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《数据结构与问题求解：Java语言描述》笔记

《数据结构与问题求解：Java语言描述》阅读笔记目录一、内容综述 (2)1. 本书简介与背景介绍 (3)2. 本书阅读目的与预期成果 (4)二、基础概念与预备知识 (5)1. 数据结构定义与重要性 (7)2. 算法概念及其与数据结构的关系 (9)3. Java语言基础语法回顾 (9)4. 预备知识 (11)三、数据结构概述 (13)1. 数据结构的分类与特点介绍 (14)2. 数据结构的选择与应用场景分析 (16)四、线性数据结构 (18)1. 数组的概念与应用 (20)2. 链表的概念与应用 (20)3. 队列和栈的概念与应用 (22)4. 线性数据结构的性能分析 (23)五、非线性数据结构 (25)1. 树形数据结构概述 (26)2. 二叉树及其相关操作与应用实例分析讲解 (27)3. 图论中的基本概念及图的表示方法介绍等 (28)一、内容综述《数据结构与问题求解：Java语言描述》是一本关于数据结构和算法的经典教材，作者是Robert Sedgewick和Kevin Wayne。

本书以Java语言为实现工具，详细介绍了数据结构的基本概念和常用算法，以及如何将这些概念和算法应用于实际问题。

全书共分为5章，分别是基本数据结构、排序算法、图论、动态规划和高级数据结构。

第1章主要介绍了基本数据结构，包括线性表、栈和队列等。

线性表包括顺序表、链表和树表等，讲解了它们的基本操作和应用场景。

栈和队列分别介绍了它们的抽象数据类型、操作方法和应用实例。

第2章主要介绍了排序算法，包括冒泡排序、选择排序、插入排序、快速排序、归并排序和堆排序等。

每种排序算法都详细讲解了其原理、实现过程和优缺点，以及在不同场景下的应用。

第3章主要介绍了图论的基本概念和常用算法，如图的表示、遍历、最短路径算法(Dijkstra算法、FloydWarshall算法)、最小生成树算法(Kruskal算法、Prim算法)等。

还介绍了图的一些扩展概念，如带权有向图、带权无向图、加权图等。

atomic structure

Terms:
Atomic mass unit(amu): 原子质量单位；Atomic number: 原子序数；Atomic weight: 原子重量； Bohr atomic model:波尔原子模型；Quantum mechanics:量子理论；Quantum number:量子数；
Bonding force:原子间连接力；Bonding energy: 原子连接能；Ionic bond:离子键；
•
A wave-mechanical model
• • • The electron is considered to exhibit both wave-like and particle-like; An electron is no longer treated as a particle moving in a discrete orbital; Position is described by a probability distribution or electron cloud.
Fundamental concepts
• Each atom (原子) consists of a very small nucleus (原子核) composed of protons (质子) and neutrons (中子), which is encircled (围绕) by moving electrons (电子). • Both electrons and protons are electrically charged, the charge magnitude being which is negative (负电荷) in sign for electrons and positive (正电荷) for protons; neutrons are electrically neutral (电中性).

C++第三章课后答案

C++第三章课后答案C++第三章习题及答案1、什么是结构化程序设计？它有什么优缺点？所谓结构化程序设计，是一种自顶而下、逐步求精的模块化程序设计方法。

2、什么是对象？什么是类？简述对象与类之间的关系！对象是系统中用来描述客观事物的一个实体，它是用于构成系统的一个基本单位，而系统可以看作是由一系列相互作用的对象组成。

类定义了同类对象的公共属性和行为，属性用数据结构表示，行为用函数表示！《类=数据结构+对数据进行操作的函数》。

对象和类的关系相当于元素和集合的关系、变量和变量的“数据类型”的关系。

从程序设计的角度来说，类是一种复杂的自定义数据类型，对象是属于这种数据类型的变量。

3、什么是面向对象程序设计？面向对象程序设计方法具有哪些基本特征？请比较面向对象程序设计和面向对象过程程序设计有何异同？4、何谓成员变量？何谓成员函数？C++将对象的属性抽象为数据成员，将对象的行为抽象为成员函数。

5、C++中结构和类之间有何异同？结构在默认情况下的成员是公共的,而类在默认情况下的成员是私有的。

在C++中,结构是特殊的类。

6、在C++中如何定义类？如何实现定义的类？如何利用类申明对象？7、类的成员的访问权限有哪几种？请说明它们分别有什么作用？三种，privte:类的私有成员，只能被本类的成员函数访问或调用。

Public：公有成员，可以被本类的成员或其他类的成员函数（通过对象）访问或调用。

Protected:保护成员，可以被本类的成员函数或派生类的成员函数访问或调用。

8、何谓构造函数？何谓析构函数？请说明它们分别有什么作用？构造函数：不需要用户程序调用，就能在创建对象时由系统自动调用，其作用是在对象被创建时利用初始值去构造对象，使得在声明对象时就能自动完成对象的初始化。

析构函数：在对象的生存周期即将结束时由系统自动调用的，其作用是用来在对象被删除前做一些清理工作和数据保存工作。

9、如何定义一个内联成员函数？内联函:内联函数必须是和函数体申明在一起，才有效。

数据结构与程序设计C++描述(Kruse著)高等教育出版社_课后答案.

Programming Principles 11.2 THE GAME OF LIFEExercises 1.2Determine by hand calculation what will happen to each of the configurations shown in Figure 1.1 overthe course of five generations. [Suggestion: Set up the Life configuration on a checkerboard. Use one color of checkers for living cells in the current generation and a second color to mark those that will be born or die in the next generation.]Answer(a)Figure remains stable.(b)(c)(d)Figure is stable.12 Chapter 1 _ Programming Principles(e)(f)Figure repeats itself.(g)(h)(i)Figure repeats itself.(j)(k)(l)Figure repeats itself.Section 1.3 _ Programming Style 31.3 PROGRAMMING STYLEExercises 1.3E1. What classes would you define in implementing the following projects? What methods would your classespossess?(a) A program to store telephone numbers.Answer The program could use classes called Phone_book and Person. The methods for aPhone_bookobject would include look_up_name, add_person, remove_person. The methods for a Person object would include Look_up_number. Additional methods to initialize and print objects ofboth classes would also be useful.(b) A program to play Monopoly.Answer The program could use classes called Game_board, Property, Bank, Player, and Dice. In additionto initialization and printing methods for all classes, the following methods would be useful. The class Game_board needs methods next_card and operate_jail. The class Property needs methods change_owner, look_up_owner, rent, build, mortgage, and unmortgage. The class Bank needs methods pay and collect. The class Player needs methods roll_dice, move_location, buy_property and pay_rent. The class Dice needs a method roll.(c) A program to play tic-tac-toe.Answer The program could use classes called Game_board and Square. The classes need initialization and printing methods. The class Game_board would also need methods make_move andis_game_over. The class Square would need methods is_occupied, occupied_by, and occupy. (d) A program to model the build up of queues of cars waiting at a busy intersection with a traffic light. Answer The program could use classes Car, Traffic_light, and Queue. The classes would all need initializationand printing methods. The class Traffic_light would need additional methods change_statusand status. The class Queue would need additional methods add_car and remove_car.E2. Rewrite the following class definition, which is supposed to model a deck of playing cards, so that itconforms to our principles of style.class a { // a deck of cardsint X; thing Y1[52]; /* X is the location of the top card in the deck. Y1 lists the cards. */ public:a( );void Shuffle( ); // Shuffle randomly arranges the cards.thing d( ); // deals the top card off the deck};Answerclass Card_deck {Card deck[52];int top_card;public:Card_deck( );void Shuffle( );Card deal( );};4 Chapter 1 _ Programming PrinciplesE3. Given the declarationsint a[n][n], i, j;where n is a constant, determine what the following statement does, and rewrite the statement to accomplishthe same effect in a less tricky way.for (i = 0; i < n; i..)for (j = 0; j < n; j..)a[i][j] = ((i . 1)/(j . 1)) * ((j . 1)/(i . 1));Answer This statement initializes the array a with all 0’s except for 1’s down the main diagonal. A less tricky way to accomplish this initialization is:for (i = 0; i < n; i..)for (j = 0; j < n; j..)if (i == j) a[i][j] = 1;else a[i][j] = 0;E4. Rewrite the following function so that it accomplishes the same result in a less tricky way.void does_something(int &first, int &second){first = second −first;second = second −first;first = second . first;}Answer The function interchanges the values of its parameters:void swap(int &first, int &second)/* Pre: The integers first and second have been initialized.Post: The values of first and second have been switched. */{int temp = first;first = second;second = temp;}E5. Determine what each of the following functions does. Rewrite each function with meaningful variablenames, with better format, and without unnecessary variables and statements.(a) int calculate(int apple, int orange){ int peach, lemon;peach = 0; lemon = 0; if (apple < orange)peach = orange; else if (orange <= apple)peach = apple; else { peach = 17;lemon = 19; }return(peach);}Answer The function calculate returns the larger of its two parameters.int larger(int a, int b)/* Pre: The integers a and b have been initialized.Post: The value of the larger of a and b is returned. */{if (a < b) return b;return a;}Section 1.3 _ Programming Style 5(b) For this part assume the declaration typedef float vector[max];float figure (vector vector1){ int loop1, loop4; float loop2, loop3;loop1 = 0; loop2 = vector1[loop1]; loop3 = 0.0;loop4 = loop1; for (loop4 = 0;loop4 < max; loop4..) { loop1 = loop1 . 1;loop2 = vector1[loop1 −1];loop3 = loop2 . loop3; } loop1 = loop1 −1;loop2 = loop1 . 1;return(loop2 = loop3/loop2); }Answer The function figure obtains the mean of an array of floating point numbers. float mean(vector v)/* Pre: The vector v contains max floating point values.Post: The mean of the values in v is returned. */{float total = 0.0;for (int i = 0; i < max; i..)total += v[i];return total/((float) max);}(c) int question(int &a17, int &stuff){ int another, yetanother, stillonemore;another = yetanother; stillonemore = a17;yetanother = stuff; another = stillonemore;a17 = yetanother; stillonemore = yetanother;stuff = another; another = yetanother;yetanother = stuff; }Answer The function question interchanges the values of its parameters.void swap(int &first, int &second)/* Pre: The integers first and second have been initialized.Post: The values of first and second have been switched. */{int temp = first;first = second;second = temp;}(d) int mystery(int apple, int orange, int peach){ if (apple > orange) if (apple > peach) if(peach > orange) return(peach); else if (apple < orange)return(apple); else return(orange); else return(apple); elseif (peach > apple) if (peach > orange) return(orange); elsereturn(peach); else return(apple); }Answer The function mystery returns the middle value of its three parameters.6 Chapter 1 _ Programming Principlesint median(int a, int b, int c)/* Pre: None.Post: Returns the middle value of the three integers a, b, c. */{if (a > b)if (c > a) return a; // c > a > belse if (c > b) return c; // a >= c > belse return b; // a > b >= celseif (c > b) return b; // c > b >= aelse if (c > a) return c; // b >= c > aelse return a; // b >= a >= c}E6. The following statement is designed to check the relative sizes of three integers, which you may assumeto be different from each other:if (x < z) if (x < y) if (y < z) c = 1; else c = 2; elseif (y < z) c = 3; else c = 4; else if (x < y)if (x < z) c = 5; else c = 6; else if (y < z) c = 7; elseif (z < x) if (z < y) c = 8; else c = 9; else c = 10;(a) Rewrite this statement in a form that is easier to read.Answerif (x < z)if (x < y) // x < z and x < yif (y < z) c = 1; // x < y < zelse c = 2; // x < z <= yelse // y <= x < zif (y < z) c = 3; // y <= x < zelse c = 4; // impossibleelse // z <= xif (x < y) // z <= x < yif (x < z) c = 5; // impossibleelse c = 6; // z <= x < yelse // z <= x and y <= xif (y < z) c = 7; // y < z <= x// z <= y <= xif (z < x) // z <= y <= x, z < xif (z < y) c = 8; // z < y <= xelse c = 9; // z == y < x, impossibleelse c = 10; // y <= z == x, impossible(b) Since there are only six possible orderings for the three integers, only six of the ten cases can actuallyoccur. Find those that can never occur, and eliminate the redundant checks.Answer The impossible cases are shown in the remarks for the preceding program segment. After their removal we have:if (x < z)if (x < y) // x < z and x < yif (y < z) c = 1; // x < y < zelse c = 2; // x < z <= yelse c = 3; // y <= x < zelse // z <= xif (x < y) c = 6; // z <= x < yelse // z <= x and y <= xif (y < z) c = 7; // y < z <= xelse c = 8; // z <= y <= xSection 1.3 _ Programming Style 7(c) Write a simpler, shorter statement that accomplishes the same result.Answerif ((x < y) && (y < z)) c = 1;else if ((x < z) && (z < y)) c = 2;else if ((y < x) && (x < z)) c = 3;else if ((z < x) && (x < y)) c = 6;else if ((y < z) && (z < x)) c = 7;else c = 8;E7. The following C++ function calculates the cube root of a floating-point number (by the Newton approximation),using the fact that, if y is one approximation to the cube root of x, thenz . 2y . x=y23cube roots is a closer approximation.float function fcn(float stuff){ float april, tim, tiny, shadow, tom, tam, square; int flag;tim = stuff; tam = stuff; tiny = 0.00001;if (stuff != 0) do {shadow = tim . tim; square = tim * tim;tom = (shadow . stuff/square); april = tom/3.0;if (april*april * april −tam > −tiny) if (april*april*april −tam< tiny) flag = 1; else flag = 0; else flag = 0;if (flag == 0) tim = april; else tim = tam; } while (flag != 1);if (stuff == 0) return(stuff); else return(april); }(a) Rewrite this function with meaningful variable names, without the extra variables that contributenothingto the understanding, with a better layout, and without the redundant and useless statements. Answer After some study it can be seen that both stuff and tam play the role of the quantity x in the formula, tim plays the role of y, and tom and april both play the role of z. The object tiny is asmall constant which serves as a tolerance to stop the loop. The variable shadow is nothing but2y and square is y2 . The complicated two-line if statement checks whether the absolute valuejz3 − xj is less than the tolerance, and the boolean flag is used then only to terminate the loop. Changing all these variables to their mathematical forms and eliminating the redundant ones gives:const double tolerance = 0.00001;double cube_root(double x) // Find cube root of x by Newton method{double y, z;y = z = x;if (x != 0.0)do {z = (y . y . x/(y * y))/3.0;y = z;} while (z * z * z −x > tolerance || x −z * z * z > tolerance);return z;}(b) Write a function for calculating the cube root of x directly from the mathematical formula, by startingwith the assignment y = x and then repeatingy = (2 * y . (x/(y * y)))/3until abs(y * y * y −x) < 0.00001.8 Chapter 1 _ Programming PrinciplesAnswer const double tolerance = 0.00001;double formula(double x) // Find cube root of x directly from formula{double y = x;if (x != 0.0)do {y = (y . y . x/(y * y))/3.0;} while (y * y * y −x > tolerance || x −y * y * y > tolerance);return y;}(c) Which of these tasks is easier?Answer It is often easier to write a program fromscratch than it is to decipher and rewrite a poorly writtenprogram.E8. The mean of a sequence of numbers is their sum divided by the count of numbers in the sequence. Thestatistics (population) variance of the sequence is the mean of the squares of all numbers in thesequence, minusthe square of the mean of the numbers in the sequence. The standard deviation is the square root of the variance. Write a well-structured C++ function to calculate the standard deviation of a sequence of n floating-point numbers, where n is a constant and the numbers are in an array indexed from 0 to n−1, which is a parameter to the function. Use, then write, subsidiary functions to calculate the mean and variance.Answer #include <math.h>double variance(double v[], int n);double standard_deviation(double v[], int n) // Standard deviation of v[]{return sqrt(variance(v, n));}This function uses a subsidiary function to calculate the variance.double mean(double v[], int n);double variance(double v[], int n)// Find the variance for n numbers in v[]{int i;double temp;double sum_squares = 0.;for (i = 0; i < n; i..)sum_squares += v[i] * v[i];temp = mean(v, n);return sum_squares/n −temp * temp;}This function in turn requires another function to calculate the mean.double mean(double v[], int n) // Find the mean of an array of n numbers{int i;double sum = 0.0;for (i = 0; i < n; i..)sum += v[i];return sum/n;}Section 1.3 _ Programming Style 9E9. Design a program that will plot a given set of points on a graph. The input to the program will be a textfile, each line of which contains two numbers that are the x and y coordinates of a point to be plotted. The program will use a function to plot one such pair of coordinates. The details of the function involve plotting the specificmethod of plotting and cannot be written since they depend on the requirements of the plottingequipment, which we do not know. Before plotting the points the program needs to know the maximum and minimum values of x and y that appear in its input file. The program should therefore use another function bounds that will read the whole file and determine these four maxima and minima. Afterward,another function is used to draw and label the axes; then the file can be reset and the individual points plotted.(a) Write the main program, not including the functions.Answer #include <fstream.h>#include "calls.h"#include "bounds.c"#include "draw.c"int main(int argc, char *argv[])// Read coordinates from file and plot coordinate pairs.{ifstream file(argv[1]);if (file == 0) {cout << "Can not open input points file" << endl;cout << "Usage:\n\t plotter input_points " << endl;exit(1);}double xmax, xmin; // bounds for x valuesdouble ymax, ymin; // bounds for y valuesdouble x, y; // x, y values to plotbounds(file, xmax, xmin, ymax, ymin);draw_axes(xmax, xmin, ymax, ymin);file.seekg(0, ios :: beg); // reset to beginning of filewhile (!file.eof( )) {file >> x >> y;plot(x, y);}}(b) Write the function bounds.Answer void bounds(ifstream &file, double &xmax, double &xmin,double &ymax, double &ymin)// determine maximum and minimum values for x and y{double x, y;file >> x >> y;xmax = xmin = x;ymax = ymin = y;while (!file.eof( )) {file >> x >> y;if (x < xmin)xmin = x;if (x > xmax)xmax = x;if (y < ymin)ymin = y;10 Chapter 1 _ Programming Principlesif (y > ymax)ymax = y;}}(c) Write the preconditions and postconditions for the remaining functions together with appropriate documentationshowing their purposes and their requirements.Answer void draw_axes(double xmax, double xmin,double ymax, double ymin)/* Pre: The parameters xmin, xmax, ymin, and xmax give bounds for the x and y co-ordinates. Post: Draws and labels axes according to the given bounds. */{}void plot(double x, double y)/* Pre: The parameters x and y give co-ordinates of a point.Post: The point is plotted to the ouput graph. */{}1.4 CODING, TESTING, AND FURTHER REFINEMENT Exercises 1.4E1. If you suspected that the Life program contained errors, where would be a good place to insert scaffoldinginto the main program? What information should be printed out?Answer Since much of the program’s work is done in neighbor_count, a good place would be within the loops of the update method, just before the switch statement. The values of row, col, and neighbor_count could be printed.E2. Take your solution to Section 1.3, Exercise E9 (designing a program to plot a set of points), and indicategood places to insert scaffolding if needed.Answer Suitable places might be after draw_axes (printing its four parameters) and (with a very small test file to plot) after the call to plot (printing the coordinates of the point just plotted).E3. Find suitable black-box test data for each of the following:(a) A function that returns the largest of its three parameters, which are floating-point numbers. Answereasy values: .1; 2; 3., .2; 3; 1., .3; 2; 1..typical values: .0; 0:5;−9:6., .1:3; 3:5; 0:4., .−2:1;−3:5;−1:6..extreme values: .0; 0; 0., .0; 1; 1., .1; 0; 1., .0; 0; 1.(b) A function that returns the square root of a floating-point number.Answereasy values: 1, 4, 9, 16.typical values: 0.4836, 56.7, 9762.34.extreme value: 0.0.illegal values: −0.4353, −9.Section 1.4 _ Coding, Testing, and Further Refinement 11(c) A function that returns the least common multiple of its two parameters, which must be positive integers.(The least common multiple is the smallest integer that is a multiple of both parameters. Examples: The least common multiple of 4 and 6 is 12, of 3 and 9 is 9, and of 5 and 7 is 35.)Answereasy values: .3; 4., .4; 8., .7; 3..typical values: .7; 8., .189; 433., .1081; 1173..illegal values: .7;−6., .0; 5., .0; 0., .−1;−1..(d) A function that sorts three integers, given as its parameters, into ascending order.Answereasy values: .5; 3; 2., .2; 3; 5., .3; 5; 2., .5; 2; 3., .−1;−2;−3.extreme values: .1; 1; 1., .1; 2; 1., .1; 1; 2..typical values: .487;−390; 20., .0; 589; 333..(e) A function that sorts an array a containing n integers indexed from 0 to n −1 into ascending order, where a and n are both parameters.Answer For the number n of entries to be sorted choose values such as 2, 3, 4 (easy values), 0, 1, maximumsize of a (extreme values), and −1 (illegal value). Test with all entries of a the same value, the entries already in ascending order, the entries in descending order, and the entries in randomorder.E4. Find suitable glass-box test data for each of the following:(a) The statementif (a < b) if (c > d) x = 1; else if (c == d) x = 2;else x = 3; else if (a == b) x = 4; else if (c == d) x = 5;else x = 6;Answer Choose values for a and b, such as .1; 2., .2; 1., and .1; 1., so that each of a < b, a > b, and a == b holds true. Choose three similar pairs of values for c and d, giving nine sets of test data.(b) The Life method neighbor_count(row, col).Answer Set row in turn to 1, maxrow, and any intermediate value, as well as 0 and maxrow . 1 (as illegalvalues). Choose col similarly. For each of the legal (row, col) pairs set up the Life object so thatthe number of living neighbors of (row, col) is each possible value between 0 and 8. Finally, makethe cell at (row, col) itself either living or dead. (This process gives 98 sets of test data, provided maxrow and maxrow are each at least 3.)Programming Projects 1.4P1. Enter the Life program of this chapter on your computer and make sure that it works correctly. Answer The complete program is implemented in the life subdirectory for Chapter 1.#include "../../c/utility.h"#include "life.h"#include "../../c/utility.cpp"#include "life.cpp"int main() // Program to play Conway's game of Life./*Pre: The user supplies an initial configuration of living cells.Post: The program prints a sequence of pictures showing the changes in the configuration of living cells according to the rules forthe game of Life.Uses: The class Life and its methods initialize(), print(), andupdate(); the functions instructions(), user_says_yes().*/12 Chapter 1 _ Programming Principles{Life configuration;instructions();configuration.initialize();configuration.print();cout << "Continue viewing new generations? " << endl;while (user_says_yes()) {configuration.update();configuration.print();cout << "Continue viewing new generations? " << endl;}}const int maxrow = 20, maxcol = 60; // grid dimensionsclass Life {public:void initialize();void print();void update();private:int grid[maxrow + 2][maxcol + 2]; // Allow two extra rows and columns. int neighbor_count(int row, int col);};void Life::print()/*Pre: The Life object contains a configuration.Post: The configuration is written for the user.*/{int row, col;cout << "\nThe current Life configuration is:" <<endl;for (row = 1; row <= maxrow; row++) {for (col = 1; col <= maxcol; col++)if (grid[row][col] == 1) cout << '*';else cout << ' ';cout << endl;}cout << endl;}int Life::neighbor_count(int row, int col)/*Pre: The Life object contains a configuration, and the coordinates row and col define a cell inside its hedge.Post: The number of living neighbors of the specified cell is returned. */{int i, j;int count = 0;for (i = row - 1; i <= row + 1; i++)for (j = col - 1; j <= col + 1; j++)count += grid[i][j]; // Increase the count if neighbor is alive.count -= grid[row][col]; // A cell is not its own neighbor.return count;}Section 1.4 _ Coding, Testing, and Further Refinement 13void Life::update()/*Pre: The Life object contains a configuration.Post: The Life object contains the next generation of configuration. */{int row, col;int new_grid[maxrow + 2][maxcol + 2];for (row = 1; row <= maxrow; row++)for (col = 1; col <= maxcol; col++)switch (neighbor_count(row, col)) {case 2:new_grid[row][col] = grid[row][col]; // Status stays the same. break;case 3:new_grid[row][col] = 1; // Cell is now alive.break;default:new_grid[row][col] = 0; // Cell is now dead.}for (row = 1; row <= maxrow; row++)for (col = 1; col <= maxcol; col++)grid[row][col] = new_grid[row][col];}void Life::initialize()/*Pre: None.Post: The Life object contains a configuration specified by the user.*/{int row, col;for (row = 0; row <= maxrow+1; row++)for (col = 0; col <= maxcol+1; col++)grid[row][col] = 0;cout << "List the coordinates for living cells." << endl;cout << "Terminate the list with the the special pair -1 -1" << endl;cin >> row >> col;while (row != -1 || col != -1) {if (row >= 1 && row <= maxrow)if (col >= 1 && col <= maxcol)grid[row][col] = 1;elsecout << "Column " << col << " is out of range." << endl;elsecout << "Row " << row << " is out of range." << endl;cin >> row >> col;}}void instructions()/*Pre: None.Post: Instructions for using the Life program have been printed.*/14 Chapter 1 _ Programming Principles{cout << "Welcome to Conway's game of Life." << endl;cout << "This game uses a grid of size "<< maxrow << " by " << maxcol << " in which" << endl;cout << "each cell can either be occupied by an organism or not." << endl;cout << "The occupied cells change from generation to generation" << endl;cout << "according to the number of neighboring cells which are alive."<< endl;}P2. Test the Life program with the examples shown in Figure 1.1.Answer See the solution to the exercise in Section 1.2.P3. Run the Life program with the initial configurations shown in Figure 1.4. Several of these go throughmany changes before reaching a configuration that remains the same or has predictable behavior. Answer This is a demonstration to be performed by computer.1.5 PROGRAM MAINTENANCEExercises 1.5E1. Sometimes the user might wish to run the Life game on a grid smaller than 20_60. Determine how it ispossible to make maxrow and maxcol into variables that the user can set when the program is run. Try to make as few changes in the program as possible.Answer The integers maxrow and maxcol should become data members of the class Life. The method initialize, must now ask for input of the two data members maxrow and maxcol. Upper boundsof 20 and 60 for these integers should be stored in new constants called maxrowbound and maxcolbound. The amended file life.h now takes the form.const int maxrowbound = 20, maxcolbound = 60;// bounds on grid dimensionsclass Life {public:void initialize( );void print( );void update( );private:int maxrow, maxcol;int grid[maxrowbound . 2][maxcolbound . 2];// allows for two extra rows and columnsint neighbor_count(int row, int col);};As noted above, the method initialize needs minor modifications.E2. One idea for speeding up the function Life :: neighbor_count(row, col) is to delete the hedge (the extrarows and columns that are always dead) from the arrays grid and new_grid. Then, when a cell is on the boundary, neighbor_count will look at fewer than the eight neighboring cells, since some of these are outside the bounds of the grid. To do this, the function will need to determine whether or not the cell(row, col) is on the boundary, but this can be done outside the nested loops, by determining, before the loops commence, the lower and upper bounds for the loops. If, for example, row is as small as allowed, then the lower bound for the row loop is row; otherwise, it is row −1. Determine, in terms of the size of the grid, approximately how many statements are executed by the original version of neighbor_count and by the new version. Are the changes proposed in this exercise worth making?Section 1.5 _ Program Maintenance 15Answer We need four if statements at the beginning of neighbor_count to determine whether (row, col) is on the boundary. This gives a total of 4 _ maxrow _ maxcol extra statements. If the cell is inthe first or last row or column, but not in a corner, then the nested loops would iterate 3 fewertimes. There are 2_maxrow.2_maxcol−8 such positions. With the cell in one of the 4 cornerpositions, the nested loops would iterate 5 fewer times. Hence the total number of statementssaved is−4 _ maxrow _ maxcol . .2 _ maxrow . 2 _ maxcol −8..20:Thus this proposed change actually costs additional work, except for very small values of maxrow and maxcol.The modified function could be coded as follows.int Life :: neighbor_count(int row, int col)/* Pre: The Life object contains a configuration, and the coordinates row and col define a cell inside its hedge.Post: The number of living neighbors of the specified cell is returned. */{int i, j;int count = 0;int rowlow = row −1, rowhigh = row . 1,collow = col −1, colhigh = col . 1;if (row == 1) rowlow..;if (row == maxrow) rowhigh−−;if (col == 1) collow..;if (col == maxcol) colhigh−−;for (i = rowlow; i <= rowhigh; i..)for (j = collow; j <= colhigh; j..)count += grid[i][j]; // Increase the count if neighbor is alivecount −= grid[row][col]; // Reduce count, since cell is not its own neighborreturn count;}Programming Projects 1.5P1. Modify the Life function initialize so that it sets up the initial Life :: grid configuration by accepting occupied positions as a sequence of blanks and x’s in appropriate rows, rather than requiring the occupiedpositions to be entered as numerical coordinate pairs.Answer The following program includes all the changes for projects P1–P6. The changes required for Projects P7 and P8 are system dependent and have not been implemented.#include "../../c/utility.h"#include "life.h"#include "../../c/utility.cpp"#include "life.cpp"int main() // Program to play Conway's game of Life./*Pre: The user supplies an initial configuration of living cells.Post: The program prints a sequence of pictures showing the changes inthe configuration of living cells according to the rules forthe game of Life.Uses: The class Life and methods initialize(), print(), and update();。

c++语言基础教程详细解析(吕凤翥著)

第一章习题一、选择填空1.下列各种高级语言中，（）是面向对象的程序设计语言。

A.BASIC;B.PASCAL;C.C++D.Ada2.下列各种高级语言中，（）是最早提出了对象的概念。

A.Algol 60;B.Simula 67;C.Smalltalk;D.C++3.下述面向对象抽象的原理中，（）是不对的。

A. 数据抽象;B. 行为共享;C.进化;D. 兼容;4.（）不是面向对象系统所包含的要数。

A. 重载;B. 对象;C. 类;D. 继承;5.关于C++与C语言的关系的描述中，（）是错误的。

A. C语言是C++的一个子集;B. C语言与C++是兼容的;C. C++对C语言进行了一些改进;D. C++和C语言都是面向对象的;6.下面关于对象概念的描述中，（）是错误的。

A.对象就是C语言中的结构变量;B.对象代表着正在创建的系统中的一个实体；C. 对象是一个状态和操作（或方法）的封装体;D.对象之间的信息传递是通过消息进行的;7.下面关于类概念的描述中，（）是错误的。

A.类是抽象数据类型的实现;B.类是具有共同行为的若干对象的统一描述体;C.类是创建对象的样板;D.类就是C语言中的结构类型;8.C++对C语言作了很多改进，下列描述中（）使得C语言发生了质变，即从面向过程变成为面向对象。

A.增加了一些新的运算符;B.允许函数重载，并允许设置缺省参数;C.规定函数说明必须用原型;D.引进了类和对象的概念;9.按照标识符的要求，（）符号不能组成标识符。

A.连接符;B. 下划线;C.大小写字母;D.数字字符;10.下列符号中，（）不可作为分隔符。

A.，;B.：;C.？;D.;二、判断下列描述的正确性，对者划√，错者划×。

1．C++引进了引用的概念，对编程带来了很多方便。

√2．C++允许使用友元，但是友元会破坏封装性。

√3．C++中使用了新的注释符（//），C语言中注释符（/*…*/）不能在C++中使用。

北京理工大学2013级数据结构B试题(A卷)_答案模板

、选择题1从逻辑结构上可以把数据结构分为【 A 、动态结构和静态结构 C 、线性结构和非线性结构2、在一个长度为n 的顺序存储的线性表中，向第i 个元素（n+1）之前插入一个新元素时，需要从后向前依次后移【 B 】个元素。

4、在一个单链表中，已知q 所指结点是p 所指结点的前驱结点，若在q 和p 之间插入s 结点, 则执行【C 】。

A 、 s->next = p->next; p->next = s; B 、 p->next = s->next; s->next = p; C 、 q->next = s; s->next = p; D p->next = s; s->next = q;5、一个栈的入栈序列是1, 2, 3, 4, 5,则栈不可能输出的序列是【C 】。

A 、54321B 、45321C 、43512D 123456、判断一个队列Q （元素最多为Mt ）为空的条件是【C 】。

A 、Q->rear —Q->front = M B 、Q->rear —Q->front -1 ==M C 、Q->rear == Q->frontD Q->rear + 1 == Q->front7、在一个链队列中，假设f 和r 分别指向队首和队尾，贝朋入s 所指结点的运算是【A 】A 、r->next = s; r=s;B 、f->next = s; f=s;C 、s->next = r; r=s;D s->next = f; f=s;8深度为5的二叉树至多有【A 】个结点。

A 、31B 、32C 16D 109、在一非空二叉树的中序遍历序列中，根结点的右边【 A 】。

A 、只有右子树上的所有结点B 、只有右子树上的部分结点C 、只有左子树上的所有结点B 、只有左子树上的部分结点10、如果一棵完全二叉树有1001个结点，则其叶子结点个数为【D 】。

高级开发人员架构师面试题

高级开发人员架构师面试题此文中所提到的面试题是园子中Tony Qu兄弟翻译过来的，特此感谢高级开发人员/架构师1 有问题，当myString丌能满足时间格式要求的时候，会引发异常，建议使用 2PDB PDB 是用于保存调试和项目状态信息的文件，在debug的时候将产生pdb文件，调试的时候应该放在和对应应用程序集相同目录。

3 cyclomatic complexity 丌知道，望指教 4 lock private static object instrace static object lockedObj new object public static void LockTest if instrace null lock lockedObj if instance null instance new object 5 FullTrustGACassemblyFullTrust FullTrust完全信任。

放入GAC中的Assembly是否FullTrust我的理解丌是。

我理解FullTrust是可以通过代码设定的 6 可以更加灵活的设置对代码的访问权限，实现代码级保护。

这点丌是特清楚，有明白的给讲解下 7 gacutil /l find /i Corillian 全局程序集缓存中如果有Corillian就更新该程序集，没有就安装8 sn -t 显示程序集的公钥标记 9 DCOM135 135端口，因为DCOM的端口号是随机分配的，默认情况下，会分配1024以上的端口号，所以默认情况下DCOM丌能穿越防火墙。

因为根本丌晓得开哪个端口。

但有解决办法可以使DCOM分配的端口号固定，有关内容我在有过一些描述。

135是远程过程调用RPC的默认端口 10OOPSOA 我想OOP和SOA应该没有对比性吧。

OOP是一种编程模型，强调将复杂的逻辑分解出小的模块，特性是继承，封装和多态。

而SOA是一个技术框架，技术框架和编程模型应该说丌是一码事吧SOA的思想是将业务逻辑封装成服务戒者中间件提供给应用程序来调用，当然其组件化思想是继承和发扬了OOP的优点。

南京大学计算机系面试资料

不难。

一个小时的机试，去年的题目是用C++（规定只能用C++）写几个关于链表函数；之后是面试，一个教室，三位老师，老师会问跟他们自己领域有关的内容，我面试的第一位老师搞软件工程的，问的几乎都是软件工程，第二位老师问得比较多的是跟课程相关的，比如操作系统，体系结构的基础知识，第三位问的是Java相关的，线程产生有哪些方式，不同方式的区别，让我用英语描述MVC及其作用。

大概就这样，淡定就好，两天后拿到offer，最后选择了中科院。

说说如果让你设计一个“智能书架（或者智能冰箱之类的）”，你会使用什么样的技术。

为什么？没有其他提示。

2、算法题，上台阶问题：一共n级台阶，你一次只能上一级或者两级，请问有几种不同的走法？提示，用递归实现。

其实就是斐波那契数列的应用。

3、用英语叙述一下你在研究生期间有什么打算，准备从事哪个方向的研究，请详述。

4、翻译一段英文论文。

其实就一段话，但不认识的词儿超多。

5、介绍一下图灵机6、介绍一下图像处理方面的几个相关算法。

什么都不知道。

7、介绍一下你的毕设情况，越详细越好。

8、用英语介绍一门你比较熟悉的语言。

9、算法题：一次性遍历出一个单链表的中间节点。

10、介绍一下有关嵌入式的东西，譬如说嵌入式系统、软件之类的。

11、智力题：17min过桥问题。

同学的一个是黑帽子白帽子问题。

用英语介绍下你的专业和计算机专业的关系（本人本科专业是信息管理与信息系统）2、整数1-2n中任取n+1个数，必定有两个互素。

是和不是都要说明理由3、给了一段英文文章，关于图形方面的，翻译一下4、现在让你做一个预测股票走势的软件，你会怎样设计5、英文自我介绍6、概率上面的中心极限定理t1：英语成绩怎么样，要看考研成绩单英文详细描述你最善长的一件事（没有准备，不知道用什么词，所以换了个问题今天早上发生的事）为什么跨考计算机，具体地说说将来想从事的方向介绍毕设（知道我跨考的也问，好奇为主吧）t2(貌似是做软件工程，因为我没学过软件工程，所以没问这方面的知识点):介绍一下静态和动态多态快排的时间、空间复杂度，包括一般和最坏情况设计一个实现搜索引擎的算法，概要地描述一下英文描述今天的天气t3:英文描述印象最深刻的老师，包括事件和感想（给了一两分钟准备）指针传递与值传递的不同说说平时接触过的软件中用到malloc的地方谈谈对自己本专业和计算机专业的认识1、淘米老鼠的第2题。

原子结构如何在元素中找到它 Atomic Structure 英语作文论文

原子结构如何在元素中找到它Atomic StructureHow to find it in an elementWe all have studied the atom and the atomic structure in our academics. The atomic structure of an element is all about the composition of its nucleus as well as the array of its electrons around the nucleus. This mainly constitutes of electron, proton, and neutron. It has been rightly said that the foundation of chemistry is an atom. Since all the living as well non-living material is made up of atom; this is why it is important to know about its structure. Atom is a tiny portion of the element that participates in the chemicalreaction. Atoms are made up of elements called electron, proton, and neutron. The organized way of elements of an atom is called the atomic structure of an element.An atom consists of a nucleus and a series of orbitals. The nucleus is situated in the centre of the atom and orbitals are surrounding them. Further, the atom of an element is made up of three different types of sub-atomic particles namely electron, proton and neutrons. The nucleus of the atom consists of protons which are positively charged and neutral neutrons. Electrons which are negatively charged elements of an atom revolve around the nucleus in many separate paths which are called orbitals.In simple words, it can be said that the atomic structure of an element consists of positivelycharged protons and neutral neutrons in the nucleus and negatively charged electrons surrounding the nucleus. The atomic structure of various elements varies from each other because of the diverse types of elements present in it. Now we will tell you about the Atomic Structure.Elements of Atomic Structure:NucleusIt is the central part of an atom which constitutes of two sub-particles namely proton and neutron. It is held together by the “strong force”. The elements of the nucleus are combined together to form the overall atomic structure. Discussions on these sub-particles will be made in this article with an acute difference among them.The ProtonsThe protons of an atom are found inside the nucleus of the atom which is in the centre. The protons are positively charged particles. They have a mass of one atomic unit or it can also be known as one dalton. Rutherford discovered proton. It is the element having a positive charge and indicated by “P’. It is having a fixed charge of 1.6*10-^-19 Columbus and the relative mass of this element is 1.6*10^-24 g. In the nucleus of every atom, one or more protons are present there. This is one of the necessary elements of the nucleus to present an atomic structure. The number of this element presented in the nucleus of an atom is considered as the significant property of each element.The total number of protons presented in the atom is called the atomic number of that element and repres ented by “z”. Every element has different numbers of protons and this is why every element has its different atomic number. For example, Hydrogen has only one proton, carbon has a total number of six protons, and oxygen has eight protons. The presented number of protons in an element decides what kind of element is. The numbers of these particles are also used to determine the chemical nature of the elements.The ElectronThe electron was discovered by J.J Thompson who was a British physicist. An electron is indicated by “e” and a negatively charged element of an atom. Electrons are found to beorbiting the nucleus. The fixed charge of this element is a negative charge having 1.6 *10^-19 Columbus as well as a comparative mass of it that is very small and calculated as 0 is 1/1836. Electrons spin surround the nucleus of an atom in orbital paths. These are smaller than neutron as well as a proton. These particles are more than 1800 times tiny compared to either a neutron or proton. The nature of electron is used by the physicist to determine the properties of an atom such as boiling point, stability, and conductivity. The electrons in an atom are arranged in a very significant way.The electrons are found in the electron shells outside the nucleus. The electrons are negatively charged particles of an atom. The atomic structure of an element also takes into account the mass of the atom as well as thecharge of the atom. The arrangement of electrons in an atom referred to the orbital definition of the location of these electrons in an atom. They are arranged depending on the potential energy in various orbits. The inner orbitals revolving the nucleus are spherical whereas the outer orbitals have a much-complicated configuration. The energy level is represented by 1,2,3,4… and the orbits are by K, L, M, N AND so on. They can change their orbits. The electrons are attracted to protons. The number of electrons in an atom is always equal to the number of protons presented in that element. The numbers of the electron in an atom can change and determine the ion of the atom.The NeutronThe Neutron was discovered by James Chadwick. The neutrons are also found in the nucleus of the atom. In fact, its particles have an atomic mass of one unit which is also known asone dalton. Neutrons are neither positively charged nor are they negatively charged. These are neutral. It is represented by the “n” and has no charge. This element of the atomic structure is neutral having no charged. The relative mass of this element of an atom is the same as of a proton which is 1.6*10^-24. This element is available in the nucleus of an atom. The whole number of neutrons in an atom is called the neutron number. The mass of the neutron is a little larger compared to a proton. The chemical, as well as the nuclear properties of an element, are examined by both the number of protons and the number of neutrons presented in thenucleus. This is a significant element necessary for the stability of nuclei.Neutrons present in an atom can change its relative mass because the weights of the neutrons are equal to the weight of the proton along with electron together. The total number of the neutrons together with the protons is considered as an atomic mass number. The number of neutrons available in the nucleus of the atom can be different from that of others which differentiates the isotopes of an element to others. Now we will tell you about the Atomic Mass.Atomic MassThe mass of an atom is determined by the total mass of the neutrons and protons in the atom.This is because the neutrons and protons in the atom have a mass of one atomic unit, also known as one dalton. Mathematically speaking, the mass of a neutron or a proton is around 1.67X10^-24 grams. On the other hand, the mass of an electron is ignored while calculating the mass of an atom. This is because the mass of an electron is around 1/8000th</sup> of one atomic mass unit. Its mass is much insignificant in terms of protons and neutrons and it only weighs around 9.11×10^-28 grams. Hence, the greatest contribution while determining the mass of an atom is that of a neutron while the least contribution while determining the total mass of an atom is that of an electron.Atomic ChargeThe atomic structure of an element also includes the atomic charge. Electrons have really high contributions towards the charge of an atom. This is because the electrons are positively charged equal to the negative charge of the protons. The charge of the electrons is shown as -1 whereas the charge of the electrons is shown as +1. Neutrons have no charge and consequently, they make no contributions towards the charge of the atom. Studies hold that, in an atom which is ought to be neutrally charged, the number of protons will be equal to the number of neutrons. Hence, the positive, as well as the negative charges, will cancel out each other.Volume of AtomsIf the total size of protons, electrons and neutrons are taken into account, it would be evident that almost the entire volume of the atom is empty. Despite that, solid objects are impenetrable. This can be attributed to the fact that the electrons present in the atom are negatively charged and hence they cause repulsion to one another. This is what keeps the atoms from occupying the same place. The mass number of an atom, as well as the atomic number of an atom, also determines the atomic structure.Atomic NumberThe numbers of protons that are present determine the atomic number of an element. It is the number of protons or the atomic number is what distinguishes one element from the other.However, the number of neutrons and electrons are variables. The neutrons can vary in number and consequently they produce isotopes. The electrons of an element can also vary in number in the atoms and they end up producing ions. The example of carbon can be taken which has an atomic number of 6 and that is because the number of protons present is 6.Mass NumberThe mass number of an atom is determined by the sum total of the number of protons with the sum total of the number of neutrons. As mentioned earlier, the protons and neutrons of an atom have a similar mass and due to the negligible mass that electrons possess that cannot be taken into consideration. However, it is worth mentioning that the isotopes of anelement will have the same atomic number but they will have varying mass numbers.For determining or calculating the mass of an atom, the arithmetic mean of the mass numbers of the isotopes of the element which occur naturally is calculated. To cite an example, the mass of an atom of the element chlorine if found out to be 35.45 atomic mass unit and this is because chlorine has a number of naturally occurring isotopes most of whose atomic mass is either 35 amu with a combination of 17 protons and 18 neutrons or an atomic mass of 37 amu with a combination of 17 protons and 20 neutrons.The atomic structure of any element also comprises of Isotopes. Isotopes are defined as the atoms of the same element but they vary inthe number of neutrons however they have the same number of protons. Even though the isotopes have a different number of neutrons, they do not have significant differences in their physical properties. Isotopes, however, have largely differentiated stabilities. The isotopes which are unstable generally emit electrons as well as alpha rays. In addition, protons, neutrons and positrons are also subject to be emitted with the electrons being subjected to capture to form a relatively stable configuration of atoms and it is done through a process of radioactive decay. The atoms which are newly created are high in energy, cause the emission of gamma rays and are called radioisotopes. Now we will tell you about different elements of Atomic Structure.The net charge of an atomThere is no net charge of the atom. Electrons are negatively charged particles though protons are positively charged particles. The positive charge equivalent to the proton and the negative charge of the electron can drop one another. In this way, the atom does not have any net charge. In an atom that is unbiased, the number of electrons revolving around the core and the number of protons inside the core.There were two critical disadvantages in the planetary model of the atom. The first is that unlike planets that rotate around a Sun, electrons are charged particles. In the traditional electromagnetism, as indicated by the Larmor formula, electric charge is known to discharge electromagnetic waves.The circling charge must be continually lost on the nucleus to the energy and the spiral, with it must collide in the little piece of one another. The other issue was that the models of planets couldn’t clarify the exceptionally high emissions and spectra of atoms.。

原子类atomic原理

原子类atomic原理概述Atomic，在化学中原子指的是不可分割的实体。

同样的在并发体系中，原子类则是所有操作都具有原子性的，也就是说它的一个操作一旦开始，就不会被其他线程干扰。

同时原子类是"并java并发体系"中，无锁方案的重要组成部分。

在之前的文章中–“并发知识梳理”，这篇文章中我们提出了累加器问题：当多个线程同时访问下面的累加方法时，最终结果小于实际累加值，每次执行时最终结果都是不确定的。

publicclassUnsafeSequence{privateint value=0;publicintgetNext(){return++value;}}当时分析后，我们知道造成这种不确定性的原因在于我们这个累加器的加法操作++value不具有原子性导致的，如果我们的加法操作使用原子操作则可以避免这个问题。

使用在下面的代码中，我们将原来的 long 型变量 count 替换为了原子类 AtomicLong，原来的 count +=1 替换成了count.getAndIncrement()，仅需要这两处简单的改动就能使add10K() 方法变成线程安全的，原子类的使用还是挺简单的。

publicclassSafeSequence{AtomicLong count=newAtomicLong(0);publiclonggetNext(){returncount.getAndIncrement();}}❓好像很神奇，但它内部是如何保证getAndIncrement()是原子操作的呢？CAS原理其实其内部原理非常简单，都是通过CAS（Compare And Sweep，比较并交换）指令来实现。

CAS 的原理是拿期望的值和原本的一个值作比较，如果相同则更新成新的值。

可能说起来比较抽象，我们结合getAndIncrement()源码来看：publicfinallonggetAndIncrement(){returnU.getAndAddLong (this, VALUE,1L);}U.getAndAddLong()方法的源码如下，该方法首先会在内存中读取共享变量的值，之后循环调用pareAndS()方法来尝试设置共享变量的值，直到成功为止。

__atomic_test_and_set 详解 -回复

__atomic_test_and_set 详解-回复首先，让我们来详细解释并了解[__atomic_test_and_set]原子操作。

[__atomic_test_and_set]是C/C++中的一个原子操作函数，它用于设置给定内存位置的值，并返回该位置之前的值。

在多线程环境中，原子操作是一种确保多线程操作不会互相干扰的机制。

使用这种机制，可以确保在多个线程同时访问一块共享内存时，每个线程都可以在没有干扰的情况下完成其操作。

现在，让我们逐步分解[__atomic_test_and_set]原子操作的实现和功能。

一、原子操作的定义和需求：原子操作是指能够在一个不可分割的操作中完成更新的操作。

它可以作为一种同步机制，确保在多个线程之间进行共享数据的读取和写入时，不会出现数据竞争的情况。

在多线程编程中，线程间的竞争条件可能会导致程序运行的不确定性，例如，当两个线程同时进行写操作时，可能导致数据的覆盖或不一致性。

因此，原子操作是必不可少的。

二、[__atomic_test_and_set]的功能：[`__atomic_test_and_set]函数用于设置给定内存位置的值，并返回该位置之前的值。

它有两个参数：第一个参数是对一个内存位置的指针，第二个参数是对一个内存模型标识符的引用。

这个函数首先读取给定内存位置的值，然后将指定内存位置的内容设置为非零值。

最后，它返回之前的值。

这个过程是原子的，不会被其他线程干扰。

三、使用[__atomic_test_and_set]的示例：下面是一个使用`__atomic_test_and_set]的示例程序，它演示了两个线程同时访问共享内存的情况：cpp#include <stdio.h>#include <stdlib.h>#include <pthread.h>int shared_value = 0;void *thread_function(void *arg) {int thread_id = *((int *)arg);int old_value;do {old_value = __atomic_test_and_set(&shared_value,__ATOMIC_SEQ_CST);} while (old_value != 0);printf("Thread d modified shared value\n", thread_id);__atomic_clear(&shared_value, __ATOMIC_SEQ_CST);pthread_exit(NULL);}int main() {pthread_t threads[2];int thread_ids[2] = {1, 2};for (int i = 0; i < 2; i++) {pthread_create(&threads[i], NULL, thread_function, (void *)&thread_ids[i]);}for (int i = 0; i < 2; i++) {pthread_join(threads[i], NULL);}printf("Shared value: d\n", shared_value);return 0;}在这个示例中，两个线程`thread_function` 同时访问共享变量`shared_value`。

操作系统课程设计

计算机与信息学院操作系统课程设计专业班级信息安全10-02班学生姓名及学号刘禹 20103046课程教学班号0002任课教师田卫东老师实验指导教师田卫东老师实验地点逸夫实验楼5072012~2013学年第2学期一、题目：多进程/线程编程：哲学家问题。

设置进程/线程，描述哲学家；随机启动哲学家，显示进程/线程执行状态；随着线程的执行，更新显示；编写正确的哲学家程序，设法延迟线程的执行，使之出现死锁；编写正确的哲学家程序，保证不出现死锁；题目描述：哲学家问题：由Dijkstra提出并解决，是典型的同步问题。

问题描述有多个（设为number）哲学家共用一张圆桌，分别坐在周围number个椅子上，在圆桌上有number个碗和number个筷子，他们的生活方式是交替地进行思考和进餐。

平时，一个哲学家进行思考，饥饿时便试图取用其左右最靠近他的筷子，只有在他拿到两只筷子时才能进餐，进餐完毕，放下筷子继续思考。

二、开发环境：开发工具：Visual C++ 6.0 ；开发环境：Windows下多进程/线程编程开发语言：C++三、总体算法：设哲学家人数为number（程序中赋值为5），采用哲学家问题的一种解法：（1）采用p、v操作，哲学家拿起筷子后就持有该资源，等到吃完放下筷子后其他哲学家才能拿起该筷子；同样，哲学家需要的某筷子被其他哲学家持有时，等待，进程因而阻塞。

（2）第一个到第number-1个哲学家采用一种进程描述，先左手拿筷子，之后右手才能拿。

（3）只有最后一个哲学家定义另一进程：必须拿右手边筷子，之后再拿左手边的筷子。

（4）所采用方法其实质是编号资源的方法，把筷子看为临界资源，按其序号编号，哲学家只能由小到大申请资源即为本算法。

（5）出现死锁的方法：最后一个哲学家也是先拿左手边的筷子，再拿右手边的筷子时就容易出现死锁。

四、详细设计1.若干常量、变量的定义：const int number=5;//哲学家人数const int sleep_time=10;//显示时间const int eating_time=50;//吃饭显示的时间CRITICAL_SECTION chopstics[number];//临界资源：筷子CRITICAL_SECTION cs;//控制打印的临界区变量，由于一段时间只能允许一个进程打印到屏幕，所以屏幕也为临界资源2.状态显示函数：定义了若干函数显示当前状态：thingking(int i) （哲学家i正在思考）,hungry(int i) （饥饿）,pickup(int i,int j) （哲学家i拿起了筷子j）,eating(int i) （正在吃饭）,putdown(int i,int j) （哲学家i放下筷子j）。

Atomic Structure e of Si-Rich 6H-SiC( 000 1 )-2×2 Surface

Atomic Structure of Si-Rich 6H-SiC(1000)-2×2 SurfaceY. Hoshino1,3*, R. Fukuyama1, Y. Matsubara1, T. Nishimura1, S.Tanaka2,M. Kohyama2, and Y. Kido1Abstract000)-2×2 surface has been determined by The atomic structure of the Si-rich 6H-SiC(1high-resolution medium energy ion scattering (MEIS) and photoelectron spectroscopy using synchrotron-radiation-light (SR-PES). The MEIS analysis reveals the fact that Si adatoms (0.2-0.3 ML) overlie a Si-adlayer (0.8-0.9 ML) sitting on the bulk-truncated surface (1ML=1.22×1015 atoms/cm2). In fact, we observed two surface-related components in the Si 2p core level spectra corresponding to the adatom and the adlayer Si atoms and the intensity ratio of the former to the latter was ~1/3. On the other hand, the C 1s core level observed has the bulk component only. We propose an adatom-adlayer model satisfying the 2×2 surface reconstruction, three-fold symmetry and the results obtained by MEIS and SR-PES. Further MEIS analysis using focusing/blocking effect clearly shows that the Si-adatoms take an H-site. The surface structure predicted by the ab initio molecular dynamics calculation coincides with the above structure model except for slight lateral displacements of the Si-adlayer and reproduces well the present experimental results of MEIS and SR-PES.1 Dept Physical Sciences, Faculty of Science. and Engineering, Ritsumeikan University, Kusatsu, Shiga 525-85772 National Institute of Advanced Industrial Sci. and Technol, Ikeda, Osaka 563-85773 Dept. Electronic Sci. and Eng., Kyoto University, Nishikyo-ku, Kyoto 610-8510I. INTRODUCTIONSilicon carbide has attracted much attention as the best candidates for high temperature, high power and high frequency electronic devices. In the device fabrication, the atomic structure of crystal surfaces influences the quality of epitaxial layers and the reaction processes with various gases and metal films. So far, a lot of investigations[1-5] have been done for the Si-terminated (0001) surfaces mainly from practical aspect of its low interfacial defect densities and good quality of the epilayers. In contrast, we have only a few reports on the C-terminated ()1000surfaces because of its large defect levels at oxide/semiconductor interfaces and difficulties in controlling dopant concentrations. Recently, Fukuda et al.[6] succeeded in considerable reduction of the interfacial levels by a pyrogenic oxidation followed by annealing under hydrogen ambient. In addition, the oxidation rate for the C-terminated SiC surface is much larger than that for the Si-terminated one. Therefore, now it is strongly required to characterize the oxidation and the metal/SiC contact formation for the C-terminated surface.It is known that the SiC()1000surface takes several surface reconstructions dependent on sample preparation. Bernhardt et al. reported that a (3×3) surface reconstruction appeared by heating at 1050°C for 15 min in an ultra-high vacuum (UHV) and prolonged heating at 1075°C led to a Si-depleted 2×2 surface ([2×2]C)[7,8]. These two phases tend to coexist presumably due to nearly equal surface energies of the two structures. Scanning tunneling microscope (STM) and quantitative low energy electron diffraction (LEED) analyses showed that the [2×2]C surface consisted of Si adatoms of 0.25 ML (1 ML = 1.22×1015 atoms/cm2) taking an H3-site. Heating at temperatures above 1150°C led to a (1×1) graphitized surface. On the other hand, annealing at 1150°C in Si-flux of about 1 ML/min for 20 min formed a 000(-2×2 surface ([2×2]Si). Johansson et al.[9] also prepared the 3×3 Si-enriched Si)1reconstructed surface by heating at 1050°C and concluded from C 1s core level analysis that the surface consisted of at least two carbon overlayers.000-2×2 surface by heating at 950°C In this work, we first prepared the Si-enriched SiC()1for 5 min in UHV after pre-deposition of Si (3 ML). It is shown that this 2×2 surface is the Si-enriched [2×2]Si surface whose existence was first reported by Bernhardt et al.[7,8]. Unfortunately, however, the surface structure has not been clarified yet. High-resolution medium energy ion scattering (MEIS) combined with photoelectron spectroscopy using synchrotron-radiation-light (SR-PES) is a powerful tool to determine the atomic configuration of surfaces and interfaces. The structural analysis was carried out in situ at the beamline named SORIS allowing both MEIS and SR-PES. We also performed the ab initio calculations based on the density-functional theory (DFT) using the pseudopotential method. Finally, an energetically stable and most probable surface structure satisfying the experimental results is proposed.II. EXPERIMENTWe purchased N-doped 6H-SiC(1000wafers from CREE Inc. and cut them into small pieces with a typical size of 10×10 mm2. After cleaning the surface by a modified RCA method[10], the sample was introduced into an UHV chamber and degassed at 600°C for 5 h. Then cooling it down to room temperature (RT), a small amount of Si (3 ML) was deposited by molecular beam epitaxy (MBE) and after that the sample was annealed at 950°C for 5 min. Reflection high-energy electron diffraction (RHEED) showed a sharp 2×2 pattern with strong Kikuchi lines. This surface is stable without coexistence of other surface phases.The experiment was performed at beam line 8 constructed at Ritsumeikan SR Center, which combined MEIS with SR-PES. All the systems were working under UHV conditions. The samples were heated by infrared-radiation and the temperature was monitored with a Pt-Rh thermocouple set about 1mm above the sample. A well collimated He+ beam was incident on a sample and backscattered He+ ions were detected by a toroidal electrostatic analyzer (ESA) with an excellent energy resolution (E∆) of 9.0×10-4. On the other hand,E/emitted photoelectrons were analyzed by a hemispherical ESA with an energy resolution of about 10 meV at a typical pass energy of 2.95 eV. The total energy resolution was estimatedto be 100-150 meV including contributions from a Doppler broadening and energy spreads of incident photons. The details of the experimental setup are described in the previous reports[11,12]. The present MEIS analysis determined the elemental depth profiles and the atomic configuration using the ion blocking and focusing effects. Complementally, the information on the chemical bonding and the electronic structure was obtained by SR-PES analysis.It must be noted that the toroidal ESA detected only He+ ions. In order to determine the absolute amount of atoms of interest, we need the scattered He+ fractions as a function of scattered He+ velocity. For He ions scattered from low Z atoms (Z < ~20) located near a top-surface, the He+ fraction does not reach an equilibrium[13]. So we measured in advance the surface peaks for amorphous-Si and graphite targets and obtained the He+ fractions, which was enhanced by 120 % and 200 %, respectively compared with the equilibrium He+ fractions which were derived from the scattering yields from the deeper layers.III. AB INITIO CALCULATIONSThe ab initio molecular dynamics (MD) calculations were performed by the ab initio pseudopotential method based on the DFT within the local density approximation[14]. The Norm-conserved pseudopotentials developed by Troullier-Martins[15] were employed. In order to get the electronic ground state, we used the residual minimization method and direct inversion in the iterative subspace (RMM-DIIS)[16,17] and the conjugate-gradient method18 coupled with the efficient charge-mixing scheme[19,20]. Here, a plane-wave cutoff energy of 40 Ry was selected based on the tests of total energy convergence. We developed an appropriate supercell of a 6H-SiC()1000-2×2 surface. The supercell consists of a Si-adatom, three Si-adlayer atoms, six C-Si bilayers with a surface C layer, four hydrogen atoms for termination of dangling bond of the backside surface C atoms. In addition, the supercell includes six vacuum layers, which sufficiently separates each 2×2 surface slab. Three and six sampling k-points for self-consistent calculations were used. Thus we obtainstable atomic configurations through relaxation processes according to Hellmann- Feynman forces, which converged within 0.001 eV/Å per each atom. The bottom C-Si bilayer was fixed in the bulk configuration of the theoretical lattice constants (a=3.08 Å, c=15.08 Å). The present scheme was successfully applied previously to adhesive energies and Schottky barrier-heights of 3C-SiC/Ti interfaces[21,21] and 3C-SiC/Al interfaces[23,24].IV. RESULTS AND DISCUSSIONFigure 1 shows the MEIS spectrum observed for 120 keV He+ ions incident at 54.7° and scattered to 85.1° in the (0211) plane with respect to surface normal. Such a grazing emergence geometry makes it possible to improve the depth resolution and thus to separate the scattering components from each atomic layer. The surface peak from Si consists of two components, from a Si-adlayer and the top C-Si bi-layer. The C front edge shifts by ~1 keV to the lower energy side, indicating the existence of a Si-adlayer of about 1 ML on the top C-Si bilayer. The absolute amount of the Si-adlayer is derived to be 1.1±0.1 ML from the area of the deconvoluted surface peak and from the knowledge of the integrated beam current, the solid angle subtended by the toroidal ESA, and the He+ fraction. Taking a further glancing emergent geometry (88.0°) clearly resolves the previous Si-adlayer into two scattering components, Si-adatoms of ~0.25 ML and an underlying Si-adlayer of ~0.95 ML (see Fig. 2).Figure 3 shows the Si 2p core level spectra observed for incident photon energy of 140 and 280 eV at emission angles of 0° and 60° with respect to surface normal. The spectra are decomposed into three components, bulk and two surface-related (S1 and S2) ones, assuming Gaussian shapes, the singlet/triplet branching ratio of 1/2 and the energy interval of 0.60 eV. From the bottom to top, the spectra are more surface sensitive. The binding energy of the bulk component measured from the Fermi level was determined precisely by taking higher photon energy of 280 and 420 eV which gives a larger escape depth. The surface-related components denoted by S1 and S2 have lower binding energies of –1.21±0.1 and –0.41±0.1eV, respectively relative to that of the bulk. From the relative intensity ratios dependent on photon energy and emission angle, the components S1 and S2 are assigned to the Si-adatoms and the Si-adlayer, respectively. The intensity ratio of S1/S2 at normal emission is estimated to be about 1/3. This is consistent with the previous MEIS result. We also observed the C 1s spectra at photon energy of 420 eV and found the bulk component only. This also supports the surface structure taking Si-adatoms on a Si-adlayer overlying the top C-Si bilayer.Now we propose a probable surface structure satisfying the above MEIS and PES results considering the 2×2 reconstruction with three fold symmetry, as shown in Fig. 4. The shaded area indicates the 2×2 unit cell. In this model, three Si atoms of the Si-adlayer making a trimer bonded to one Si-adatom and the amounts of the Si adatoms and Si-adlayer are 0.25 and 0.75 ML, respectively. There are two possibilities concerning the location of the Si adatoms, a T-site (left side of Fig. 4) or an H-site (right side of Fig. 4). In Fig. 4, the adatom S is located at the center of the triangle PQR of the adlayer in the T-site model, and is located at the center of the triangle PQ’R’ in the H-site model. Note that the adlayer atoms P, Q, R, Q’ and R’ are common in both the models. According to this model, the amount of 0.25 ML of C atoms of the top C-Si bilayer is visible from the surface normal direction. To confirm this situation, we measured the MEIS spectrum at normal incidence. As a result, it was found that almost C atoms of the top C-Si bilayer are visible from the normal direction. This suggests a significant lateral distortion of the Si-adlayer.Considering the above structural model and the C3v symmetry, we performed the ab initio MD calculations. We determined the atomic configurations minimizing the total energy for three specific terminations of the stacking sequence, i.e., S-1(CACBAB), S-2(BCACBA) and S-3(ABCACB) against the bulk stacking sequence ABCACB of 6H-SiC25. The most stable surface structure is shown in Fig. 5. The surface structure does not depend on the termination of the stacking sequence as previous calculation of silicate adlayers on 33×surface[25], and no significant preference is seen for H- or T-site of the Si-adatoms. TheSi-adlayer consists of two types of trimers, PQR and PQ’R’ in Fig. 5, holding C3v symmetry and the larger one (PQ’R’) is bonded to one Si adatom (S in Fig. 5, where Si-adatoms take an H-site). The presence of the two types of trimers on the Si-adlayer means the lateral displacement of each Si atom above the C atom, which is consistent with the MEIS experiment. It is interesting that the triangular bonding of Si occurs stably at the smaller trimer (PQR) with enough bond charges. In contrast, the Si-adatoms take a perfectly symmetric position with respect to the bulk crystal structure.The bond length between the adlayer Si atom and the C atom is 1.88 Å, which is close to the bulk SiC bond length, 1.87 Å in theoretical calculations and 1.89 Å in experiments. The interlayer distance between the Si-adlayer and the top C-Si bilayer is slightly smaller than this Si-C bond length, because of the lateral displacement of each Si-adlayer atom, 0.35 Å, as shown in Fig. 5. The bond length between the adlayer Si atoms is 2.48 Å, which is slightly larger than that of 2.41 Å between the adlayer Si atom and the Si-adatom. These bond lengths are larger than the bulk Si bond length of 2.33 Å in theoretical calculations and 2.35 Å in experiments. The angle between the two adatom-adlayer bonds is 99.7°, which is about 9 % smaller than the tetrahedral angle of the bulk Si. The origin of this distortion may have some relation with the electronic structure of the dangling bond of the adatom.In any case, the present structure minimizes the number of dangling bonds and also minimizes the total energy, although there exist bond length and bond angle distortions associated with peculiar bonding network. There remain two kinds of dangling bonds at the top C-Si bilayer and the Si-adatom, which should cause some surface electronic states. Experimentally, we have recently observed the valence band spectra and found non-dispersive two surface states with energies of 1.5 and 2.2 eV below the Fermi level probably originating from the dangling bonds of the Si adatoms and of the top C-Si bilayer visible from the normal direction[26]. The detailed theoretical and experimental results of the electronic structure will be given in the near future.In order to determine whether the Si-adatoms take the H-site or T-site, we performed anazimuth scan for the scattering component from Si of the top C-Si bilayer at fixed incident and emergent angles of 54.7° and 85.1°, respectively. Such a grazing-angle emission condition enhances the focusing and blocking effects due to the contribution from a large number of atoms lying in a string not from a single atom only. If the Si-adatoms take an H-site, the He ions scattered from Si of the top C-Si bilayer would undergo a pronounced focusing effect at the [211]-azimuth (see Fig. 4). On the other hand, a T-site location givesa blocking effect in this scattering geometry. At the [001]-azimuth, both H- and T-sites1block the He ions scattered from Si of the top C-Si bilayer and lead to reduction of the scattering yield. Figure 6 shows the observed azimuth scan profile for the scattering component mainly from Si of the top C-Si bilayer. Here, 0° and 30° correspond to the [0011]-azimuth angle, respectively. We also performed Monte Carlo 1]- and [021simulations of ion trajectories assuming the atomic configurations predicted by the ab initio calculations. Apparently, the observed profile has a pronounced focusing effect at the [0211]-azimuth and thus supports the H-site location of the Si-adatoms.V. CONCLUSIONThe 6H-SiC(1000)-2×2 surface was prepared by heating at 950ºC in UHV after Si deposition of 3 ML. This surface corresponds to just the Si-rich [2×2] surface ([2×2]Si) which was first reported by Bernhardt et al.[7,8]. Our high-resolution MEIS and SR-PES analyses have revealed the fact that there exist the Si-adatoms (~1/3 ML) on the Si-adlayer (~0.8-0.9 ML) sitting on the top C-Si bilayer. The present MEIS and PES results predict the surface structure consisting of the Si-adlayer which forms hexagons in the unit of a trimer bonded to a Si-adatom (H- or T-site). However, the amount of the C atoms visible from the normal direction was significantly larger than that expected from the above structure model, suggesting a slight lateral distortion of the Si-adlayer. So, we performed the ab initio MD calculations using the plane waves as a basis function and employing the pseudopotential method. The surface atomic configuration minimizing the total energy basically supports the Si-adaom/Si-adlayer model but the bonding between the Si-trimer and Si-adatom issymmetrically elongated in the lateral plane. This structure model explains completely the MEIS and SR-PES results. However, there are no significant differences between the total energies and also the structures of the Si-adlayer calculated assuming the H- and T-site location of the Si-adatoms. The azimuth scan for the scattering component from Si of the top C-Si bilayer under a grazing emission condition clearly shows preference of the H-site location.ACKNOWLEDGEMENTSThe authors would like to thank Prof. H. Namba and Dr. K. Ogawa for maintaining the SR-PES system of BL-8 at Ritsumeikan SR Center. Special thanks are also due to Dr. T. Okazawa for his help in the MEIS experiment.References[1] J. R. Waldrop, J. Appl. Phys. 75, 4548 (1994).[2] L. I. Johansson, F. Owman, and P. Mårtensson, Phys. Rev. B 53, 13793 (1996).[3] H. Matsunami and T. Kimoto, Mater. Sci. Eng., R. 20, 125 (1997).[4] Y. Hoshino, T. Nishimura, T. Yoneda, K. Ogawa, H. Namba, and Y. Kido, Surf. Sci. 505, 234 (2002).[5] Y. Hoshino, S. Matsumoto, and Y. Kido, Phys. Rev. B 69, 155303 (2004).[6] K. Fukuda, S. Suzuki, T. Tanaka, and K. Arai, Appl. Phys. Lett. 76, 1585 (2000).[7] J. Bernhardt, M. Nerdling, U. Starke, and K. Heinz, Mater. Sci. Eng., B61, 207 (1999).[8] J. Bernhardt, A. Seubert, M. Nerdling, U. Starke, and K. Heinz, Mater. Sci. Forum, 338-342, 345 (2000).[9] L.I. Johansson, P.-A. Glans, N. Hellgren, Surf. Sci. 405, 288 (1998).[10] W. Kern and D.A. Poutinen, RCA Rev. 31, 187 (1970).[11] Y. Kido, H. Namba, T. Nishimura, A. Ikeda, Y. Yan, and A. Yagishita, Nucl. Instrum. Methods Phys. Res. B135–138, 798 (1998).[12] T. Nishimura, A. Ikeda, and Y. Kido, Rev. Sci. Instrum. 69, 1671 (1998).[13] Y. Kido, T. Nishimura, and F. Fukumura, Phys. Rev. Lett. 82, 3352 (1999).[14] J. P. Perdew and A. Zunger, Phys. Rev. B 23, 5048 (1981).[15] N. Troullier and J. L. Martins, Phys. Rev. B 43, 1993 (1991).[16] P. Pulay, Chem. Phys. Lett. 73, 393 (1980).[17] G. Kresse and J. Furthmüller, Phys. Rev. B 54,11169(1996).[18] D. M. Bylander, L. Kleinman and S. Lee, Phys. Rev. B 42,4021 (1990).[19] P. Pulay: J. Comp. Chem. 3,556 (1982).[20] G. P. Kerker: Phys. Rev. B 23, 3082 (1981).[21] S. Tanaka and M. Kohyama, Phys. Rev. B 64, 235308 (2001).[22] M. Kohyama and J. Hoekstra, Phys. Rev. B 61, 2672 (2000).[23] S. Tanaka and M. Kohyama, Appl. Surf. Sci. 216, 471 (2003).[24] J. Hoekstra and M. Kohyama, Phys. Rev. B 57, 2334 (1998).[25] W. Lu, P. Krüger and J. Pollmann, Phys. Rev. B 61, 13737 (2000).[26] Y. Hoshino, R. Fukuyama and Y. Kido, Phys. Rev. B 70, 165303 (2004).Figure CaptionsFIG .1. MEIS spectrum observed for 120keV He + ions incident on the 6H-SiC(1000)- 2×2 surface at an angle of 54.7º and scattered to 85.1º in the (0211) plane. Open circles indicate the observed spectrum and the solid curves are the best-fitted ones (bold: total, light: deconvoluted) assuming a Si-adlayer of 1.1 ML sitting on the top C-Si bilayer. The bold and thin curves correspond to total and the scattering component from each atomic layer, respectively.FIG . 2. MEIS spectrum observed for 120keV He + ions incident at incident and detection angles of 54.7° and 88.0°,respectively in the (0011) plane. The bright and deep gray areas correspond to the scattering components from theSi-adtoms and the Si-adlayer, respectively.Scattered He +Energy (keV)1st bilayer Si1st bilayer CC1.1( 0.1)ML Extra-layer SiScattered He +Energy (keV)FIG .4. Top and sideviews of a probable surface structure for the 6H-SiC(1000)-2×2. Open and closed circles denote Si and C atoms, respectively. The larger the circles size, the upper the lattice positions. The Si-adatoms take a T-site (left side) and an H-site (right side). P, Q, R, Q’ and R’ indicate adlayer atoms common to the two models, although strictly all the atoms are equivalent by C 3v symmetry in each model. S indicates the adatom.Binding Energy (eV)FIG . 5. Top and side views of the surface structure predicted by the ab initio calculations. Open and closed circles show Si and C atoms, respectively. P, Q, R, Q’ and R’ are the adlayer atoms and S indicates the adatom.FIG . 6. Azimuth scan profile observed for the scattering component mainly from Si of the top C-Si bilayer (open circles). The incident and detection angles were fixed at 54.7° and 85.1° with respect to surface normal. The azimuth angles of 0° and 30° correspond to the crystal planes of [0011] and [211], respectively.-202468101214161820222426283032N o r m a l i z e d Y i e l d (a .u .)Azimuth angle (deg.)。

java面试多线程中知识点 Atomic原子类总结整理

个人觉得这一节掌握基本的使用即可！• 1 Atomic 原子类介绍• 2 基本类型原子类– 2.1 基本类型原子类介绍– 2.2 AtomicInteger 常见方法使用– 2.3 基本数据类型原子类的优势– 2.4 AtomicInteger 线程安全原理简单分析• 3 数组类型原子类– 3.1 数组类型原子类介绍– 3.2 AtomicIntegerArray 常见方法使用• 4 引用类型原子类– 4.1 引用类型原子类介绍– 4.2 AtomicReference 类使用示例– 4.3 AtomicStampedReference 类使用示例– 4.4 AtomicMarkableReference 类使用示例• 5 对象的属性修改类型原子类– 5.1 对象的属性修改类型原子类介绍– 5.2 AtomicIntegerFieldUpdater 类使用示例1 Atomic 原子类介绍Atomic 翻译成中文是原子的意思。

在化学上，我们知道原子是构成一般物质的最小单位，在化学反应中是不可分割的。

在我们这里 Atomic 是指一个操作是不可中断的。

即使是在多个线程一起执行的时候，一个操作一旦开始，就不会被其他线程干扰。

所以，所谓原子类说简单点就是具有原子/原子操作特征的类。

并发包java.util.concurrent的原子类都存放在java.util.concurrent.atomic 下,如下图所示。

JUC原子类概览根据操作的数据类型，可以将JUC包中的原子类分为4类基本类型使用原子的方式更新基本类型•AtomicInteger：整型原子类•AtomicLong：长整型原子类•AtomicBoolean ：布尔型原子类数组类型使用原子的方式更新数组里的某个元素•AtomicIntegerArray：整型数组原子类•AtomicLongArray：长整型数组原子类•AtomicReferenceArray ：引用类型数组原子类引用类型•AtomicReference：引用类型原子类•AtomicMarkableReference：原子更新带有标记的引用类型。

《C++程序设计》(第2版)教学资源主教材习题解答 C++第六章习题解答

第六章模板与数据结构习题一、.基本概念与基础知识自测题6.1 填充题6.1.1 模板是为了实现代码的（1），它把数据类型改为一个（2），称为（3）程序设计。

模板包括（4）和（5）。

答案：（1）重用（2）设计参数（3）参数化(parameterize)（4）函数模板(function template)（5）类模板(class template)6.1.2 调用函数模板时，可以显式指定模板参数类型，也可以隐式进行，称为（1），这是根据（2）来决定的。

答案：（1）模板实参推演(template argument deduction)（2）一组实际类型或（和）值6.1.3 顺序查找可以用于（1）线性表，而对半查找可以用于（2）线性表。

答案：（1）无序的（所有）（2）有序的6.1.4 最常见的排序方式有（1）、（2）和（3）。

如果现有一个已排好序的线性表，在表尾添加了一个元素，采用（4）排序法使它重新成为有序的所需工作量最小。

答案：（1）选择（2）插入（3）交换（4）交换（可利用原来的有序性）6.1.5 给出以下指针的说明方式：指向一个4元素整型数组的指针为（1）；指向一个返回整型数，参数为两个整型数的函数的指针（2）；指向一个数组的指针，而该数组元素都是指向一个返回整型指针的无参函数（3）。

答案：（1）int(*p)[4]（2）int(*p)(int,int)（3）以指向6元素数组为例：int*(*)() (*p)[6]简答题6.2.1需要编写一个对多维数组通用的算法（即各维的大小未定），怎样才能把实参多维数组的信息全部传递到函数中去？答：最佳方法是用函数模板，多维数组用模板类型参数传递，各维的大小作为参数传递。

也可以用一维数组加各维的大小都作为参数传递。

6.2.2什么叫函数模板？什么叫模板函数？什么叫类模板？什么叫模板类？答：不受数据类型限制的通用型的函数使代码的可重用性大大提高。

把数据类型改为一个设计参数是一个可行的方案。

C++面向程序对象语言（双语）选择题+答案

C++⾯向程序对象语⾔（双语）选择题+答案mChapter I: Principles of Object-Oriented Programming1. Which of the following languages is not a procedure-oriented programming language?a) ALGOL(算法语⾔)b) COBOL（⾯向商业通⽤语⾔）c) FORTRAN（公式翻译程序语⾔）d) None of the above2. Which of the following programming approach （程序设计⽅法）used functions as a key concept to perform action-oriented tasks?(指导任务？？)a) Structured（结构化）programmingb) Modular （模块化）programmingc) Procedure-oriented programming（⾯向对象）d) Object-oriented programming3. Identify the drawback（缺点）of using procedure-oriented programming, if any:a) Data is hidden（隐藏）from external functions(外部函数)b) New functions can be added whenever necessaryc) Does not reflect real world problemsd) All of the above4. Which is not associated（联系）with Object-oriented programming?a) Data abstraction（数据提取）b) Automatic initialization（⾃动初始化）c) Dynamic binding（动态绑定）d) None5. The term operator overloading（操作符重载）in C++ refers to:a) Inheritance（继承）b) Message passing （信息传递）c) Polymorphism（多态性）d) None6. Which one of the following OOP concepts enables reusability （再利⽤）of components（成员）?a) Inheritanceb) Encapsulation（数据分装）c) Polymorphismd) All of the above7. The concept of hierarchical classification （层次分类）is related to:a) Abstractionb) Inheritancec) Function overloading（重载）d) None8. Object-based （基于对象）programming languages do not support:i. Inheritanceii. Dynamic bindingiii. Encapsulationiv. All of the abovea) Both i and iib) iii onlyc) iv onlyd) i, ii, and iii9. C++ does not supporti. Genericity（泛型C#）ii. Early bindingiii. Garbage collection（碎⽚收集）iv. Multiple Inheritance（多重继承）a) i onlyb) ii onlyc) iii onlyd) ii, iii and iv10. Which of the following is an Object-oriented programming language?i. Smalltalkii. Object Pascaliii. Javaiv. All of the abovea)Both ii and iiib)i onlyc) i ii onlyd) iv onlyChapter II: Beginning With C++1. Which one of the following options is true on the topic of Simula67?i. It is the first Object-oriented programming languageii. It?s a predecessor to C++iii. It was designed for doing simulationsiv. All of the abovea) Both i and iib) i onlyc) iii onlyd) iv only2. Which of the following features （特征）that distinguish（区别）object oriented programming from other conventional（常规的）programming?i. Structural design（结构设计）ii. Inheritanceiii. Modular programming（模）iv. bottom-upa) i onlyb) Both ii and iiic) Both ii and ivd) iv only3. Comments （解释）in C++ starts with _______ symbol.a) //b) \\c) **d) None of the above4. The insertion operator（插⼊操作符）is another name fora) input operatorb) output operator（输出操作符）c) extraction operatord) None of the above5. Which header file in C++ does contain function prototypes for memory（记忆）allocation （分配）?a)b)c)d)6. What is the output of the following code?int n=10;while (n<10)cout<< “Number:”<a) 10b) 11c) No outputd) None of the above7. Which of the following statements require the header file to be included?a) a=b/c;b) cout << a;c) c=sqrt(a);（平⽅根）d) Both a and c8. Identify the error, if any: char str_name …a…;a) str_name is not a valid variable nameb) Variables cannot be initialized at the time of declarationc) Missing = sign between str_name and …a?d) No error9. Name the library that must be included while using cin.a)b) （输⼊输出流）c)d)10. A C++ program structure is based on the concept ofa) Client-server modelb) N-Tier modelc) Both a and bd) None of the aboveChapter III: Tokens, Expre ssions and Control Structure s1. return is an example of aa) Keyword（关键字）b) Functionc) Statement （声明）d) Comment2. Which of the following is not a keyword?i. forii. friendiii. virtualiv. privatea) i onlyb) Both ii and iiic) Both i and ivd) ii, iii and iv3. Identify the valid （有效地）variable（可变的）name from the following:i. charii. var_nameiii. _varnameiv. str_name2a) Both i and iiib) Both ii and ivc) ii, iii, and ivd) i only4. Which of the following is not a user-defined data type?a) arrayb) structurec) uniond) class5. Write the equivalent C++ statement for the following expression, X= b a +*c/da) X=sqrt(a+b) *(c/d);b) X=(squareroot(a+b)*c)/d;c) X=()(b a +* c)/d;d) None of the above6. The statement int main() is a ___________.a) function prototypeb) function callc) function header lined) None of the above7. The declaration of global variables （全局变量） must be madea) inside the functionb) outside the functionc) in a function header lined) None of the above8. Identify the error, if any:double avg=tot/n;a) Declaration should not contain any expressionb) Initialization cannot be done in the declaration statementc) Dynamic initialization is not possibled) No error9. Which of the following is true about scope resolution operator?a) Qualifies a namespace member to its namespaceb) Allows you to access a global variablec) Qualifies the hidden variabled) All of the above10. Which of the following is not a member-dereferencing operator?a) ::*b) ::c) *d) *11. Which of the following is true about new operator?i. While using new operator, sizeof() operator is not neededii. It is also not necessary to use cast operatoriii. new operator can be overloadediv. All of the abovea) Both i and iib) Both i and iiic) ii onlyd) iv only12. Identify the manipulators in C++:a) Endl（输出时换⾏）b) setw（设置域宽）c) Both a and bd) None of the above13. a = (b = 5); The C++ statement is an example ofa) Compound （合成）assignmentb) Embedded （嵌⼊）assignmentc) Chained （束缚）assignmentd) Multiple （多重）assignment14. Water-fall model is associated witha) Control structuresb) Type conversionsc) Manipulatorsd) None of the above15. Selection structure is also known bya) Branching（歧义）b) straight linec) iterationd) None of the above16. Consider the following code:cout<< “Welcome”;This is an example ofa) C++ statementb) Return typec) Functiond) Both a and cChapter IV: Functions in C++1. Which of the following is true about a function call in a C++ program?a) A function must be called atleast onceb) A function cannot be called from other functionsc) A function may be called whenever it is necessaryd) Both a and c2. Function prototyping definesa) The return type of the functionb) The identifier of the functionc) The number and type of argumentsd) All of the above3. Find if the following function prototype contains any error:double area(int )a) No errorb) Variable name is not included in the argument listc) Semicolon（分号）is not foundd) None of the above4. Identify which function prototype exhibits（显⽰）the following: Name of the function is sample_calc, which receives two values of type double and returns no value;a) sample_calc(double, double);b) void sample_calc(double, double);c) double sample_calc(void);d) void sample_calc(double, double)5. When you call a function by passing the address of a data variable, it is called ________.a) Call by referenceb) Call by valuec) Call by two directionsd) Both b and c6. Which of the following function calls is correct while providing default arguments:I. double calc(int a, float b=12.0);II. double calc(int a=3, float b=12.0, int c);III. double calc(int a=3, float b, int c=8);IV. double calc(int a, float b=12.0, int c=8);a) I onlyb) II onlyc) Both I and IVd) Both II, and III7. Which function call does invoke the following function prototype?float sub1(int a, float b);a) X=sub1(5.0,6.5);b) X=sub1(5,6.5);c) X=sub1(5,6);d) Both b and c8. Identify the variables, which are local to the following function:int calc(int p, int n){int q;q=pow(p,n);return(q);}a) p and nb) p,n, and qc) qd) Cannot be determined without the main() function9. Which of the following statements is true about the function that contains the const argument?a) The function should not modify the const argumentb) Const declaration is necessary only when the arguments are passed by referencec) Both b and cd) None of the above10. Which of the following functions in C++ replace the usage of macros in C?a) friend functionb) virtual functionc) inline functiond) All of the aboveChapter IX: Pointers, Virtual Functions and Polymorphism1. Which of the following is true about pointers?a) A pointer is a data typeb) A pointer is a keywordc) A special type of integer （整数）variabled) None of the above2. Which of the following statement(s) is true according to the following statement? p=*ptr;a) p must be a pointer variableb) The value of ptr is assigned to the variable pc) The address of the pointer ptr is assigned to the variable pd) The value of the variable that the pointer ptr is pointing to is assigned to the variable p3. Compile time polymorphism is also known asa) early bindingb) static bindingc) static linkingd) All of the above4. C++ supports a mechanism virtual function to achievea) compile time polymorphismb) function overloadingc) run time polymorphismd) operator overloading5. Which of the following is NOT true about virtual functions?I. They cannot be static membersII. A virtual function can be a friend of another classIII. We can have virtual constructors, but we cannot have virtual destructors IV. They are accessed by using object pointersa) II onlyb) III onlyc) Both II and IIId) I, II and IV6. Consider the following code segment:int main(){int x, *x_ptr=&x;x=5;x_ptr=NULL;cout<< x << “ “ << *x_ptr;return 0;}What is the output of the above code?a) 0b) but addresses are samec) but having different addressesd) 57. Consider the following code segment:int main(){double f, *f_ptr=&f;f=5.25;f_ptr=4.5;cout<< “Value of f is:” << “ “ << f;return 0;}What is the output of the above code?a) Value of f is: 5.25b) Value of f is:c) Value of f is: 4.5d) Compiler error8. How will you assign value …5? to the variable …x? inside a member function using this pointer?a) this->x=5;b) this.x=5;c) x=5;d) None of the above9. Consider a class X, which includes a virtual function called X_output. The virtual function receives a float value and returns nothing. Find out the function prototype for the same.a) virtual void X_output(float );b) X:: virtual void X_output(float );c) virtual X:: void X_output(float );d) virtual :: void X_output(float );10. What would be the output of the following code?#include#includeclass A{public:virtual void disp(){cout<<"This is from class A";}};class B : public A{public:void disp(){cout<<"This is from class B";}};void main(){clrscr();A *s;s = new B();s->disp();}a) This is from class Ab) This is from class Bc) This is from class A This is from class Bd) None of the above11. Which of the following is NOT true about pure virtual function?a) It is also called do-nothing functionb) It cannot declare its own objectsc) A virtual function, which is not equated to zero is called a pure virtual functiond) None of the aboveChapter V: Classe s and Objects1. The declaration of a class includesa) Declaration of data membersb) Declaration of function prototypec) Return statements of functionsd) Both a and b2. By default, the members of a class area) privateb) publicc) protectedd) static3. Which OOP feature can be enabled by using private declaration for the class members?a) Data abstractionb) Polymorphismc) Encapsulationd) Modularity4. Which of the following will assign the value to the class member variable num?void getnum(int a)a) {num=a};b) {num=a;}c) {a=num};d) {a=num;}5. Which of the following is not true about member functions?a) Can access private data variablesb) Can call another function directly without using dot ope ratorc) Both a and bd) None of the above6. Identify all the members of the following class xyz:class xyz{int x,y;public:xyz();void calc(int a, int b);void output_calc(void);};a) Data members x and yb) Data members x and y, Constructor, and member functions calc() andoutput_calc()c) Data members x and y, and member functions calc() and output_calc()d) Constructor and member functions calc() and output_calc()7. Which of the following is not true about static member variable?a) Only one instance of static member can be createdb) Visible only within the classc) It can be initialized whenever necessaryd) Both a and b8. What is the general format of calling a static member function using a class name?a) class-name :: function-nameb) function-name :: class-namec) class-name :: function-name;d) function-name :: class-name;9. Which of the following is true while passing objects as function arguments? It is possiblea) to pass copy of the entire object to the functionb) to pass only the address of the object to the functionc) to pass the objects to a non-member functiond) All of the above10. A friend functionI. Can be invoked similar to other functions without using objectsII. Cannot access to other member functions directlyIII. Cannot be called using the object of the class to which it has been declared as friendIV. Can be declared only in the public part of a classa) I, II and IIIb) I, II and IVc) I, II, III and IVd) IV onlyChapter VI: Constructors and Destructors1. Which of the following is not true about constructors and destructors?a) They must have the same name as classb) They cannot return valuesc) They cannot be called more than once in a programd) They are called automatically2. The default constructor for class A isa) A :: A()b) A :: A(int)c) A :: A(int);d) A :: A();3. Which of the following statements do correctly describe the charac teristics of constructors?I. They cannot be inheritedII. They cannot be virtualIII. They need not be declared in the public sectionIV. We cannot refer to their addressesa) Both I and IIb) I, II, and IVc) Both II and IIId) I, III and IV4. Which of the following is called an implicit constructor for the class xyz?a) xyz(){ }b) xyz(int){}c) xyz(){ };d) None of the above5. Consider the following code segment:class simple{int a,b;public:simple();simple(char[]);};Which of the following would assigns the string “welcome” to the second c onstructor?a) simple s(“welcome”);b) simple s(welcome);c) simple s=”welcome”;d) None of the above6. Identify if any error in the following code segment1. class example2. {3. f loat x;4. p ublic:5. void example();6. example(int, float);7. };a) Line 7 should not include the semicolonb) Line 6 is an incorrect statementc) Line 5 cannot include voidd) No error7. Which of the following statements are true about copy constructors?I. It declares and initializes an object from another objectII. It will not be useful when arguments are passed by valueIII. It can be used to pass arguments by referenceIV. The compiler provides its own copy constructor, if it is not defined.a) I, II, IVb) All are correctc) Both II and IIId) III only8. The following statement creates a constant object of a class simple:const simple m(a,b);Which the following is true regarding the above statement?a) The compiler generates compile-time error if you try to change the values of …a?and …b?b) Const member is a function prototypec) The compiler generates an error if m tries to invoke non-const member functionsd) All of the above9. Which of the following statements do incorrectly describe the characteristics of destructors?I. A destructor is a member function, which has the same name as classnameII. A destructor receives only one argumentIII. A destructor does not return any valueIV. Usage of destructors in a program clean up memory space that is not useda) Both I and IIb) All are incorrectc) Both II and IIId) I, II and IV10. Which of the following is a correct description of the destructor …sample? and incl udes output statements (if possible)?a) sample :: ~sample() {cout << “Welcome”; }b) sample :: ~sample() { };c) void ~sample() { }d) ~sample() { };Chapter VII: Operator Overloading and Type Conversions1. Which of the following operators cannot be overloaded?a) ?:b) ::c) .*d) All of the above2. Operator Overloading is also known by the terma) runtime polymorphismb) compile-time polymorphismc) Both a and bd) None of the above3. Operator overloading is necessary becausea) C++ attempts to make the user-defined classes act like built-in typesb) To provide new definitions for most of the C++ operatorsc) To add user-defined data type like basic data typesd) All of the above4. Which of the following is required to overload an operator in C++?a) operator functionb) built-in functionc) Both a and bd) None of the above5. Identify the correct definition of an operator function for the class sample, which overloadsa unary minus operator and returns no values.a) sample :: oper-() { };b) void sample :: operator-() { }c) void sample :: operator-() { };d) sample :: operator-() { }6. Which of the following is not true about the code segment given below?float sample :: operator +(float x) {// code }a) It receives only one float type argument explicitlyb) It returns a float type valuec) It is a member function of sampled) None of the above7. Which of the following operators cannot use friend functions for overloading?I. ==II. ( )III. [ ]IV. ->a) I onlyb) II onlyc) II, III and IVd) I, II and III8. Which of the following is not true about operator overloading?a) Only existing operators can be overloadedb) Binary arithmetic operators (+,-,*,/) （⼆进制算⼦）need not return a valuec) It is impossible to redefine an operatord) All of the above9. Which of the following code segments will convert （改变）a class object namely sample totype double?a) sample :: operator double() { }b) operator double() { }c) sample :: operator double();d) operator double();10. Which of the following statements does correctly describe the casting operator function?a) It must not specify return typeb) It must be a class memberc) It must not have any argumentsd) All of the aboveChapter VIII: Inheritance: Extending Classe s1. Which of the following OOP concepts is supported by Inheritance?a) Abstractionb) Encapsulationc) Reusability （再利⽤）d) None of the above2. Which of the following does a derived class inherit from a base class?（从基类继承）a) public and protected（受保护）class membersb) public and private class membersc) only public datad) Everything3. Which of the following statement(s) is true, if a derived class is publicly inherited from abase class?I. The public members of the base class become public members of thederived classII. The public members of the base class become private members of the derived classIII. The public members of the base class are inaccessible（不可达到）to the objects of the derived class IV. All of the abovea) I onlyb) Both I and IIc) Both I and IIId) IV only4. Consider the following code segment:{int a;public:int b;void inp();}class B : A{// members of B}Which of the following statement(s) is NOT true regarding the above code?a) The public members, namely …b? and inp() of class A become private members of class Bb) The public members, namely …b? and inp() of class A can be accessed by the member functions of class Bc) The public members, namely …b? and inp() of class A are inaccessible to the objects of class Bd) None of the above5. According to the following code, which of the following class members does the derived class inherit from the base class?class base{float x;int y;public:int a;void get_a();void put_a();}class derived : public base{// members of B}a) float xc) get_a(), void put_a()d) All of the above6. Which of the following statement(s) is true about the visibility（能见度？）of inherited members?a) When a class is inherited in protected mode, the private members of the baseclass become protected members of the derived classb) When a class is inherited in private mode, only the public members of the base class can be inherited to the derived classc) When a class is inherited in public mode, the private members of the base class cannot be inheritedd) None of the above7. What are the visibility modifiers（修饰语）available in C++?a) publicb) privatec) protectedd) All of the above。

原子操作类atomicinteger详解

原子操作类atomicinteger详解原子操作是多线程编程中常用的一种技术，它能够确保数据的一致性并且避免竞态条件。

在Java中，java.util.concurrent.atomic包中提供了一套原子操作类，其中包括AtomicInteger、AtomicBoolean、AtomicLong、AtomicReference等，本文主要介绍其中的AtomicInteger类。

一、AtomicInteger的基本概念AtomicInteger是一个用来进行原子操作的整数类，它可以用来解决并发环境下数据竞争的问题，从而提升程序的并发性和性能。

其主要特点有：1.原子性：整个读取、修改、写入的操作过程是原子化的，即整个过程要么全部执行成功，要么全部执行失败，不会出现中间状态。

2.性能高效：它采用了一些硬件级别的机制来保证其性能，可以充分利用现代CPU的多级缓存结构，避免了锁的开销。

3.兼容性好：AtomicInteger提供了对volatile关键字的支持，可以保证多个线程之间的可见性。

二、AtomicInteger的使用方法AtomicInteger类提供了一系列的方法，包括原子的加减、读取、设置等操作。

1.构造方法在使用AtomicInteger类之前必须先创建一个实例，该类提供了两种构造方法public AtomicInteger()public AtomicInteger(int initialValue)其中AtomicInteger()构造方法会使用默认的初始值0，而AtomicInteger(int initialValue)构造方法则会使用指定的初始值。

2.原子加、减和CAS操作AtomicInteger类提供了三个方法：addAndGet、incrementAndGet、decrementAndGet。

这些方法能够保证对于操作的原子性，保证每一次操作都是完整的。

另外，类中也提供了CAS(Compare and Set)操作，方法名为compareAndSet，其原理是先判断当前值是否为预期值，如果是，则将新值设置进去，否则不做处理并返回false。

分布式数据库试题及答案

4.4.1. 找出发生故障时系统中的活动事务，确定出反做和重做事务集。 ........ 36 4.4.2. 用C或其他语言定义出数据库记录（D记录）和检查点记录（K记录）的数据结构。 36 4.5. 设数据项x,y存放在S1场地，u,v存放在S2场地，有分布式事务T1和T2,T1在S1场地的操作为 R1(x)W1(x)R1(y)W1(y),T2 在 S1 场地的操作为 R2(x)R2(y)W2(y);T1 在 S2 场地上的操作作为R1(u)R1(v)W1(u),T2在S2场地上的操作作为W2(u)R2(v)W2(v)。对下述2种情况，各举一种可能的局部历程（H1和H2），并说明理由 ......................... 36 4.5.1. 局部分别是可串行化，而全局是不可串行化的 ........................ 37 4.5.2. 局部和全局都是可串行化的。 ...................................... 37 4.5.3. 要求按照严格的2PL协议，加上适当的加锁和解锁命令，（注意，用rl(x)表示加读锁，wl(x)表示加对x加写锁，ul(x)表示解锁） .......................... 37 5. 二零年秋试题 .............................................................. 38 5.1. 概念题 ............................................................. 38 5.1.1. 解释对象数据库系统中面向对象的相关概念 .......................... 38 5.1.2. 从概念上比较对象数据库模型与对象关系模型 ........................ 38 5.1.3. 利用左深树、右深树、浓密树来进行查询优化的各自特点 .............. 38 5.1.4. 试解释影响并行数据库系统中并行算法性能的三个因数 ................ 39 5.1.5. 简述用爬山算法进行查询优化的基本思想 ............................ 39 5.2. 下面是某个公司一个人事关系数据库的全局模式： EMP={ENO*,ENAME,POSITION,PHONE} PAY={POSITION*,SALARY} ENO 为职员号， POSITION为岗位。SALARY表示岗位对应的工资，*对应的属性表示主关键字。该公司分布在两个场地上，其中，在场地1经常处理所有职员数据，而场地2只处理工资低于1000的职员数据，为了节省磁盘空间和增大处理局部性： .............................. 40 5.2.1. 将以上全局关系进行分片设计，写出分片定义和分片条件。 ............ 40 5.2.2. 指出分片的类型，并画出分片树。 .................................. 40 5.2.3. 给出分配设计。 .................................................. 40 5.3. 对题二所确定的分片模式，要求查询岗位为“salesman”的所有职员的姓名和工资，写出的在全局模式上的SQL查询语句，并要求转换成相应的关系代数表示，画出全局查询树。假设“salesman”的工资为800元。要求给出中间转换过程。 ............. 41 5.3.1. 进行全局优化，画出优化后的全局查询树。 .......................... 41 5.3.2. 进行分片优化，画出优化后的分片查询树。 .......................... 42 5.4. 按如下给出的条件，求出半连接优化计划和执行场地，并作后优化处理 ..... 42 5.5. 下面是当一个数据库系统出现故障时，日志文件中的信息 ................. 48 5.5.1. 画出对应的事务并发执行图。 ...................................... 49 5.5.2. 找出发生故障时系统中的活动事务，确定出反做和重做事务集。 ........ 49 5.5.3. 指出需要undo的和redo的数据记录。 ................................ 49 5.6. 设数据项x,y存放在S1场地，u,v存放在S2场地，有分布式事务T1和T2。T1在S1场地的操作为R1(x)W1(x)R1(y)W1(y)，T2在S1场地的操作为R2(x)R2(y)W2(y)；T1在S2场地上的操作作为R1(u)R1(v)W1(u)，T2在S2场地上的操作作为W2(u)R2(v)W2(v)。对下述2种情况，各举一种可能的局部历程（H1和H2），如果是可串行化的，指出事务的执行次序。对第3种情况，给出符合基本2PL协议的调度。（T1 加锁命令用L1(X)表示，开锁命令U1(X) 表示。对任何数据的加锁可在事务开始后立即进行）。 ........................... 49 5.6.1. 局部是不可串行化的。 ............................................ 50 5.6.2. 局部是可串行化的，而全局是不可串行化的。 ........................ 50 5.6.3. 局部是可串行化的，全局也是可串行化的。 .......................... 51 5.7. 设计一种满足下列要求的索引结构。 ...............................象的数据库是如何产生的，其基本原理是什么？有哪些创新特性？ ... 18 r i ∝ r j 一定等于r j ∝ r i 吗？在什么条件下r i ∝ r j = r j ∝ r i 成立? ......... 18

信息学奥赛一本通-第3章--第3节-堆与应用(C++版)精选全文

void work() {
int i, x, y, ans = 0; cin >> n; for(i = 1 ; i <= n ; i++) {
cin >> x; h.push(x); }
//建堆
for(i = 1 ; i < n ; i++) {
x = h.top();h.pop(); y = h.top();h.pop(); ans += x + y; h.push(x + y); } cout << ans << endl; }
//大根堆
pop_heap(heap + 1, heap + heap_size + 1, greater<int>()); //小根堆
return heap[heap_size--];
}
五、堆的应用
▪ 例1、合并果子(fruit) 【问题描述】在一个果园里，多多已经将所有的果子打了下来，而且按果子的不同种类分成了不同的堆。多多决定把所有的果子合成一堆。每一次合并，多多可以把两堆果子合并到一起，消耗的体力等于两堆果子的重量之和。可以看出，所有的果子经过n-1次合并之后，就只剩下一堆了。多多在合并果子时总共消耗的体力等于每次合并所耗体力之和。因为还要花大力气把这些果子搬回家，所以多多在合并果子时要尽可能地节省体力。假定每个果子重量都为1，并且已知果子的种类数和每种果子的数目，你的任务是设计出合并的次序方案，使多多耗费的体力最少，并输出这个最小的体力耗费值。例如有3种果子，数目依次为1，2，9。可以先将 1、2堆合并，新堆数目为3，耗费体力为3。接着，将新堆与原先的第三堆合并，又得到新的堆，数目为12，耗费体力为 12。所以多多总共耗费体力 =3+12=15。可以证明15为最小的体力耗费值。