S08例程：本程序是I/O模塊的例程，所用資源為單片機通用IO接口，所接器件為LED燈， 其中D2與PTA0，D3與PTA1相連接！
上傳時間： 2013-12-27
上傳用戶：zhoujunzhen
希爾排序(縮小增量法) 排序過程：先取一個正整數d1<n，把所有相隔d1的記錄放一組，組內進行直接插入排序；然后取D2<d1，重復上述分組和排序操作；直至di=1，即所有記錄放進一個組中排序為止
上傳時間： 2013-12-30
上傳用戶：waizhang
For solving the following problem: "There is No Free Lunch" Time Limit: 1 Second Memory Limit: 32768 KB One day, CYJJ found an interesting piece of commercial from newspaper: the Cyber-restaurant was offering a kind of "Lunch Special" which was said that one could "buy one get two for free". That is, if you buy one of the dishes on their menu, denoted by di with price pi , you may get the two neighboring dishes di-1 and di+1 for free! If you pick up d1, then you may get D2 and the last one dn for free, and if you choose the last one dn, you may get dn-1 and d1 for free. However, after investigation CYJJ realized that there was no free lunch at all. The price pi of the i-th dish was actually calculated by adding up twice the cost ci of the dish and half of the costs of the two "free" dishes. Now given all the prices on the menu, you are asked to help CYJJ find the cost of each of the dishes.
標簽： Limit following solving problem
上傳時間： 2014-01-12
上傳用戶：362279997
編制具有如下原型的函數prime，用來判斷整數n是否為素數：bool prime(int n) 而后編制主函數，任意輸入一個大于4的偶數d，找出滿足d=d1+D2的所有數對，其中要求d1與D2均為素數（通過調用prime來判斷素數）。如偶數18可以分解為11+7以及13+5；而偶數80可以分解為：43+37、61+19、67+13、73+7。 提示：i與d-i的和恰為偶數d，而且只有當i與d-i均為奇數時才有可能成為所求的“數對”。
上傳時間： 2017-01-02
上傳用戶：Breathe0125
一、KOL是什么?KOL能做什么? MCK是什么?MCK能做什么? KOL是一套對象庫，免費而且開放源代碼，大家可以http://bonanzas.rinet.ru/去下載到 它。它能夠使Delphi編出尺寸非常小的32位Windows圖形用戶界面的應用程序。 目前支持的Delphi版本:Delphi6,Delphi5,Delphi4,Delphi3和Delphi2。 使用KOL在D2-D5中編寫的最小圖形界面程序(只有一個窗體的空應用程序)，只有13K左右。 在D6中為20K左右。 MCK一套以KOL為基礎的可視化組件，用它就像用VCL一樣的方便省事而且可以達到和KOL一 樣的效果。
上傳時間： 2013-12-18
上傳用戶：1966640071
簡單的無線遙控音樂制作 采用HC2272，HC2262接收， 對應管腳連接： 1、無線遙控部分 D0----P1^0 D1----P1^1 D2----P1^2 D3----P1^3 VT----P3^2 2、顯示部分： 74LS48驅動共陰數碼管 A-----P3^0 B-----P3^1 C-----P3^2 D-----P3^3 3、蜂鳴器（或喇叭） P3^4
上傳時間： 2014-12-01
上傳用戶：txfyddz
This is a source of 13.560MHz RFID card reader for TRH031M as ATMEGA8-16AU MPU. The title is 3Alogics TRH031M 13.56MHz RFID Reader V1.0. project : RFID Reader V2.0 Target : MEGA8-16AU Crystal: 16.000 Mhz Input : TRH031M RFID 13.56MHz Output : RS232C 115200,N,8,1 Compiler : ICC-AVR V6.26C // Pin configuration for ATMEGA8-16AU TQFP32 // #30 RXD <--- RXD (SP3232) // #31 TXD ---> TXD (SP3232) // #32 PD2/INT0 <--- INTR (TRH031M) // # 1 PD3/INT1 ---> RST (TRH031M) // # 2 PD4 ---> D4 (TRH031M) // # 9 PD5 ---> D5 (TRH031M) // #10 PD6 ---> D6 (TRH031M) // #11 PD7 ---> D7 (TRH031M) // #12 PB0 ---> D0 (TRH031M) // #13 PB1 ---> D1 (TRH031M) // #14 PB2 ---> D2 (TRH031M) // #15 PB3 ---> D3 (TRH031M) // #16 PB4 ---> CSB (TRH031M) // #17 PB5 ---> LED // #23 PC0 ---> A0 (TRH031M) // #24 PC1 ---> A1 (TRH031M) // #25 PC2 ---> A2 (TRH031M) // #26 PC3 ---> ALE (TRH031M) // #27 PC4 ---> /RD (TRH031M) // #28 PC5 ---> /WR (TRH031M)
標簽： 13.560 ATMEGA 3Alogi source
上傳時間： 2017-03-31
上傳用戶：s363994250
program to solve a finite difference discretization of Helmholtz equation : (D2/dx2)u + (D2/dy2)u - alpha u = f using Jacobi iterative method. COMMENTS: OpenMP version 3: 1 PR outside the iteration loop, 4 Barriers Directives are used in this code to achieve paralleism. All do loops are parallized with default static scheduling.
標簽： discretization difference Helmholtz equation
上傳時間： 2014-01-11
上傳用戶：bruce5996
-- Hamming Decoder -- This Hamming decoder accepts an 8-bit Hamming code (produced by the encoder above) and performs single error correction and double error detection. -- download from: www.pld.com.cn & www.fpga.com.cn LIBRARY ieee USE ieee.std_logic_1164.ALL ENTITY hamdec IS PORT(hamin : IN BIT_VECTOR(0 TO 7) --d0 d1 D2 d3 p0 p1 p2 p4 dataout : OUT BIT_VECTOR(0 TO 3) --d0 d1 D2 d3 sec, ded, ne : OUT BIT) --diagnostic outputs END hamdec ARCHITECTURE ver1 OF hamdec IS BEGIN
標簽： Hamming produced Decoder decoder
上傳時間： 2017-07-15
上傳用戶：520
Two scripts are included here. 1. convsys.m - combines the state space representation of two systems connected in series. [Ao,Bo,Co,Do]=convsys(A1,B1,C1,D1,A2,B2,C2,D2) This algorithm gives the convolution of two state space representations | A1 B1 | | A2 B2 | u ==> | | ==> | | ==> y | C1 D1 | | C2 D2 | The algorithm also accepts state space objects as inputs and gives out a state space object as output. 2. sysfeedbk.m [Ao,Bo,Co,Do]=convsys(A1,B1,C1,D1,A2,B2,C2,D2) Gives the closed loop state space representation for two systems connected with negative feedback in the following manner. | A1 B1 | u ==> | | ==> y + o | C1 D1 | | - | | | | A2 B2 | | |= | |= | | C2 D2 | The zip file also contains checkcompatibility.m , which checks the compatibility of matrix dimensions in the system and cleanss.m which can be used to clean a state space representation.
標簽： representation included combines scripts
上傳時間： 2017-07-25
上傳用戶：semi1981
