Location:
Search - 1100
Search list
Description: cc1100模块源代码，C51单片机控制 双向收发代码-This program series is used for a chipcon1100 module controled by C51 MCU.
Including RX & TX.
Platform: |
Size: 73728 |
Author: 朱多辉 |
Hits:
Description: Old English literature, or Anglo-Saxon literature, encompasses literature written in Old English in Anglo-Saxon England, in the period after the settlement of the Saxons and other Germanic tribes in England as the Jutes and the Angles after the withdrawal of the Romans and "ending soon after the Norman Conquest" in 1066 that is, c. 1100–50.[1] These works include genres such as epic poetry, hagiography, sermons, Bible translations, legal works, chronicles, riddles, and others.[2] In all there are about 400 surviving manuscripts from the period.[3]-Old English literature, or Anglo-Saxon literature, encompasses literature written in Old English in Anglo-Saxon England, in the period after the settlement of the Saxons and other Germanic tribes in England as the Jutes and the Angles after the withdrawal of the Romans and "ending soon after the Norman Conquest" in 1066 that is, c. 1100–50.[1] These works include genres such as epic poetry, hagiography, sermons, Bible translations, legal works, chronicles, riddles, and others.[2] In all there are about 400 surviving manuscripts from the period.[3]
Platform: |
Size: 37888 |
Author: facundo |
Hits:
Description: Write MATLAB code that will do the following.
(a) Image resizing into sizes 2X, 5X, 10X, 0.1X, 0.25X.
(b) Image rotation with and without clipping in angles 100, 600, 900, 1100, 1500
(c) Image flipping in x and y axis.
(d) Image zooming.
Platform: |
Size: 2048 |
Author: Rokibul Ahmed |
Hits:
Description: StrongARM 1100 LCD Controller Frame Buffer Driver for Linux v2.13.6.
Platform: |
Size: 10240 |
Author: vingcaoca |
Hits:
Description: MediaQ MQ-1100/1132 LCD and Peripheral Controller Datasheet
General Release September 12, 2002
Full register level documentation and programming information.
Platform: |
Size: 2345984 |
Author: |
Hits:
Description: 基于CC1100的按键双向通信程序（C语言）-CC1100-based key-way communication program (C language)
Platform: |
Size: 5120 |
Author: sibo |
Hits:
Description: 分别用人工控制和自动搜索的方法找到迷宫出口。蹭经听说过一个走迷宫的诀窍：顺着墙沿一侧走。（一直沿左侧或一直沿右侧）。本程序实现了这一思想，小人一直沿左侧走。迷宫是随机生成的。-Were treated with manual control and automatic search methods to find the maze of export. Rub heard by a Maze of know-how:顺着墙along the side walk. (Has been traveling along the left or right). This procedure realize this idea, villains have been walking along the left side. Maze is randomly generated.
Platform: |
Size: 6144 |
Author: ntexpa |
Hits:
Description: Patolli es uno de los juegos más antiguos de América prehispánica. Lo jugaban los incas, teotihuacanos (200 a.C. - 1000 d.C.), toltecas (750 - 1000 d.C), los mayas (1100 - 1300 d.C.) y los aztecas (1168 - 1521 d.C.)
Platform: |
Size: 2863104 |
Author: alessander91 |
Hits:
Description: Device driver for the PCMCIA control functionality of StrongARM SA-1100 microprocessors.
Platform: |
Size: 3072 |
Author: luxongdi |
Hits:
Description: DVB USB framework compliant Linux driver for the DVBWorld DVB-S 2101, 2102, DVB-S2 2104, DVB-C 3101, TeVii S600, S630, S650, S660, S480, S421, S632 Prof 1100, 7500, Geniatech SU3000 Cards for Linux v2.13.6.
Platform: |
Size: 10240 |
Author: goncster |
Hits:
Description: A 4 bit twisted ring counter is a sequential circuit which produces the following sequence of
output values: 0000, 1000, 1100, 1110, 1111, 0111, 0011, 0001 and then repeats. Design a
circuit for a 4 bit twisted ring counter that uses four D flip flops. Draw a state transition
diagram, a state table and a schematic for your circuit. Design an alternate implementation
using just three flip flops and draw a state transition diagram, state table and a schematic
for your circuit. If your designs are extended to implement an n bit twisted ring counter,
how many flip flops are required using each of the two approaches. In what situations
would you prefer the first method? In what situations would you prefer the second?
Platform: |
Size: 2048 |
Author: john |
Hits:
Description: Snake program like ones on nokia -Snake program like ones on nokia 1100
Platform: |
Size: 1024 |
Author: dumeadae@emeil.ir |
Hits:
Description: 基于无线模块1100的无线计分系统(ATMEGA16)-Wireless scoring system based on wireless module 1100(ATMEGA16)
Platform: |
Size: 141312 |
Author: 徐磊 |
Hits:
Description: CC1101 tx and RX functions can be implemented within any microcontroller.
Platform: |
Size: 40960 |
Author: Malkiat |
Hits:
Description: I2C Registers common for both HDCS-1000 1100 and HDCS-1020.
Platform: |
Size: 6144 |
Author: haopinyr |
Hits:
Description: 生产调度问题，蜂群算法，不过还有BUG，1100行-Production scheduling problem, colony algorithm, but there are BUG, 1100 OK
Platform: |
Size: 74752 |
Author: 丁浩然 |
Hits:
Description: Driver GLCD Nokia 3310 for PIC
Platform: |
Size: 8192 |
Author: Faouzi |
Hits:
Description: Nokia 1100 LCD simulation
Platform: |
Size: 22528 |
Author: Faouzi |
Hits:
Description: DVB USB framework compliant Linux driver for the DVBWorld DVB-S 2101, 2102, DVB-S2 2104, DVB-C 3101, TeVii S600, S630, S650, S660, S480, S421, S632 Prof 1100, 7500,.
Platform: |
Size: 9216 |
Author: tuewaxv |
Hits:
Description:
XOR是二进制数的运算符,实行non-carry除了两个数字数字的位数。
例如,1010 XOR 0110 = 1100。
现在我们XOR扩展到m-based数字。
例如,基于对5号,40302 XOR 44312 = 34114。
也就是说,两个m-based数字x和y,x XOR y =(x + y) 。
给定两个m-based数字n个数字,你可以重新排列数字交换任意两个数字在这个号码。
重新安排两个数字后,你XOR两个数字和得到一个新的号码。
例如,您可以重新安排40302到20034年,重新安排20034到24431年,然后你得到20034 XOR 24431 = 44410。
请查收最大数量,可以在上一步获得的。
-XOR is an operator for binary numbers which applies an non-carry addition on two numbers digit by digit.
For example, 1010 XOR 0110 = 1100.
Now we extend XOR into m-based numbers.
For example, for 5-based number, 40302 XOR 44312 = 34114.
That is, for two m-based digit x and y, x XOR y = (x+ y) m.
Given two m-based numbers with n digits, you can rearrange the number by swapping any two digits in this number.
After rearranging two numbers, you XOR two numbers and get a new number.
For example, You can rearrange 40302 into 20034 and rearrange 44312 into 24431, then you get 20034 XOR 24431 = 44410.
Please find the maximum possible number that can be obtained in the previous step.
Platform: |
Size: 1024 |
Author: secret |
Hits: