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byy
- 编程验证[x+1]是GF(28)=GF(2)[x]/(x8+x4+x3+x+1)的本原元，并给出GF(28)=GF(2)[x]/(x8+x4+x3+x+1)中以[x+1]为底的对数表和反对数表-Program verification [x+1] is the GF (28) = GF (2) [x]/(x8+ x4+ x3+ x+1) of the primitive element, and gives GF (28) = GF (2) [x]/(x8+ x4+ x3+ x+1) in or
DCT
- 将一幅图像先进行分成8X8块，然后对每个8X8块进行DCT变化，最后将变换后的图片进行分层，重新显示-Will an image to 8 X8 into pieces, then to every 8 X8 block DCT change, finally will transform the images after the layered, displayed again
DT-X8
- casio DTX5的升级版,DTX8，基于VS2010开发，界面更漂亮。平台是CE6.0。-the casio DTX5 the upgraded version, DTX8, VS2010 development, the interface is more beautiful. The platform is CE 6.0.
temp
- 注意：只有C代码。串行驱动led显示，一个74hc595位移寄存器驱动三极管驱动led位，两个74hc595驱动led段，方式位5位x8段x2=10个数码管5分频，每次扫描时间位1.25ms-Note: Only C code. Serial driver led display, a shift register 74hc595 drive transistor drive led, and led two drives 74hc595 paragraph, ways x8 paragraph
DirectX
- direx8.1是应用与xp给vc添加数据库支持-direx8.1
pci_express_decrypt
- Decrypted PCIexperss ip core x1 x4 x8
fcs_gen
- 用于产生CRC校验的matlab源码，生成多项式g(x) = x32 + x26 + x23 + x22 + x16 + x12 + x11 + x10 + x8 + x7 + x5 + x4 + x2 + x + 1，按比特输入按字节输出-Matlab code for CRC-16 generation
16X16LED
- 16*16点阵LED显示中文程序/由左向右流动.8X8显屏4个组成.-16* 16 dot matrix LED display English Program/from left to right movement of .8 X8 screen has four integral.
crcasm.asm
- 8 bit very fast CRC generator for polynom x8+x2+x1+x (0x07), fits any AVR processor and any type of 8-bit data trasfer with 8 bit CRC code written in AVR assembler with AVRStudio 4. Author: Jens Hallgren, Sweden-8 bit very fast CRC generator for po
XY
- 设有数组X和Y。X数组中有X1,…,X10；Y数组中有Y1,…,Y10。试编制程序计算Z1=X1+Y1,Z2=X2+Y2,Z3=X3-Y3,Z4=X4-Y4,Z5=X5-Y5,Z6=X6+Y6,Z7=X7-Y7,Z8=X8-Y8,Z9=X9+Y9,Z10=X10+Y10,结果存入Z数组。-Equipped with an array of X and Y. X array are X1, ..., X10 Y array are Y1, ..., Y10. Test preparation p
shuzuyunsuan
- 设在内存区域中有数组X和Y。X数组有X1，X2，…，X10；Y数组有Y1，Y2，…，Y10，编程序实现下面的计算： Z1=X1－Y1，Z2=X2＋Y2，Z3=X3－Y3，Z4=X4＋Y4，Z5=X5＋Y5，Z6=X6－Y6，Z7=X7－Y7，Z8=X8＋Y8，Z9=X9－Y9，Z10=X10＋Y10。计算结果Z存放在数组Z中。 为了便于检查结果，设数组X的内容为db 22,11,22,11,11,22,22,11,22,11,0,0,0,0,0,0存放于ds：0的内存中，为了便于计算，把
8Queens
- 八皇后问题：设8皇后问题的解为 (x1, x2, x3, …,x8), 约束条件为：在8x8的棋盘上，其中任意两个xi 和xj不能位于棋盘的同行、同列及同对角线。要求用一位数组进行存储，输出所有可能的排列-Eight queens problem: the solution set for the 8 queens problem (x1, x2, x3, ..., x8), constraints are: 8x8 chessboard in which any two xi and xj c
H27UAG8T2A
- 现代2G的nandflash资料，网上找了很久都没找到，最后还是从朋友那里要到的-NAND INTERFACE - x8 bus width. - Address/Data Multiplexing - Pin-out compatibility for all densities DATA RETENTION - 5,000 Program/Erase cycles (with 12 bit/528
CRCDecoding
- CRC检错程序。只能检错不能纠错。（40,32）的分组码检错，反馈函数：x8+x7+x4+x3+x+1-CRC error detection process. Not only error detection correction. (40,32) and block code error detection, feedback function: x8+ x7+ x4+ x3+ x+1
Dx8.1
- Direct X8库文件,方便各位同行开发-Direct X8 library
PCK_CRC3_D4
- CRC校验码生存程序 校验序列码生成多项式： X16+X13+X12+X11+X10+X8+X6+X5+X2+1 输入数据为16个字节（128位），输出16bit校验序列-CRC, the survival program check sequence code generator polynomial: X16+ X13+ X12+ X11+ X10+ X8+ X6+ X5+ X2+1 input data is 16 bytes (128 bits), output 16bit
CRC8_Dev
- 多项式为g(x)=x8+x5+x4+1的CRC-8 检验开发.绝对正确且易懂-Polynomial g (x) = x8+ x5+ x4+1 of the CRC-8 test development. Absolutely correct and easy to understand
bahuanghou
- 八皇后：在8X8格的国际象棋上摆放八个皇后，使其不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上，问有多少种摆法。-8 queens: in 8 X8 s international chess placed on eight queen, to make it can t attack each other, that is, any two queen not at all with one line, one and the same column or the same
Eight-empresses
- 八皇后问题是一个古老而著名的问题，是回溯算法的典型例题。该问题是十九世纪著名的数学家高斯1850年提出：在8X8格的国际象棋上摆放八个皇后，使其不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上，问有多少种摆法。-Eight the empress problems are old but famous a thou problems, is the typical model example that returns to trace calculate way.The pro
FrameEncap
- 要求实现的程序为图形化界面（如下图所示），可以输入以太帧目的MAC地址、源MAC地址和数据字段(以十六进制输入)，输出封装的帧的相关信息（前导码、帧前定界符、目的地址、源地址、帧长度、数据和帧校验和）, 生成多项式G（X）=X8+X2+X1+1。-Requests to realize the program for the graphic interface (shown below), can enter the etheric frame purpose MAC address, sour