,xm}，则另一序列Z={z1,z2,…,zk}，是X 的子序列是指存在一个严格递增下标序列{i1,i2,…,ik}使得对于所有j=1,2,…,k 有：zj=xij.例如，序列 Z={a,b,f,c}是序列X={a,b,c,f,b,c}的子序列，相应的递增下标序列为{1,2,4,6}。给定2 个序列X 和Y，当另一序列Z 既是X 的子序列又是Y 的子序列时，称Z 是序列X 和Y 的公共子序列.给定2 个序列X={x1,x2,…,xm}和Y={y1,y2,…,yn}，找出X 和Y 的最长公共子序列. 分析：设系列X={x1,x2,…,xm}和Y={y1,y2,…,yn}的最长公共子序列为Z={z1,z2,…,zk} ，则 (1)若xm=yn，则zk=xm=yn，且zk-1 是xm-1 和yn-1 的最长公共子序列. (2)若xm≠yn 且zk≠xm，则Z 是xm-1 和Y 的最长公共子序列。 (3)若xm≠yn 且zk≠yn，则Z 是X 和yn-1 的最长公共子序列。由最长公共子序列问题的最优子结构性质建立子问题最优值的递归关系。用c[i][j]记录序列Xi 和Yj 的最长公共子序列的长度。其中， Xi={x1,x2,…,xi}；Yj={y1,y2,…,yj}。当 i=0 或j=0 时，空序列是Xi 和Yj 的最长公共子序列。故此时C[i][j]=0。
Date : 2008-10-13 Size : 973.37kb User : rcponder
[Other] TwoMain DL : 0: 自解密的加密程序的制作，是为了保证自己的数据不被别人非法地窃取，有必要对数据文件进行适当地加密。本程序采取一定的算法，对密码和数据进行加密，生成一个自解密的EXE文件。这个自解密文件是由两部分组成的。第一部分为可执行代码部分，其作用是对密码进行判断，对数据进行解密；后一部分是被加过密的密码和数据。　　本程序分两个模块进行编写。第一个是main.c，它是用来进行加密，产生EXE文件的；第二个是main2.c，它是用来解密的。分别对它们进行编译生成main.exe和main2.exe，再对其进行合并即可。　　合并方法如下：　　 COPY /B main.EXE＋main2.EXE Key.EXE(注意main.EXET和main2.EXE的顺序) 　　使用时请按下列格式键入。　　 Key〈欲加密文件〉〈产生后的EXE文件名〉　　按提示输入密码后将生成指定的可执行文件。解密时运行这个可执行文件。输入密码后(若输入三次都不正确，将退出解密程序)，即把原文件内容还原到你指定的文件中。解密程序格式如下。可执行文件名 (filename) 第一个源代码文件 main.c 第二个源代码文件main2.c 上述程序在TC2.0集成环境里通过 -since declassified encryption program production, in order to ensure that their data is not illegal to steal other people, it is necessary to conduct appropriate data file is encrypted. This program will adopt the algorithm, the passwords and data encryption, generating an EXE from the declassified documents. Since this document was declassified by the two components. The first part of the executable code, and its role is to password judgment, the data decryption; The latter part is additionally dense passwords and data. The procedure for the preparation of two modules. The first one is main.c, it is used for encryption, generated EXE file; The second is main2.c. it is used declassified. They were compiled and main2.exe main.exe generation, and its merger can be. The merger is as follows :
Date : 2008-10-13 Size : 1.6kb User : wwwwppp
[Other] binarytree-bianli DL : 0: 从键盘输入9个字符F B A D E C H G M。 • 查找结点E，并输出该结点的指针。 • 前序、中序和后序遍历输出该二叉树。 -input from the keyboard characters F 9 A B C D E G H M.
Date : 2008-10-13 Size : 5.58kb User : cara
[Other] number_theory_c++ DL : 0: 数论算法库 C++ 语言实现代码内容数论算法库，包括以下算法：欧几里德算法求a，b的最大公倍数扩展的欧几里德算法，求出gcd(a,b)和满足gcd(a,b)=ax+by的整数x和y 求解模线性方程 ax ≡ b (mod n) 其中n>0 求解模线性方程组(中国余数定理) 模取幂运算计算a^b mod n （a,b可能很大） Miller-Rabin随机性素数测试算法
Date : 2008-10-13 Size : 9.54kb User : henry
[Other] 128bitminus DL : 0: 128乘法模拟器 c M位乘N位不带符号整数的阵列乘法中加法---移位操作的被加数矩阵.每一部分乘积项ab叫做一个被加数.m*n个被加数可以用m*n个”与门”并行的产生. 以5位乘5位不带符号的阵列乘法器(m=n=5)为例(如下图): FA为一位全加器,FA的斜线方向为进位输出,竖线方向为和输出,而所有被加数项的排列和正常的A*B=P乘法过程中的被加数矩阵相同.图中用矩形围成的阵列中最后一行构成一个行波进位加法器,其时间延迟为(n-1)2T.当然,为了缩短加法时间,最后一行的行波进位加法器也可以用先行进位加法器来代替.这种乘法器实现n位×n位时,需要n(n-1)个全加器和n2个”与”门.
Date : 2008-10-13 Size : 46.25kb User : fred
[Other] Polynominal DL : 0: 用C++实现的多项式运算。程序已实现的功能有：（1）输入并建立多项式；（2）输出多项式；（3）多项式a和b相加，建立多项式a+b；（4）多项式a和b相减，建立多项式a-b；（5）计算多项式在x处的值；（6）求多项式a的导函数a’。-C realized polynomial operations. The procedure has been achieved functions : (a) input and the establishment of polynomial; (2) Output polynomial; (3) a and b polynomial sum, the establishment of a polynomial b; (4) a and b polynomial subtraction, establish a polynomial-b; (5) polynomial in x terms of value; (6) Finding a derivative of a function '.
Date : 2008-10-13 Size : 54.94kb User : 周智
[Other] c DL : 0: 通过用B树建立索引的源代码，玩玩而已的，不成可以作为参考的
Date : 2008-10-13 Size : 2.01kb User : 程明
[Other] gedebahe DL : 0: public class gedebahe { public static void main( String [] args) { for ( ) { int a=-1,b,c //b,c是分解出来的两个数 System.out.print(\"Please input a oushu:\") //输入一个偶数 try { BufferedReader in = new BufferedReader(new InputStreamReader(System.in)) String inputLine = in.readLine() a = Integer.valueOf(inputLine).intValue() }catch (Exception exc) { System.out.println(\"Sorry,please input a number.\") return } if (a%2!=0) { System.out.println(\"Sorry,please input a oushu.\") return } for (b=1 b<a b++) { c=(a-b) if(isPrime(b)&&isPrime(c)) { System.out.println(a+\"=\"+b+\"+\"+c) break } } } } public static boolean isPrime(int k)//方法用来判断一个数是否为偶数 { int a boolean t = true if (k==1) return t for (a=2 a<k a++) { if (k%a==0) break } if (a!=k) t=false return t-gedebahe public class (public static void main (String [] args) (for () (int a = a, b, c / / b, c is the decomposition of the two several System.out.print ( "Please input a oushu : ") / / input an even try (BufferedReader in = new BufferedReader (new InputStreamReader (System.in)) = String inputLine in.readLine () a = Integer.valueOf (inputLine). intValue ()) catch (Exception exc) ( System.out.println ( "Sorry, please input a number.") return) if (a 2%! = 0) (System.out.println ( "Sorry, please input a oushu.") return) for (b = a blt; a b) (c = (a-b) if (isPrime (b) isPrime (c)) (System.out.println (a "=" b "" c) break)))) public static boolean isPrime (int k) / / method used to determine whether a number is even (int a boolean true if t =
Date : 2008-10-13 Size : 881byte User : 康健伟
[Other] C++ 实现B树 DL : 0: C++ 实现B树
Date : 2009-04-24 Size : 2.71kb User : qianwenji1988@hotmail.com

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