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Description: -8mrm67ub5be56b
Platform: |
Size: 1865 |
Author: 安全飞车 |
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Description: //24c01-24c16读写驱动程序， sbit a0=ACC^0 //定义ACC的位，利用ACC操作速度最快 sbit a1=ACC^1 sbit a2=ACC^2 sbit a3=ACC^3 sbit a4=ACC^4 sbit a5=ACC^5 sbit a6=ACC^6 sbit a7=ACC^7 -/ / 24c01 - 24c16 literacy drivers, ACC sbit a0 = ^ 0 / / ACC definition of the position, using the speed of the fastest ACC sbit ACC a1 = ^ 1 = a2 sbit ACC sbit a3 ^ 2 ^ 3 = ACC sbit a4 = ^ ACC a5 = 4 sbit ACC sbit a6 ^ 5 ^ 6 = ACC sbit a7 ACC = ^ 7
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Size: 958 |
Author: 哟 乁 |
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Description: //24c01-24c16读写驱动程序， sbit a0=ACC^0 //定义ACC的位，利用ACC操作速度最快 sbit a1=ACC^1 sbit a2=ACC^2 sbit a3=ACC^3 sbit a4=ACC^4 sbit a5=ACC^5 sbit a6=ACC^6 sbit a7=ACC^7 -/ / 24c01 - 24c16 literacy drivers, ACC sbit a0 = ^ 0 / / ACC definition of the position, using the speed of the fastest ACC sbit ACC a1 = ^ 1 = a2 sbit ACC sbit a3 ^ 2 ^ 3 = ACC sbit a4 = ^ ACC a5 = 4 sbit ACC sbit a6 ^ 5 ^ 6 = ACC sbit a7 ACC = ^ 7
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Size: 958 |
Author: kkysun |
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Description: 非常好用的大容量EEPROM存储器，带2线或3线SPI接口，带128字节高速缓存。-very handy, large capacity EEPROM memory, with two lines or three-wire SPI interface, with 128 byte cache.
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Size: 187827 |
Author: dh |
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Description: 系统辨识的输入信号为一个周期的M序列，从 中分离出并显示a1 、a2、 b1、 b2，画出输出观测值z的经线图形，并显示坐标网格-system identification of the input signal for a period of the M sequence, isolated from the show and a1, a2, b1, b2, paint output observation z warp graphics, and display Grid
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Size: 1099 |
Author: 小精豆 |
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Description: 最佳矩阵连乘 给定n个矩阵｛A1,A2,…An｝，其中Ai与A i+1是可乘的，i=1，2…，n-1。考察这n个矩阵的连乘积A1A2…An。矩阵A和B可乘的条件是矩阵A的列数等于矩阵B的行数。若A是一个p×q矩阵，B是一个q×r矩阵，则其乘积C=AB是一个p×r矩阵,需要pqr次数乘。
由于矩阵乘法满足结合律，故计算矩阵的连乘积可以有许多不同的计算次序。例如，设3个矩阵{A1,A2,A3}的维数分别为10×100，100×5，和5×50。若按加括号方式((A1A2)A3)计算，3个矩阵连乘积需要的数乘次数为10×100×5+10×5×50=7500。若按加括号方式(A1(A2A3))计算，3个矩阵连乘积总共需要10×5×50+10×100×50=75000次数乘。由此可见，在计算矩阵连乘积时，加括号方式，即计算次序对计算量有很大影响。
矩阵连乘积的最优计算次序问题，即对于给定的相继n个矩阵｛A1,A2,…An｝(其中矩阵Ai的维数为pi-1×p，i=1,2，…，n)，确定计算矩阵连乘积A1,A2,…An的计算次序，使得依此次序计算矩阵连乘积需要的数乘次数最少。
-best matrix continually multiply given n matrix (A1, A2, ... An), Ai and A is a mere i, i = 1, 2 ..., n-1. N explore the link matrix product ... An A1A2. Matrices A and B can either condition is out of the matrix A few matrix B is the number of rows. If A is a p q matrix B is a q-r-matrix, its product C = AB is a p r matrix, the number required by pqr. Because matrix multiplication meet the law of combination, it's even calculated matrix product can be calculated in many different priorities. For example, the matrix-based 3 (A1, A2, A3) dimension of 10 100, 100 5 5 and 50. If bracketed by the way ((A1A2) A3), even three product matrix multiplication in the number of 10 100 10 5 50 = 7,500. If bracketed by the way (A1 (A2A3)), three matrix product even need a total of 10 5 50
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Size: 6148 |
Author: 肿事右 |
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Description: 抛物线法求解
方程的构造方法:给出[0,1]区间上的随机数(服从均匀分布)作为方程的根p*.
设你的班级数为a3,学号的后两位数分别为a2与a1,从而得到你的三次方程
例如:你的31班的12号,则你的方程是21x3+60x2+2x+a0=0的形式.
方程中的系数a0由你得到的根p*来确定.
-parabolic equation method Construction Methods : given interval [0,1] on the random number (subject to uniform distribution) as the root equation p *. set up your classes at a3, after learning of the double-digit for a1 and a2, so you get the three equations for example : your 31 classes on the 12th, then your equation is 21x3 60x2 2x a0 = 0 forms. the equation coefficients a0 you get from the root of p * to determine.
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Size: 883 |
Author: 董方 |
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Description:
改进的牛顿法求解：
方程的构造方法:给出[0,1]区间上的随机数(服从均匀分布)作为方程的根p*.
设你的班级数为a3,学号的后两位数分别为a2与a1,从而得到你的三次方程
例如:你的31班的12号,则你的方程是21x3+60x2+2x+a0=0的形式.
方程中的系数a0由你得到的根p*来确定.
-improve Newton's method : Construction of the equation : given interval [0,1] on the random number (subject to uniform distribution) as the root equation p *. set up your classes at a3, after learning of the double-digit for a1 and a2, so you get the three equations for example : Your 31 classes on the 12th, then your equation is 21x3 60x2 2x a0 = 0 forms. the equation coefficients a0 you get from the root of p * to determine.
Platform: |
Size: 801 |
Author: 董方 |
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Description: 抛物线法求解
方程的构造方法:给出[0,1]区间上的随机数(服从均匀分布)作为方程的根p*.
设你的班级数为a3,学号的后两位数分别为a2与a1,从而得到你的三次方程
例如:你的31班的12号,则你的方程是21x3+60x2+2x+a0=0的形式.
方程中的系数a0由你得到的根p*来确定.
-parabolic equation method Construction Methods : given interval [0,1] on the random number (subject to uniform distribution) as the root equation p *. set up your classes at a3, after learning of the double-digit for a1 and a2, so you get the three equations for example : your 31 classes on the 12th, then your equation is 21x3 60x2 2x a0 = 0 forms. the equation coefficients a0 you get from the root of p * to determine.
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Size: 790 |
Author: 董方 |
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Description: 最短路径算法源码，很多人需要的。本人载网站开发gis,游自编的最短路径查询程序，速度特快，3万节点，35000条路全部遍历，只需1秒。现将最短路径的思路告诉大家，希望大家在优化，并用不同语言编制，我正在学delphi,准备用delphi做成库，本例以由拓扑关系的arc/info 文件为数据源。其中a1,b1,c1是以fnode排序生成的数组，a1对应fnode,b1对应tnode,c1对应length,同样a2,b2,c2,是以tnode 生成的数组。Indexa1是对应某一起点与其相连的终点的个数，indexb1时对应某一终点与其相连的起点的个数，即其拓扑关系。
-shortest path algorithm source, a lot of people in need. I gis contained web site development, Yu directed the shortest path query procedures, speed express routes, 30,000 nodes, all 35,000 traverse the road, only a seconds. Now Shortest Path tell you the idea that we should optimization, and use different language, delphi is studying, I wanted to use delphi caused basement, in the cases from the arc topology / info file as the data source. Which a1, b1, c1 fnode ranking is generated by the array, a1 fnode counterparts, b1 counterparts tnode, c1 corresponding length, the same a2, b2, c2, tnode is generating array. Indexa1 counterparts is a starting point, linked with the end of the number, indexb1 counterparts at a terminal connected to its starting point the number, which is its topologic
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Size: 2416 |
Author: x |
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Description: Matlab光学设计程序
给出了一种光学设计的程序,对光学薄膜设计的人很有用-Matlab source code for optical design
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Size: 2048 |
Author: peng |
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Description: HT24 系列的EEPROM 总共8 个管脚，三个为芯片地址脚A0、A1、A2，在单片机对它进行操作时，从SDA 输入A0、A1、A2 数据和芯片外部A0、A1、A2 所接地址需一一对应。一个为芯片写保护脚WP，WP 脚接低电平时，芯片可进行读写操作；WP 脚接高时，芯片只可进行读，不可进行写。
另外两个管脚为电源脚VCC，VSS。
用单片机对HT24 系列的EEPROM 进行控制时，HT24 系列的EEPROM 的外部管脚VCC、VSS、WP、A0、A1、A2 根据需要，对应接上，SDA、SCL 接到单片机控制脚上。-HT24 Series EEPROM total of 8 pins, three pins for the chip address A0, A1, A2, in the microcontroller to operate it, from SDA input A0, A1, A2, and chip external data A0, A1, A2 of the Then one correspondence address must be in. One for the chip, write-protect pin WP, WP feet then low, the chip can be read and write operations WP pin is high then, the chip can only be read, not write. The other two pins for the power pin VCC, VSS. With the microcontroller on the HT24 Series EEPROM control when, HT24 Series EEPROM external pin VCC, VSS, WP, A0, A1, A2 according to the needs of the corresponding connected, SDA, SCL receives SCM control feet .
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Size: 1024 |
Author: 张钧 |
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Description: Ambarella A2-A1-RH-A288S datasheet (The A288S is a state-of-the-art high performance and highly integrated security and surveillance camera processor capable of video capture and playback up to 1920x1080i60 or 1280x720p60. Also used in the Matrox CompressHD card.)
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Size: 238592 |
Author: tv84 |
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Description: Source Code Explorer DBISAM Whit A2 Function
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Size: 24576 |
Author: senshidesu1 |
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Description: LSD-STBD-A001-A2实验指导书-LSD-STBD-A001-A2 experiment instructions
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Size: 569344 |
Author: 董明 |
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Description: Diseñ o Digital de Complemento A2 simulado en Proteus
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Size: 14336 |
Author: Carlos Ortega |
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Description: Yokogawa signal converter LRX-01 A2
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Size: 133120 |
Author: 阳生 |
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Description: LMS算法下的关于双参数a1、a2的收敛曲线(The convergence curves of parameters a1 and a2 under LMS algorithm.)
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Size: 9216 |
Author: 话梦 |
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Description: RLS算法下关于双参数a1、a2的收敛曲线(The convergence curves of parameters a1 and a2 under RLS algorithm.)
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Size: 11264 |
Author: 话梦 |
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