Search - c B - CodeBus

Search - c B - List

[File Format] gauss-C DL : 0: * 高斯列主元素消去法求解矩阵方程AX=B,其中A是N*N的矩阵,B是N*M矩阵 * 输入: n----方阵A的行数 * a----矩阵A * m----矩阵B的列数 * b----矩阵B * 输出: det----矩阵A的行列式值 * a----A消元后的上三角矩阵 * b----矩阵方程的解X -out the main elements of Gaussian elimination method for solving matrix equations AX = B, where A is N* N matrix, B is N* M matrix* Input : n---- A phalanx of a few-a matrix* A* m-matrix shown in a few B* b---- Matrix B* output : det-A matrix of a determinant value---- A* Elimination of upper triangular matrix* b---- Matrix The X-Solutions
Date : 2025-12-21 Size : 3kb User : xuyan
[File Format] eXmodem DL : 0: 本文件描述了一个基于串口实现文件传输的协议eXmodem，该协议基于标准Xmodem File Transfer Protocol，是一个独立标准，完全功能，全新的文件传输协议。 1)该协议除实现Xmodem File Transfer基本功能，即已实现了在两个设备相互之间，通过串口把一个设备内文件传输至另一个设备内；其增强功能所列参考如下： a)两个设备相互之间传输文件能够以一致的文件目录存放； b)两个设备相互之间可以传输任意大小的文件，而Xmodem File Transfer只能传送128 *n字节大小的文件； c)具备避免一个文件在两个设备相互之间重复传输的功能，提高了文件传输的效率。 -This document describes the realization of a document based on serial transmission agreement eXmodem. Based on the agreement Xmodem standard File Transfer Protocol, Standard is an independent, fully functional, new file transfer protocol. 1) Implementation of the agreement except Xmodem File Transfer basic functions, is realized in equipment between the two, put through a serial devices to another document transmission equipment; its reference listed enhancements are as follows : a) between two equipment to transfer files to the same directory storage; b) equipment between two transmission can be any size document, and Xmodem File Transfer can only transmit 128* n byte size of the document; c) avoid a document with the equipment between two transmission repeat function, improve the file t
Date : 2025-12-21 Size : 14kb User : 6756
[File Format] VGA DL : 0: 电源：稳压的＋5V电源，电流小于300mA。视频输入：RGB＋HSYNC＋VSYNC信号，取自VGA卡，刷新率与NTSC标准兼容。视频输出：混合视频和S－视频（Y/C）。支持的视频标准：PAL B、G、H和NTSCM。电路要求VGA卡能发送与PAL或NTSC标准视频时序兼容的RGB格式视频信号。 -Power: 5 V regulated power supply current of less than 300mA. Video input: RGB HSYNC VSYNC signals from the VGA card, refresh rate compatible with the NTSC standard. Video Output: Mixed video and S-video (Y/C). Supported video standards: PAL B, G, H and NTSCM. Circuit requirements of VGA cards can be sent with the PAL or NTSC standard video timing compatible RGB format video signals.
Date : 2025-12-21 Size : 32kb User :
[File Format] c_BCH DL : 0: c语言的BCH码的编译码程序，为WORD文档的c程序。-c language code BCH encoding and decoding procedures for the WORD document c procedures.
Date : 2025-12-21 Size : 9kb User : 萍果
[File Format] C++_Primer DL : 0: C++ Primer, Fourth Edition By Stanley B. Lippman, Josée Lajoie, Barbara E. Moo Englisht version-C++ Primer, Fourth Edition By Stanley B. Lippman, Josée Lajoie, Barbara E. Moo Englisht version
Date : 2025-12-21 Size : 1.58mb User : 吕婉辰
[File Format] xml DL : 0: The W3C DOM Core interfaces defines a minimal set of: A. interfaces for accessing and manipulating document objects B. Java object implementations for use with XML parsers. C. Conventions and processes for creating live HTML pages. D. Mutable document -The W3C DOM Core interfaces defines a minimal set of: A. interfaces for accessing and manipulating document objects B. Java object implementations for use with XML parsers. C. Conventions and processes for creating live HTML pages. D. Mutable document
Date : 2025-12-21 Size : 43kb User : yatou
[File Format] BSPLINE.ZIP DL : 0: C++ src for simple b-spline curve algorithm
Date : 2025-12-21 Size : 3kb User : v2k
[File Format] 0809_5 DL : 0: ad转换的C语言程序，硬件部分由AT89C51,MAX232，ADC0809，NF555组成，555用来产生ADC0809的CLK，现在好像是300多K，输入电压直接加在ADC0809的IN0，因为我只用一路所以ADC0809的 A、B、C三个进址我直接接地。ADC0809的ALE和StartART我连在一起由单片机的P3.4控制。-ad conversion C-language program, the hardware part by AT89C51, MAX232, ADC0809, NF555 composition, 555 used to generate the ADC0809' s CLK, it seems that it more than 300 K, the input voltage directly added to the ADC0809' s IN0, because I only use all the way so ADC0809 The A, B, C three sites I went straight into the ground. ADC0809 of ALE and StartART I linked to the P3.4 controlled by the microcontroller.
Date : 2025-12-21 Size : 1kb User : nanfang
[File Format] wuxianchui DL : 0: 无线词汇解释1. 香农定理类比：城市道路上的汽车的车速和什么有关系？和道路的宽度有关系，和自己车的动力有关系，也其他干扰因素有关系（如：车量的多少和红灯的数量）。香农定理是所有通信制式最基本的原理。 C=Blog2(1＋S/N)：其中C是可得到的链路速度，B是链路的带宽，S是平均信号功率,N是平均噪声功率，S/N即信噪比。香农定理给出了链 -Wireless glossary 1. Shannon theorem analogy: City of the speed of vehicles on the road and what a relationship? And road width of a relationship, and their own vehicle dynamics, relations, and also other factors that interfere with relationships (such as: how much car traffic and the number of red light). Shannon' s theorem is a basic principle of all communication formats. C = Blog2 (1+ S/N): where C is the available link speed, B is the bandwidth of the link, S is the average signal power, N is the average noise power, S/N or signal to noise ratio. Shannon' s theorem gives the chain
Date : 2025-12-21 Size : 48kb User :
[File Format] EDA DL : 0: 算术逻辑单元设计5位的操作数X和Y输入后暂存在寄存器A和B中，两位的操作控制码control暂存在寄存器C中，按照control码的不同，分布实现下列操作-no
Date : 2025-12-21 Size : 57kb User : 护墙
[File Format] org DL : 0: 实现大部分OGC规范，主要是B/S模式，可以根据具体修改为C/S-Achieve most of the OGC specifications, primarily B/S mode, you can modify depending on C/S
Date : 2025-12-21 Size : 4.3mb User : zhang wuji
[File Format] SINISAS-HOME-PAGE.files DL : 0: The learning process involves three steps: (a) learning the random forests, (b) constructing the second-level training set, and (c) learning the stacked classifier.
Date : 2025-12-21 Size : 6.37mb User : 魏雅娟
[File Format] Discrete-Mathematics-report(1) DL : 0: 1. 从键盘输入两个命题变元P和Q的真值，求它们的合取、析取、条件和双条件的真值。（A） 2. 求任意一个命题公式的真值表（B），并根据真值表求主范式（C）注意：题目类型分为A，B，C三类，其中A为基本题，完成A类题目可达到设计的基本要求，其他均为加分题，并按字母顺序分数增加越高。-Input two propositional variables P and Q is the true value from the keyboard, find their conjunctive, disjunctive, conditional, and biconditional true value. (A) Lord paradigm (C) (2) seek any propositional formula a truth table (B), and in accordance with the truth table Note: topic types are divided into three categories A, B, C, where A is the basic question, to complete the Class A subject to meet the design requirements, the other are the bonus, in alphabetical order the higher the score increases.
Date : 2025-12-21 Size : 750kb User : 周文俊
[File Format] java_code_1 DL : 0: 两个乒乓球队进行比赛，各出三人。甲队为a,b,c三人，乙队为x,y,z三人。已抽签决定比赛名单。有人向队员打听比赛的名单。a说他不和x比，c说他不和x,z比，请编程序找出三队赛手的名单。-Two table tennis team competition, each of the trio. A team for a, b, c trio, B teams for x, y, z trio. Draw lots to decide the game list. Inquire about the list of the game to players. a said he did not and x is greater than c said he did not x, z than programmed to identify a list of three teams racer.
Date : 2025-12-21 Size : 6kb User : 小雨李
[File Format] sadsadsassss DL : 0: 一次使用了J2EE开发平台，并学习了所需软件的安装方法，掌握运行一个JSP页面的基本步骤，了解了B/S模式和C/S模式的区别。-hello welcome
Date : 2025-12-21 Size : 9kb User : kaka
[File Format] C.S0014-D_v3.0_EVRC DL : 0: 3GPP2技术文档，EVRC-A\B\WB\NW-3GPP2 document
Date : 2025-12-21 Size : 1.77mb User : zhelmet
[File Format] fractal-use DL : 0: 分形的练习一 ①Koch曲线用复数的方法来迭代Koch曲线 clear i 防止i被重新赋值 A=[0 1] 初始A是连接（0，0）与（1，0）的线段 t=exp(i*pi/3) n=2 n是迭代次数 for j=0:n A=A/3 a=ones(1,2*4^j) A=[A (t*A+a/3) (A/t+(1/2+sqrt(3)/6*i)*a) A+2/3*a] end plot(real(A),imag(A)) axis([0 1 -0.1 0.8]) ②Sierpinski三角形 A=[0 1 0.5 0 0 1] 初始化A n=3 迭代次数 for i=1:n A=A/2 b=zeros(1,3^i) c=ones(1,3^i)/2 A=[A A+[c b] A+[c/2 c]] end for i=1:3^n patch(A(1,3*i-2:3*i),A(2,3*i-2:3*i), b ) patch填充函数 end -Fractal Exercise One The ① Koch curve Plural iteration Koch curve clear i to prevent i is reassigned A = [0 1] initial A is a connection (0,0) and (1,0) of the segments t = exp (i* pi/3) n = 2 n is the number of iterations for j = 0: n A = A/3 a = ones (1,2* 4 ^ j) A = [A (t* A+ a/3) (A/t+ (1/2+ sqrt (3)/6* i)* a) A+2/3* a] end plot (real (A), imag (A)) axis ([0 1-0.1 0.8]) ② Sierpinski triangle A = [0 1 0.5 0 0 1] initialized A n = 3 the number of iterations. for i = 1: n A = A/2 b = zeros (1,3 ^ i) c = ones (1,3 ^ i)/2 A = [A A+ [c b] A+ [c/2 c]] end for i = 1:3 ^ n patch (A (1,3* i-2: 3* i), A (2,3* i-2: 3* i), b ) patch filled function end
Date : 2025-12-21 Size : 43kb User : 郑志森
[File Format] matrix.c.tar DL : 0: first the pointer of c is a field "zurueckgerueckt" (in the first run so it shows on the last position!) This position is assigned the value of a and b, where also in these the pointer is first a "zurueckgerueckt". This is has repeatedly until c [] [] occupied all fields.
Date : 2025-12-21 Size : 1kb User : lililove01
[File Format] BNUTBDJNBOOFR DL : 0: 【程序29】题目：给一个不多于5位的正整数，要求：一、求它是几位数，二、逆序打印出各位数字。 1. 程序分析：学会分解出每一位数，如下解释：(这里是一种简单的算法，师专数002班赵鑫提供) 2.程序源代码：复制代码代码如下: #include stdio.h #include conio.h main( ) { long a,b,c,d,e,x scanf( ld ,&x) a=x/10000 分解出万位 b=x 10000/1000 分解出千位 c=x 1000/100 分解出百位 d=x 100/10 分解出十位 e=x 10 分解出个位 if (a!=0) printf( there are 5, ld ld ld ld ld\n ,e,d,c,b,a) else if (b!=0) printf( there are 4, ld ld ld ld\n ,e,d,c,b) else if (c!=0) printf( there are 3, ld ld ld\n ,e,d,c) else if (d!=0) printf( there are 2, ld ld\n ,e,d) else if (e!=0) printf( there are 1, ld\n ,e) getch() } -Topic: no more than five to one positive integer requirements: First, find it is several orders, two, reverse print out the digits. 1. Program analysis: the decomposition of each digit Society 2. Source Code: Code is as follows: #include stdio.h #include conio.h main () { long a, b, c, d, e, x scanf ( ld , & x) a = x/10000 break out ten thousand* / b = x 10000/1000 break out one thousand* / c = x 1000/100 break out one hundred* / d = x 100/10 break out ten* / e = x 10 break out bits* / if (! a = 0) printf ( there are 5, ld ld ld ld ld \ n , e, d, c, b, a) else if (! b = 0) printf ( there are 4, ld ld ld ld \ n , e, d, c, b) else if (! c = 0) printf ( there are 3, ld ld ld \ n , e, d, c) else if (! d = 0) printf ( there are 2, ld ld \ n , e, d)
Date : 2025-12-21 Size : 3kb User : 张欣欣
[File Format] Byangtiao DL : 0: Hermite曲线和Beizer曲线基本是一致的，虽然原理上有着一定的差异，但从性质以及参数的影响程度来说，大致上是相似的。而且这两种曲线在表达复杂的曲线时，都是利用低次曲线拼接的方式来表达。相比这两种曲线，B样条曲线就显得比较具有优势，具体优势开头也有提到，这里就不重复了，缺点也非常明显，控制点不在曲线上导致不好容易控制。-Produce their own creative language with c b-spline curves procedures, hope that helps
Date : 2025-12-21 Size : 13kb User : 掉毛

« 12 »