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[assembly language] 4.rar DL : 0: 题目：利用条件运算符的嵌套来完成此题：学习成绩>=90分的同学用A表示，60-89分之间的用B表示，60分以下的用C表示。 1.程序分析：(a>b)?a:b这是条件运算符的基本例子。
Date : 2025-12-21 Size : 19kb User :
[assembly language] 精致小闹钟 DL : 1: 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 : 13.57kb User : 李文
[assembly language] baoshixiaoshizhong DL : 1: 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 : 13.52kb User : 蔡志斌
[assembly language] CLOCK_LED_Show DL : 0: 89S51 与数码管显示器以及键盘SW 对应引脚的联机为： P0.0-------------------------数码管显示器的a脚 P0.1-------------------------数码管显示器的b脚 P0.2-------------------------数码管显示器的c脚 P0.3-------------------------数码管显示器的d脚 P0.4-------------------------数码管显示器的e脚 P0.5-------------------------数码管显示器的f脚 P0.6-------------------------数码管显示器的g脚 P0.7-------------------------数码管显示器的dp脚； P2.2-------------------------数码管显示器1的驱动脚 P2.3-------------------------数码管显示器2的驱动脚 P2.4-------------------------数码管显示器3的驱动脚 P2.5-------------------------数码管显示器4的驱动脚 P2.6-------------------------数码管显示器5的驱动脚 P2.7-------------------------数码管显示器6的驱动脚； P2.2-------------------------键盘SW1 P2.3-------------------------键盘SW2 P2.4-------------------------键盘SW3 P2.5-------------------------键盘SW4 P2.6-------------------------键盘SW5 P2.7-------------------------键盘SW6-devised with the digital control display and keyboard SW corresponding pin online : P0.0 - ------------------------ Digital Display of a foot P0 .1 ------------------------- LED display b feet P0.2 ------------------------- Digital Display foot width of the c-- ----------------------- LED display d feet P0.4 - --- --------------------- LED display e feet P 0.5------ ------------------- Digital Display f Encl foot-------- ----------------- LED display g feet P0.7 - --------- --------------- LED display dp feet; P2.2 - ------------------------ Digital Display Driver feet of a P2.3 - ------------------------ Digital Display Driver 2 feet P2.4 - ------------------------ Digital Display Driver 3 feet dry weight------------------------- Digital Display Driver 4 feet P2.6 - ----------------------
Date : 2008-10-13 Size : 1.79kb User : 恒学
[assembly language] 8255chengxu DL : 0: 可编程并行接口８２５５Ａ完成的交通灯实验　　　用８２５５Ａ的Ｂ端口和Ｃ端口控制１２个ＬＥＤ的亮和灭（输出为０则亮，输出为１则灭），模拟十字路口的交通灯。 -programmable parallel interface 8255A completed, the traffic lights experimental 8255A port B and C - I control 12 LED bright and methomyl (output of 0-liang, the output of an anti), the simulation of traffic lights at a crossroads.
Date : 2008-10-13 Size : 780byte User : 许明
[assembly language] bbbbbbbbb DL : 0: 这是一个音乐程序，按大写字母“A”，唱乐曲“玛丽有只小羔羊”；按大写字母“B”，唱乐曲“太湖船” 按大写字母“C”，唱乐曲“祝福歌” 按\"Q\"鍵，退出-This is a music procedures, capital letters "A", singing songs, "Mary has only a small lamb." by the capital letter "B", singing songs, "Taihu ship" by the capital letter "C", singing songs, "Song of Blessing" by the "Q" key, exit
Date : 2008-10-13 Size : 1.77kb User : KARAZHAN
[assembly language] yinyeyanzou DL : 0: 这是一个音乐程序，按大写字母“A”，唱乐曲“玛丽有只小羔羊”；按大写字母“B”，唱乐曲“太湖船” 按大写字母“C”，唱乐曲“祝福歌” 按\"X\"鍵，退出-This is a music procedures, capital letters "A", singing songs, "Mary has only a small lamb." by the capital letter "B", singing songs, "Taihu ship" by the capital letter "C", singing songs, "Song of Blessing" by the "X" button exit
Date : 2008-10-13 Size : 2.34kb User : 史明
[assembly language] huisu DL : 0: 回溯（b a c k t r a c k i n g）是一种系统地搜索问题解答的方法。为了实现回溯，首先需要为问题定义一个解空间（ solution space），这个空间必须至少包含问题的一个解（可能是最优的）。在迷宫老鼠问题中，我们可以定义一个包含从入口到出口的所有路径的解空间；在具有n 个对象的0 / 1背包问题中（见1 . 4节和2 . 2节），解空间的一个合理选择是2n 个长度为n 的0 / 1向量的集合，这个集合表示了将0或1分配给x的所有可能方法。当n= 3时，解空间为{ ( 0 , 0 , 0 )，( 0 , 1 , 0 )，( 0 , 0 , 1 )，( 1 , 0 , 0 )，( 0 , 1 , 1 )，( 1 , 0 , 1 )，( 1 , 1 , 0 )，( 1 , 1 , 1 ) }。-retrospective (b a c k t r a c k i n g) is a systematic search to answer the question. To achieve retrospective, the first issue of the need for a definition of the solution space (solution space), The space must contain at least one solution to the problem (which may be optimal). Rats in a maze problem, we can contain a definition from the entrance to the export of the solution space trails; n is the object of the 0 / 1 knapsack problem (see 1. 4 and 2. 2) The solution space is a reasonable choice of two n length of the 0 n / a vector set, this assembly that it will 0 or 1 x allocated to all possible ways. When n = 3, the solution space for the ((0, 0, 0), (0, 1, 0), (0, 0, 1), (1, 0, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1)).
Date : 2008-10-13 Size : 28.91kb User : john
[assembly language] sgj2 DL : 0: 假定已经有许多应用采用了程序1 - 1 5中所定义的C u r r e n c y类，现在我们想要对C u r r e n c y类的描述进行修改，使其应用频率最高的两个函数A d d和I n c r e m e n t可以运行得更快，从而提高应用程序的执行速度。由于用户仅能通过p u b l i c部分所提供的接口与C u r r e n c y类进行交互，-assumption has been used for many applications of the program 1 - 1 5 as defined in the C u r n c e r y category, right now we want C u r n c e r y kind of description changes make it the highest frequency of application function A 2 d d and I n c e r m e n t can run faster, thereby enhancing the application's execution. Because users only via p u b l i c parts of the interface and C u r n c e r y interactive category,
Date : 2008-10-13 Size : 961byte User : 史桂金
[assembly language] masmspecialll DL : 0: 直线加法的实验报告实现任意二位加法 (a+b)-(c+d)-straight Adder report on the achievement of arbitrary two Adder (a, b) - (c d)
Date : 2008-10-13 Size : 8.96kb User : jiang
[assembly language] Hanogkhk DL : 0: 用递规子程序的方法实现HANOI塔问题.子程序模块个数不限. 要求: 盘子个数可以输入,第一个塔为A,第二个塔为B,第三个塔的名称为C.打印出移动过程. -regulation subroutine with the delivery method HANOI Tower. Open-number subroutine modules. Requirements : Number plate can be imported, a Tower A, the second tower B, the third tower in the name of C. Print Mobile process.
Date : 2008-10-13 Size : 1.22kb User : 陈于
[assembly language] TRAFFICLIGHTS8255 DL : 0: 交通信号灯的控制： 1．通过8255A并口来控制LED发光二极管的亮灭。 2． A口控制红灯，B口控制黄灯，C口控制绿灯。 3．输出为0则亮，输出为1则灭。 4．用8253定时来控制变换时间。要求：设有一个十字路口，1、3为南，北方向，2、4为东西方向，初始态为4个路口的红灯全亮。之后，1、3路口的绿灯亮，2、4路口的红灯亮，1、3路口方向通车。延迟30秒后，1、3路口的绿灯熄灭，而1，3路口的黄灯开始闪烁（1HZ）。闪烁5次后，1、3路口的红灯亮，同时2、4路口的绿灯亮，2、4路口方向开始通车。延迟30秒时间后，2、4路口的绿灯熄灭，而黄灯开始闪烁。闪烁5次后，再切换到1、3路口方向。之后，重复上述过程。 -traffic signal control : 1. A parallel port by 8255 to control the LED light emitting diodes-destruction. 2. A red light population control, population control yellow B and C population control green. 3. Output-0, the output of a destruction. 4. Use 8253 to control transform regular time. Requirements : there is a crossroads, 1,3 to the south, to the north, the east-west direction 2,4. Initial state for four junctions in the whole bright red light. After 1,3 junctions green light, red light junctions in the 2,4-, 1,3 junctions opening direction. Delayed 30 seconds after the green light junctions 1,3 out and the junction of 1,3 begin blinking yellow light (1HZ). Stars 5, 1,3 junctions in the red light, green light junctions in the 2,4-, 2,4 direction was opened to traffic junctions. Delayed
Date : 2008-10-13 Size : 259.45kb User : gaodebo
[assembly language] 汇编 DL : 0: 4.asm…… 响铃程序，输入一个数字字符N，响铃N次。（完成）ysk3.asm ……显示一个星型倒三角。m1.asm ………编程将键盘输入的8位无符号二进制数转化为十六进制数和十进制数，并输出结果form.asm ……采用子程序编程按以下三种格式（██，◣，◥）打印九九乘法表：（完成）char.asm ……小写字母a b c d ……x y z的ASCII码分别为61H 62H 63H 64H……78H 79H 7AH, 而大写字母A B C D ….X Y Z的ASCII码分别为41H 42H 43H 44H …58H 59H 5AH，使用串处理指令编程从键盘输入16个字符（大小写字母及其它字母均有），存入以BUF1开始的一片存储区中，并将其传送到以BUF2开始的一片存储区中，在传送是将其中的小写字母均改为大写字母，并将第一个小写字母在串中的位置（距串头BUF1的相对位移量）以十六进制形式输出。（完成）-4.asm ... ... beep procedures, the importation of a number of characters N, N-beep. (Completed) ysk3.asm ... shows a Star inverted triangle. ... ... M1.asm programming to the keyboard input of eight unsigned binary number into a hexadecimal number and decimal number, and the output form.asm ... using subroutine program by the following three formats ( , TT, Short) Print Jiujiuchengfabiao : ( completed) char.asm ... lowercase letters a b c d ... x y z ASCII respectively 61H 62H 63H 64H 78H 79H ... 7AH and capital letters A B C D .... X Y Z ASCII respectively 41H 42H 43H 44H 58H 59H L1.5AH ..., the use of string processing programming instructions from the keyboard 16 characters (letters and other case-sensitive alphanumeric both), credited to BUF1 started a storage area and its transmission
Date : 2008-10-13 Size : 2.83kb User : 冯萍
[assembly language] simple DL : 0: 单纯形法算法,int K,M,N,Q=100,Type,Get,Let,Et,Code[50],XB[50],IA,IAA[50],Indexg,Indexl,Indexe float Sum,A[50][50],B[50],C[50]
Date : 2008-10-13 Size : 5.63kb User : zd
[assembly language] music DL : 0: 用汇编语言实现的音乐演奏这是一个音乐程序，按大写字母“A”，唱乐曲“玛丽有只小羔羊”；按大写字母“B”，唱乐曲“太湖船” 按大写字母“C”，唱乐曲“祝福歌” 按\"X\"鍵，退出
Date : 2008-10-13 Size : 1.23kb User : hdd
[assembly language] music DL : 0: 有3种音乐选择演奏,按A\\B\\C选择音乐,按Q推出程序.
Date : 2008-10-13 Size : 3.21kb User : linlin
[assembly language] crc2345 DL : 0: crc任意位生成多项式任意位运算自适应算法循环冗余校验码(CRC，Cyclic Redundancy Code)是采用多项式的编码方式，这种方法把要发送的数据看成是一个多项式的系数，数据为bn-1bn-2…b1b0 (其中为0或1)，则其对应的多项式为： bn-1Xn-1+bn-2Xn-2+…+b１X+b０例如：数据“10010101”可以写为多项式 X７+X４+X２+1。循环冗余校验CRC 循环冗余校验方法的原理如下： (1) 设要发送的数据对应的多项式为P(x)。 (2) 发送方和接收方约定一个生成多项式G(x)，设该生成多项式的最高次幂为r。 (3) 在数据块的末尾添加r个0，则其相对应的多项式为M(x)=XrP(x) 。(左移r位) (4) 用M(x)除以G(x)，获得商Q(x)和余式R(x)，则 M(x)=Q(x) ×G(x)+R(x)。 (5) 令T(x)=M(x)+R(x)，采用模2运算，T(x)所对应的数据是在原数据块的末尾加上余式所对应的数据得到的。 (6) 发送T(x)所对应的数据。 (7) 设接收端接收到的数据对应的多项式为T’(x)，将T’(x)除以G(x) ，若余式为0，则认为没有错误，否则认为有错-crc-generating polynomial arbitrary arbitrary Operators adaptive algorithm Cyclic Redundancy Check (CRC. Cyclic Redundancy Code) is the polynomial coder, This way the data to be sent as a polynomial coefficient data bn- 1bn-2 ... b1b0 (0 or 1), corresponding to the polynomial : bn- 1Xn-1 bn- 2Xn-2 ... b1X belts such as : data "10010101" polynomial can be written as a X7 X4 X2. Cyclic Redundancy Check Cyclic Redundancy Check method of principle as follows : (1) The data to be sent to the corresponding polynomial p (x). (2) the sender and the receiver agreed on a generator polynomial G (x), set up the generator polynomial of the highest power of r. (3) In the data block Add to the end of r-0, then the polynomial corresponding to M (x) = XrP (x). (R-bits) (4) M (x) divided by G (
Date : 2025-12-21 Size : 1kb User : yzs
[assembly language] cjj520 DL : 0: 设在A，B和C单元中分别存放着三个数。若三个数都不是0，则求出三个数之和并存放于D单元中；若其中一个数为0，则把其他两个单元也清零。-In A, B and C units were placed with three numbers. If the three numbers not 0, then find the three numbers is also stored in the D unit if one number is 0, put the other two units were also cleared.
Date : 2025-12-21 Size : 1kb User : 小鸡啄米
[assembly language] B-spline DL : 0: b样条插值的c++程序，里面东西有点多，蛋不影响，希望大家一起学习。-b-spline interpolation c++ procedures, something a little more, does not affect the eggs, I hope you learn together.
Date : 2025-12-21 Size : 583kb User : 贾青
[assembly language] One-dimensional-N-S-equation-(B) DL : 0: 一维NS方程含C代码和Fortran代码-One-dimensional compressible N-S equation (B)
Date : 2025-12-21 Size : 124kb User : 陈许鑫

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