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Multiple-<b>INPUT</b>-Multiple-Output

  • 1.有三根桿子A,B,C。A桿上有若干碟子 2.每次移動一塊碟子,小的只能疊在大的上面 3.把所有碟子從A桿全部移到C桿上 經過研究發現

    1.有三根桿子A,B,C。A桿上有若干碟子 2.每次移動一塊碟子,小的只能疊在大的上面 3.把所有碟子從A桿全部移到C桿上 經過研究發現,漢諾塔的破解很簡單,就是按照移動規則向一個方向移動金片: 如3階漢諾塔的移動:A→C,A→B,C→B,A→C,B→A,B→C,A→C 此外,漢諾塔問題也是程序設計中的經典遞歸問題

    標簽: 移動 發現

    上傳時間: 2016-07-25

    上傳用戶:gxrui1991

  • 1. 下列說法正確的是 ( ) A. Java語言不區分大小寫 B. Java程序以類為基本單位 C. JVM為Java虛擬機JVM的英文縮寫 D. 運行Java程序需要先安裝JDK

    1. 下列說法正確的是 ( ) A. Java語言不區分大小寫 B. Java程序以類為基本單位 C. JVM為Java虛擬機JVM的英文縮寫 D. 運行Java程序需要先安裝JDK 2. 下列說法中錯誤的是 ( ) A. Java語言是編譯執行的 B. Java中使用了多進程技術 C. Java的單行注視以//開頭 D. Java語言具有很高的安全性 3. 下面不屬于Java語言特點的一項是( ) A. 安全性 B. 分布式 C. 移植性 D. 編譯執行 4. 下列語句中,正確的項是 ( ) A . int $e,a,b=10 B. char c,d=’a’ C. float e=0.0d D. double c=0.0f

    標簽: Java A. B. C.

    上傳時間: 2017-01-04

    上傳用戶:netwolf

  • 輸入、輸出功能

    輸入、輸出功能,檢驗輸入的數是否為數字功能,ASCII碼與二進制數的的相互轉化功能,兩數相乘功能,因此整個程序定義了兩個宏INPUT和MULTIPLE,宏INPUT實現了輸入、檢驗和ASCII碼與二進制的相互轉換功能,MULTIPLE實現了兩數相乘功能,實現輸出功能的是子程序OUTPUT,由于在輸入的過程中多次用到了回車換行,因此定義了一個回車換行子程序CRLF。

    標簽: 輸入 輸出功能

    上傳時間: 2017-02-08

    上傳用戶:Ants

  • All of Java s Input/Output (I/O) facilities are based on streams, which provide simple ways to read

    All of Java s Input/Output (I/O) facilities are based on streams, which provide simple ways to read and write data of different types. Java provides many different kinds of streams, each with its own application. The universe of streams is divided into four large categories: input streams and output streams, for reading and writing binary data and readers and writers, for reading and writing textual (character) data. You re almost certainly familiar with the basic kinds of streams--but did you know that there s a CipherInputStream for reading encrypted data? And a ZipOutputStream for automatically compressing data? Do you know how to use buffered streams effectively to make your I/O operations more efficient? Java I/O, 2nd Edition has been updated for Java 5.0 APIs and tells you all you ever need to know about streams--and probably more.

    標簽: facilities streams provide Output

    上傳時間: 2013-12-17

    上傳用戶:6546544

  • Advances+in+Mobile+Radio+Access+Networks

    This book gives a comprehensive overview of the technologies for the advances of mobile radio access networks. The topics covered include linear transmitters, superconducting filters and cryogenic radio frequency (RF) front head, radio over fiber, software radio base stations, mobile terminal positioning, high speed downlink packet access (HSDPA), multiple antenna systems such as smart antennas and multiple input and multiple output (MIMO) systems, orthogonal frequency division multiplexing (OFDM) systems, IP-based radio access networks (RAN), autonomic networks, and ubiquitous networks. 

    標簽: Advances Networks Access Mobile Radio in

    上傳時間: 2020-05-26

    上傳用戶:shancjb

  • RT9005/RT9005A/RT9005B datasheet pdf

    The RT9005A/B is a dual-output Linear regulator for DDR-SDRAM VDDQ supply and termination voltage VTT supply.

    標簽: 9005 datasheet RT

    上傳時間: 2013-11-13

    上傳用戶:lmq0059

  • The government of a small but important country has decided that the alphabet needs to be streamline

    The government of a small but important country has decided that the alphabet needs to be streamlined and reordered. Uppercase letters will be eliminated. They will issue a royal decree in the form of a String of B and A characters. The first character in the decree specifies whether a must come ( B )Before b in the new alphabet or ( A )After b . The second character determines the relative placement of b and c , etc. So, for example, "BAA" means that a must come Before b , b must come After c , and c must come After d . Any letters beyond these requirements are to be excluded, so if the decree specifies k comparisons then the new alphabet will contain the first k+1 lowercase letters of the current alphabet. Create a class Alphabet that contains the method choices that takes the decree as input and returns the number of possible new alphabets that conform to the decree. If more than 1,000,000,000 are possible, return -1. Definition

    標簽: government streamline important alphabet

    上傳時間: 2015-06-09

    上傳用戶:weixiao99

  • We have a group of N items (represented by integers from 1 to N), and we know that there is some tot

    We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.

    標簽: represented integers group items

    上傳時間: 2016-01-17

    上傳用戶:jeffery

  • 北京大學ACM比賽題目 Write a program to read four lines of upper case (i.e., all CAPITAL LETTERS) text input

    北京大學ACM比賽題目 Write a program to read four lines of upper case (i.e., all CAPITAL LETTERS) text input (no more than 72 characters per line) from the input file and print a vertical histogram that shows how many times each letter (but not blanks, digits, or punctuation) appears in the all-upper-case input. Format your output exactly as shown.

    標簽: CAPITAL LETTERS program Write

    上傳時間: 2014-01-17

    上傳用戶:410805624

  • 道理特分解法

    #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

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