可編程并行接口８２５５Ａ完成的交通燈實驗 用８２５５Ａ的Ｂ端口和Ｃ端口控制１２個ＬＥＤ的亮和滅（輸出為０則亮，輸出為１則滅），模擬十字路口的交通燈。 -programmable parallel interface 8255A completed, the traffic lights experimental 8255A port B and C - I control 12 LED bright and methomyl (output of 0-liang, the output of an anti), the simulation of traffic lights at a crossroads.
上傳時間： 2016-08-13
上傳用戶：來茴
This utility has two views: (a) one view that will show you the entire PnP enumeration tree of device objects, including relationships among objects and all the device s reported PnP characteristics, and (b) a second view that shows you the device objects created, sorted by driver name. There is nothing like this utility available anywhere else.
標簽： enumeration utility entire devic
上傳時間： 2013-12-17
上傳用戶：zjf3110
Instead of finding the longest common subsequence, let us try to determine the length of the LCS. Then tracking back to find the LCS. Consider a1a2…am and b1b2…bn. Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
標簽： the subsequence determine Instead
上傳時間： 2013-12-17
上傳用戶：evil
A Convex Hull is the smallest convex polygon that contains every point of the set S. A polygon P is convex if and only if, for any two points A and B inside the polygon, the line segment AB is inside P. One way to visualize a convex hull is to put a "rubber band" around all the points, and let it wrap as tight as it can. The resultant polygon is a convex hull.
上傳時間： 2013-12-23
上傳用戶：it男一枚
【問題描述】 在一個N*N的點陣中，如N=4,你現在站在（1，1），出口在（4，4）。你可以通過上、下、左、右四種移動方法，在迷宮內行走，但是同一個位置不可以訪問兩次，亦不可以越界。表格最上面的一行加黑數字A[1..4]分別表示迷宮第I列中需要訪問并僅可以訪問的格子數。右邊一行加下劃線數字B[1..4]則表示迷宮第I行需要訪問并僅可以訪問的格子數。如圖中帶括號紅色數字就是一條符合條件的路線。 給定N，A[1..N] B[1..N]。輸出一條符合條件的路線，若無解，輸出NO ANSWER。（使用U，D，L，R分別表示上、下、左、右。） 2 2 1 2 （4，4） 1 （2，3） （3，3） （4，3） 3 （1，2） （2，2） 2 （1，1） 1 【輸入格式】 第一行是數m (n < 6 )。第二行有n個數，表示a[1]..a[n]。第三行有n個數，表示b[1]..b[n]。 【輸出格式】 僅有一行。若有解則輸出一條可行路線，否則輸出“NO ANSWER”。
標簽： 點陣
上傳時間： 2014-06-21
上傳用戶：llandlu
A design about 8051 (running at 12MHz) based system with 3 7-Seg displays and two buttons to implement the following functions. 1. When press the + button, the display C = A+B. 2. When press the button, the display C = A - B. “A” and “B” are 8-bit inputs when “C” is 9-bit output.
上傳時間： 2015-05-05
上傳用戶：guoxiy
#include "iostream" using namespace std; class Matrix { private: double** A; //矩陣A double *b; //向量b public: int size; Matrix(int ); ~Matrix(); friend double* Dooli(Matrix& ); void Input(); void Disp(); }; Matrix::Matrix(int x) { size=x; //為向量b分配空間并初始化為0 b=new double [x]; for(int j=0;j<x;j++) b[j]=0; //為向量A分配空間并初始化為0 A=new double* [x]; for(int i=0;i<x;i++) A[i]=new double [x]; for(int m=0;m<x;m++) for(int n=0;n<x;n++) A[m][n]=0; } Matrix::~Matrix() { cout<<"正在析構中~~~~"<<endl; delete b; for(int i=0;i<size;i++) delete A[i]; delete A; } void Matrix::Disp() { for(int i=0;i<size;i++) { for(int j=0;j<size;j++) cout<<A[i][j]<<" "; cout<<endl; } } void Matrix::Input() { cout<<"請輸入A:"<<endl; for(int i=0;i<size;i++) for(int j=0;j<size;j++){ cout<<"第"<<i+1<<"行"<<"第"<<j+1<<"列:"<<endl; cin>>A[i][j]; } cout<<"請輸入b:"<<endl; for(int j=0;j<size;j++){ cout<<"第"<<j+1<<"個:"<<endl; cin>>b[j]; } } double* Dooli(Matrix& A) { double *Xn=new double [A.size]; Matrix L(A.size),U(A.size); //分別求得U，L的第一行與第一列 for(int i=0;i<A.size;i++) U.A[0][i]=A.A[0][i]; for(int j=1;j<A.size;j++) L.A[j][0]=A.A[j][0]/U.A[0][0]; //分別求得U，L的第r行，第r列 double temp1=0,temp2=0; for(int r=1;r<A.size;r++){ //U for(int i=r;i<A.size;i++){ for(int k=0;k<r-1;k++) temp1=temp1+L.A[r][k]*U.A[k][i]; U.A[r][i]=A.A[r][i]-temp1; } //L for(int i=r+1;i<A.size;i++){ for(int k=0;k<r-1;k++) temp2=temp2+L.A[i][k]*U.A[k][r]; L.A[i][r]=(A.A[i][r]-temp2)/U.A[r][r]; } } cout<<"計算U得:"<<endl; U.Disp(); cout<<"計算L的:"<<endl; L.Disp(); double *Y=new double [A.size]; Y[0]=A.b[0]; for(int i=1;i<A.size;i++ ){ double temp3=0; for(int k=0;k<i-1;k++) temp3=temp3+L.A[i][k]*Y[k]; Y[i]=A.b[i]-temp3; } Xn[A.size-1]=Y[A.size-1]/U.A[A.size-1][A.size-1]; for(int i=A.size-1;i>=0;i--){ double temp4=0; for(int k=i+1;k<A.size;k++) temp4=temp4+U.A[i][k]*Xn[k]; Xn[i]=(Y[i]-temp4)/U.A[i][i]; } return Xn; } int main() { Matrix B(4); B.Input(); double *X; X=Dooli(B); cout<<"~~~~解得:"<<endl; for(int i=0;i<B.size;i++) cout<<"X["<<i<<"]:"<<X[i]<<" "; cout<<endl<<"呵呵呵呵呵"; return 0; }
標簽： 道理特分解法
上傳時間： 2018-05-20
上傳用戶：Aa123456789
This book is an entry-level text on the technology of telecommunications. It has been crafted with the newcomer in mind. The eighteen chapters of text have been prepared for high-school graduates who understand algebra, logarithms, and basic electrical prin- ciples such as Ohm’s law. However, many users require support in these areas so Appen- dices A and B review the essentials of electricity and mathematics through logarithms.
標簽： Telecommunications Fundamentals 1st of ed
上傳時間： 2020-05-27
上傳用戶：shancjb
This book is an entry-level text on the technology of telecommunications. It has been crafted with the newcomer in mind. The twenty-one chapters of text have been prepared for high-school graduates who understand algebra, logarithms, and the basic principles of electricity such as Ohm’s law. However, it is appreciated that many readers require support in these areas. Appendices A and B review the essentials of electricity and mathematics up through logarithms. This material was placed in the appendices so as not to distract from the main theme, the technology of telecommunication systems. Another topic that many in the industry find difficult is the use of decibels and derived units. Appendix C provides the reader a basic understanding of decibels and their applications. The only mathematics necessary is an understanding of the powers of ten
標簽： Telecommunications Fundamentals 2nd of ed
上傳時間： 2020-05-27
上傳用戶：shancjb
VIP專區-嵌入式/單片機編程源碼精選合集系列（13）資源包含以下內容：1. CPMinterrupt 860中斷管理.2. 菲利普LPC900系列寫Flash源碼.3. 前段時間做了一個AT91M55800的芯片測試.4. tms320c5402 bootloader.5. tms320c54x realtime os.6. tms320 c5416 boot code.7. MGLS-240128TA液晶點陣顯示驅動程序.8. 嵌入式WINDOWSCE的書.9. 嵌入式TCP/IP包.10. sl811hs的host源程序.11. 嵌入式產品中的osip的源代碼..12. msp430的FLASH自編程子程序.13. IGNITE開發板說明書.14. 嵌入式實時系統中的優先級反轉問題.15. 僅供參考.16. 僅供參考.17. LCD driver 程序.18. 一個能跨頁面讀寫的I2C源碼.19. 一個2051控制兩個步進電機的源碼.20. 1330液晶源碼（可直接調用漢字）.21. pcf8583 常用時鐘芯片的使用.22. 1815 LCD drive IC 類驅動測試程序.23. 內存檢測程序源代碼.24. 嵌入式系統詞匯表.25. demonstrate how to use the bulk endpoint pairing feature of the EZ-USB chip.26. A Simple isochronous transfer. Reads 8051 ports A,B and C, and continuously sends a five byte packet.27. 44b0x bootloader.28. 鍵盤只有一個鍵 b.29. tornado安裝說明及KEY.30. 6711a板程序的傅立葉變換.31. 6711開發的源程序.32. 6711開發程序例子.33. 6711開發板源程序.34. 6711開發板源程序.35. 智能樓宇自動控制系統.36. YAFFS的升級版本YAFFS2.37. 嵌入開發筆記 用ps閱讀器打開.38. 嵌入式系統的重要概念.39. 嵌入式系統的調試方法.40. 一個在mck2407板上控制無刷電機恒速運行的程序.
上傳時間： 2013-07-21
上傳用戶：eeworm
