[C++知识库]C++ Primer Plus 第6版 第12章 编程练习答案

1. 对于下面的类声明：

class Cow {
private:
?? ?char name[20];
?? ?char* hobby;
?? ?double weight;
public:
?? ?Cow();
?? ?Cow(const char* nm, const char* ho, double wt);
?? ?Cow(const Cow& c);
?? ?~Cow();
?? ?Cow& operator=(const Cow& c);
?? ?void ShowCow() const; ?
};

?给这个类提供实现，并编写一个使用所有成员函数的小程序。

#pragma once
#ifndef COW_H_
#define COW_H_
class Cow {
private:
	char name[20];
	char* hobby;
	double weight;
public:
	Cow();
	Cow(const char* nm, const char* ho, double wt);
	Cow(const Cow& c);
	~Cow();
	Cow& operator=(const Cow& c);
	void ShowCow() const;
};
#endif // !COW_H_

// cow.cpp
#define _CRT_SECURE_NO_WARNINGS

#include <cstring>
#include <iostream>
#include "cow.h"
using std::strncpy;
using std::strcpy;

// 默认构造函数
Cow::Cow() {
	hobby = nullptr;
	weight = 0.0;
}

// 构造函数
Cow::Cow(const char* nm, const char* ho, double wt) {
	strncpy(name, nm, 20);
	hobby = new char[strlen(ho) + 1];
	strcpy(hobby, ho);
	weight = wt;
}

// 复制构造函数
Cow::Cow(const Cow& c) {
	strncpy(name, c.name, 20);
	hobby = new char[strlen(c.hobby) + 1];
	strcpy(hobby, c.hobby);
	weight = c.weight;
}

// 析构函数，delete[]
// 最后将hobby置为空指针很重要
Cow::~Cow() {
	delete[] hobby;
	hobby = nullptr;
}

Cow& Cow::operator=(const Cow& c) {

	strncpy(name, c.name, 20);
	hobby = new char[strlen(c.hobby) + 1];
	strcpy(hobby, c.hobby);
	weight = c.weight;
	return *this;
}

void Cow::ShowCow() const {
	std::cout << "name: " << name
		<< ", hobby: " << hobby << ", weight: " << weight;
}

// 12-1
#include <iostream>
#include "cow.h"

int main() {
	Cow cow1;
	Cow cow2 = Cow("XiaoHua", "Run", 23.5);
	Cow cow3 = Cow("XiaoHong", "Eat", 45.7);
	Cow cow4 = cow2;
	cow1 = cow3;
	std::cout << "cow1: ";
	cow1.ShowCow();
	std::cout << "\ncow2: ";
	cow2.ShowCow();
	std::cout << "\ncow3: ";
	cow3.ShowCow();
	std::cout << "\ncow4: ";
	cow4.ShowCow();
	return 0;
}

2. 通过完成下面的工作来改进String类声明（即将String1.h升级String2.h）。

a. 对+运算符进行重载，使之可将两个字符串合并成1个。

b. 提供一个Stringlow()成员函数，将字符串中所有的字母字符转换为小写（别忘了cctype系列字符函数）。

c. 提供String()成员函数，将字符串中所有字母字符转换成大写。

d. 提供一个这样的成员函数，它接受一个char参数，返回该字符在字符串中出现的次数。

使用下面的程序测试：

// pe12-2
#include <iostream>
using namespace std;
#include "string2.h"
int main() {
?? ?String s1(" and I am a C++ student.");
?? ?String s2 = "Please enter your name: ";
?? ?String s3;
?? ?cout << s2;
?? ?cin >> s3;
?? ?s2 = "My name is " + s3;
?? ?cout << s2 << ".\n";
?? ?s2 = s2 + s1;
?? ?s2.Stringup();
?? ?cout << "The string\n" << s2 << "\ncontains " << s2.has('A')
?? ??? ?<< " 'A' characters in it.\n";
?? ?s1 = "red";
?? ?String rgb[3] = { String(s1), String("green"), String("blue") };
?? ?cout << "Enter the name of a primary color of mixing light: ";
?? ?String ans;
?? ?bool success = false;
?? ?while (cin >> ans) {
?? ??? ?ans.Stringlow();
?? ??? ?for (int i = 0; i < 3; i++) {
?? ??? ??? ?if (ans == rgb[i]) {
?? ??? ??? ??? ?cout << "That's right!\n";
?? ??? ??? ??? ?success = true;
?? ??? ??? ??? ?break;
?? ??? ??? ?}
?? ??? ?}
?? ??? ?if (success)
?? ??? ??? ?break;
?? ??? ?else
?? ??? ??? ?cout << "Try again!\n";
?? ?}
?? ?cout << "Bye\n";
?? ?return 0;
}

? ? ? ? 解答：

#pragma once
// string2.h
#ifndef STRING2_H_
#define STRING2_H_

#include <iostream>
using std::ostream;
using std::istream;

class String {
private:
	char* str;
	int len;
	static int num_strings;
	static const int CINLIM = 80;
public:
	String(const char* s);
	String();                    // 默认构造函数
	String(const String& s);     // 复制构造函数
	~String();
	int length() const { return len; }

	String& Stringlow();
	String& Stringup();
	int has(char c);       // 返回c在字符串中出现的次数
	// 重载运算符
	String& operator=(const String& s);
	String& operator=(const char* s);
	char& operator[](int i);
	const char& operator[](int i) const;
	// 友元函数
	friend bool operator<(const String& s1, const String& s2);
	friend bool operator>(const String& s1, const String& s2);
	friend bool operator==(const String& s1, const String& s2);
	friend ostream& operator<<(ostream& os, const String& s);
	friend istream& operator>>(istream& is, String& s);
	friend String operator+(const String& s1, const String& s2);
	friend String operator+(const String&, const char*);
	friend String operator+(const char*, const String&);
	// 静态函数
	static int HowMany();
};

#endif // !STRING2_H_

// string2.cpp
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstring>
#include <cctype>
#include "string2.h"
using std::strlen;
using std::strcpy;
using std::strcat;
using std::strcmp;
using std::ostream;
using std::istream;

// 初始化static类成员 ！！！
int String::num_strings = 0;

int String::HowMany() {
	return num_strings;
}

String::String(const char* s) {
	len = strlen(s);
	str = new char[len + 1];
	strcpy(str, s);
	num_strings++;
}

String::String() {
	str = nullptr;
	len = 0;
	num_strings++;
}

String::String(const String& s) {
	len = s.len;
	str = new char[len + 1];
	strcpy(str, s.str);
	num_strings++;
}

String::~String() {
	delete[] str;
	str = nullptr;
	len = 0;
	num_strings--;
}

String& String::Stringlow() {
	if (len == 0)  return *this;
	for (int i = 0; i < len; i++) {
		str[i] = std::tolower(str[i]);
	}
	return *this;
}

String& String::Stringup() {
	if (len == 0)  return *this;
	for (int i = 0; i < len; i++)
		str[i] = std::toupper(str[i]);
	return *this;
}

int String::has(char c) {
	if (len == 0)  return 0;
	int res = 0;
	for (int i = 0; i < len; i++) {
		if (str[i] == c)
			res++;
	}
	return res;
}

String& String::operator=(const String& s) {
	// 首先防止自己赋给自己
	if (this == &s)  return *this;
	len = s.len;
	delete[] str;
	str = new char[len + 1];
	strcpy(str, s.str);
	return *this;
}

String& String::operator=(const char* s) {
	len = strlen(s);
	delete[] str;
	str = new char[len + 1];
	strcpy(str, s);
	return *this;
}

// 主要为修改
char& String::operator[](int i) {
//	if(len == 0)  
//	if (i >= len)  return str[len - 1];
	return str[i];
}

const char& String::operator[](int i) const {
	return str[i];
}

bool operator<(const String& s1, const String& s2) {
	return strcmp(s1.str, s2.str) < 0;
}

bool operator>(const String& s1, const String& s2) {
	return strcmp(s1.str, s2.str) > 0;
}

bool operator==(const String& s1, const String& s2) {
	return strcmp(s1.str, s2.str) == 0;
}

ostream& operator<<(ostream& os, const String& s) {
	os << s.str;
	return os;
}

// !!!
istream& operator>>(istream& is, String& s) {
	char temp[String::CINLIM];
	is.get(temp, String::CINLIM);
	if (is)
		s = temp;
	while (is && is.get() != '\n')
		continue;
	return is;
}

String operator+(const String& s1, const String& s2) {
	int len = s1.len + s2.len;
	char* str = new char[len + 1];
	strcpy(str, s1.str);
//  strcpy(temp + s1.len, st.str);
	strcat(str, s2.str);
	return String(str);
}

String operator+(const String& s1, const char* s2) {
	int len = s1.len + strlen(s2);
	char* str = new char[len + 1];
	strcpy(str, s1.str);
	strcpy(str + s1.len, s2);
	return String(str);
}

String operator+(const char* s1, const String& s2) {
	return String(s1)+s2;
}

// pe12-2
#include <iostream>
using namespace std;
#include "string2.h"
int main() {
	String s1(" and I am a C++ student.");
	String s2 = "Please enter your name: ";
	String s3;
	cout << s2;
	cin >> s3;
	s2 = "My name is " + s3;
	cout << s2 << ".\n";
	s2 = s2 + s1;
	s2.Stringup();
	cout << "The string\n" << s2 << "\ncontains " << s2.has('A')
		<< " 'A' characters in it.\n";
	s1 = "red";
	String rgb[3] = { String(s1), String("green"), String("blue") };
	cout << "Enter the name of a primary color of mixing light: ";
	String ans;
	bool success = false;
	while (cin >> ans) {
		ans.Stringlow();
		for (int i = 0; i < 3; i++) {
			if (ans == rgb[i]) {
				cout << "That's right!\n";
				success = true;
				break;
			}
		}
		if (success)
			break;
		else
			cout << "Try again!\n";
	}
	cout << "Bye\n";
	return 0;
}

3. 新编写程序清单10.7和程序清单10.8描述的Stock类，使之使用动态内存分配的内存，而不是string类对象来存储股票名称。另外，使用重载的operator<<()定义代替show()成员函数。再使用程序清单10.9测试新的定义程序。

#pragma once
#ifndef STOCK20_H_
#define STOCK20_H_
#include <iostream>

class Stock {
private:
	char* company;
	int shares;
	double share_val;
	double total_val;
	void set_tot() { total_val = share_val * shares; }
public:
	Stock();
	Stock(const char* co, long n = 0, double pr = 0.0);
	~Stock();
	void buy(long num, double price);
	void sell(long num, double price);
	void update(double price);
	const Stock& topval(const Stock& s) const;

	friend std::ostream& operator<<(std::ostream& os, const Stock& st);
};

#endif // !STOCK20_H_

// stock20.cpp
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstring>
#include "stock20.h"
using std::strlen;
using std::strcpy;
using std::cout;

Stock::Stock() {
	company = nullptr;
	shares = 0;
	share_val = total_val = 0.0;
}

Stock::Stock(const char* co, long n, double pr) {
	int len = strlen(co);
	company = new char[len + 1];
	strcpy(company, co);
	if (n < 0)
		cout << "Number of shares can't be negative; "
		<< company << " shares set to 0.\n";
	else
		shares = n;
	share_val = pr;
	set_tot();
}

Stock::~Stock() {
	delete[] company;
}

void Stock::buy(long num, double price) {
	if (num < 0)
		cout << "Number of shares purchased can't be negative. Transaction is aborted.\n";
	else {
		shares += num;
		share_val = price;
		set_tot();
	}
}

void Stock::sell(long num, double price) {
	if (num < 0)
		cout << "Number of shares purchased can't be negative. Transaction is aborted.\n";
	else if (num > shares)
		cout << "You can't sell more than you have! Transaction is aborted.\n";
	else {
		shares -= num;
		share_val = price;
		set_tot();
	}
}

void Stock::update(double price) {
	share_val = price;
	set_tot();
}

const Stock& Stock::topval(const Stock& s) const {
	if (s.total_val > total_val)
		return s;
	else
		return *this;
}

std::ostream& operator<<(std::ostream& os, const Stock& st) {
	using std::ios_base;
	ios_base::fmtflags orig = os.setf(ios_base::fixed, ios_base::floatfield);
	std::streamsize prec = os.precision(3);
	os << "Company: " << st.company
		<< "  Shares: " << st.shares << '\n';
	os << "  Share Price: $" << st.share_val;
	os.precision(2);
	os << "  Total Worth: $" << st.total_val << '\n';
	os.setf(orig, ios_base::floatfield);
	os.precision(prec);
	return os;
}

// 12-3
#include <iostream>
#include "stock20.h"
using std::cout;

const int STKS = 4;
int main(){
    Stock stocks[STKS] = {
        Stock("NanoSmart", 12, 20.0),
        Stock("Boffo Objects", 200, 2.0),
        Stock("Monolithic Obelisks", 130, 3.25),
        Stock("Fleep Enterprises", 60, 6.5)
    };

    std::cout << "Stock holdings:\n";
    int st;
    for (st = 0; st < STKS; st++)
    {
        std::cout << stocks[st];
    }

    const Stock* top = &stocks[0];
    for (st = 0; st < STKS; st++)
    {
        top = &(top->topval(stocks[st]));
    }

    std::cout << "\nMost valuable holding:\n";
    cout << *top;
    return 0;
}

4. 请看下面程序定义的Stack类变量：

class Stack {
private:
?? ?enum { MAX = 10 };
?? ?Item* pitems; ? ? ? ? // hold stack items
?? ?int size;
?? ?int top;
public:
?? ?Stack(int n = MAX);
?? ?Stack(const Stack& st); ?// 复制构造函数
?? ?~Stack();
?? ?bool isempty() const;
?? ?bool isfull() const;
?? ?bool push(const Item& item);
?? ?bool pop(Item& item);
?? ?Stack& operator=(const Stack& st);

};

正如私有成员表明的，这个类使用动态分配的数组来保存栈项。请重新编写方法，以适应这种新的表示法，并编写一个程序来演示所有的方法，包括复制构造函数和赋值运算符。

#pragma once
#ifndef STACK2_H_
#define STACK2_H_
typedef unsigned long Item;

class Stack {
private:
	enum { MAX = 10 };
	Item* pitems;         // hold stack items
	int size;
	int top;
public:
	Stack(int n = MAX);
	Stack(const Stack& st);  // 复制构造函数
	~Stack();
	bool isempty() const;
	bool isfull() const;
	bool push(const Item& item);
	bool pop(Item& item);
	Stack& operator=(const Stack& st);

};

#endif // !STACK2_H_

#include <iostream>
#include "stack2.h"

Stack::Stack(int n) {
	pitems = new Item[n];
	size = n;
	top = 0;
}

Stack::Stack(const Stack& st) {
	size = st.size;
	pitems = new Item[size];
	top = st.top;
	for (int i = 0; i < top; i++) {
		pitems[i] = st.pitems[i];
	}
}

Stack::~Stack() {
	delete[] pitems;
	pitems = nullptr;
}

bool Stack::isempty() const {
	return  top == 0;
}

bool Stack::isfull() const {
	return top == size;
}

bool Stack::push(const Item& item) {
	if (isfull())
		return false;
	pitems[top++] = item;
	return true;
}

bool Stack::pop(Item& item) {
	if (isempty())
		return false;
	item = pitems[--top];
	return true;
}

Stack& Stack::operator=(const Stack& st) {
	if (this == &st)  return *this;
	size = st.size;
	delete[] pitems;
	pitems = new Item[size];
	top = st.top;
	for (int i = 0; i < top; i++)
		pitems[i] = st.pitems[i];
	return *this;
}

#include <iostream>
#include "stack2.h"
using std::cout;
using std::endl;

int main() {
	Stack st1(10);
	srand(time(0));
	for (size_t i = 0; i < 10; i++) {
		if (!st1.push(rand() % 100))
			cout << "Push error!" << endl;
	}
	if (!st1.push(0))
		cout << "Push 0 error!" << endl;

	Stack st2(st1);
	Stack st3 = st1;

	for (size_t i = 0; i < 11; i++) {
		Item item;
		cout << "#" << i + 1 << ": " << endl;
		if (!st1.pop(item))
			cout << "st1 pop error!" << endl;
		else
			cout << "st1: " << item << endl;
		if (!st2.pop(item))
			cout << "st2 pop error!" << endl;
		else
			cout << "st2: " << item << endl;
		if (!st3.pop(item))
			cout << "st3 pop error!" << endl;
		else
			cout << "st3: " << item << endl;

		cout << endl;
	}
}

5. Heather银行进行的研究表明，ATM客户不希望排队时间超过1分钟。使用程序清单12.10中的模拟，找出要使平均等候时间为1分钟，每小时到达的客户数应为多少（试验时间不短于100小时）？

#pragma once
// queue.h -- 12.10 -- interface for a queue
#ifndef QUEUE_H
#define QUEUE_H
class Customer {
private:
	long arrive;
	int processtime;
public:
	Customer() { arrive = processtime = 0; }
	void set(long when);
	long when() const { return arrive; }
	int ptime() const { return processtime; }
};

typedef Customer Item;

class Queue {
private:
	struct Node {
		Item item;
		struct Node* next;
	};
	enum { Q_SIZE = 10 };

	Node* front;      // 指向链表头
	Node* rear;       // 指向链表尾
	int items;        // 目前链表里的成员数量
	const int qsize;  // 链表里Item的最大数量

	Queue(const Queue& q): qsize(0){}
	Queue& operator=(const Queue& q) { return *this; }

public:
	Queue(int qs = Q_SIZE);
	~Queue();
	bool isempty() const;
	bool isfull() const;
	int queuecount() const;
	bool enqueue(const Item& item);   // 结尾加节点
	bool dequeue(Item& item);         // 链表头删节点
};

#endif // !QUEUE_H

// queue.cpp
#include <cstdlib>
#include "queue.h"

Queue::Queue(int qs) : qsize(qs) {
	front = rear = nullptr;
	items = 0;
}

Queue::~Queue() {
	Node* temp;
	while (front != nullptr) {
		temp = front;
		front = front->next;
		delete temp;
	}
}

bool Queue::isempty() const {
	return items == 0;
}

bool Queue::isfull() const {
	return items == qsize;
}

int Queue::queuecount() const {
	return items;
}

bool Queue::enqueue(const Item& item) {
	if (isfull())  return false;
	Node* add = new Node;
	// on failure, new throws std::bad_alloc exception
	add->item = item;
	add->next = nullptr;
	items++;
	if (front == nullptr)
		front = add;
	else  
		rear->next = add;
	rear = add;
	return true;
}

bool Queue::dequeue(Item& item) {
	if (front == nullptr)
		return false;
	item = front->item;
	items--;
	Node* temp = front;
	front = front->next;
	delete temp;
	if (items == 0)
		rear = nullptr;
	return true;
}

// time set to a random value in the range 1-3
void Customer::set(long when) {
	processtime = std::rand() % 3 + 1;
	arrive = when;
}

// 12-5
#include <iostream> 
#include <cstdlib>    // rand(), srand(), RAND_MAX
#include <ctime>      // time()
#include "queue.h"
const int MIN_PER_HR = 60;

bool newcustomer(double x);   // 判断是否有新用户到来

int main() {
	using std::cin;
	using std::cout;
	using std::endl;
	using std::ios_base;
	std::srand(std::time(0));  // 随机初始化rand()

	cout << "Case Study: Bank of Heather Automatic Teller\n";
	cout << "Enter maximum size of queue: ";
	int qs;
	cin >> qs;
	Queue line(qs);

	cout << "Enter the number of simulation hours (no less than 100): ";
	int hours;
	cin >> hours;
	// 循环每1分钟运行一次
	long cyclelimit = MIN_PER_HR * hours;

	//cout << "Enter the average number of customers per hour: ";
	//double perhour;
	//cin >> perhour;
	//double min_per_cust;
	//min_per_cust = MIN_PER_HR / perhour;

	Item temp;
	long turnaways = 0;           // 因满队列而拒绝接收的人数
	long customers = 0;           // 加入队列的数量
	long served = 0;              // 仿真期间服务的人数
	long sum_line = 0;            // 累计队列长度
	int wait_time = 0;            // 自动存取机的空闲时间
	long line_wait = 0;           // 累计在队列中的时间
	double avg_wait_time = 0.0;   // 顾客平均等待时间

	// 每小时使用人数初始值设为15，之后进行+1枚举测试
	double perhour = 15;
	double min_per_cust;

	// 仿真开始
	do {
		min_per_cust = MIN_PER_HR / perhour;
		turnaways = 0;           // 因满队列而拒绝接收的人数
		customers = 0;           // 加入队列的数量
		served = 0;              // 仿真期间服务的人数
		sum_line = 0;            // 累计队列长度
		wait_time = 0;            // 自动存取机的空闲时间
		line_wait = 0;           // 累计在队列中的时间
		avg_wait_time = 0.0;   // 顾客平均等待时间

	    // 清空队列
		while (!line.isempty())
			line.dequeue(temp);

		for (int cycle = 0; cycle < cyclelimit; cycle++) {
			if (newcustomer(min_per_cust))   // 有新顾客到来
			{
				if (line.isfull())
					turnaways++;
				else {
					customers++;
					temp.set(cycle);          // cycle = time of arrival
					line.enqueue(temp);
				}
			}
			if (wait_time <= 0 && !line.isempty()) {
				line.dequeue(temp);  // 下一个客户
				wait_time = temp.ptime();
				line_wait += cycle - temp.when();
				served++;
			}
			if (wait_time > 0)
				wait_time--;
			sum_line += line.queuecount();
		}

		// 报告结果
		if (customers > 0) {
			cout << "custormers accepted: " << customers << endl;
			cout << "   customers served: " << served << endl;
			cout << "          turnaways: " << turnaways << endl;
			cout << " average queue size: ";
			cout.precision(2);
			cout.setf(ios_base::fixed, ios_base::floatfield);
			cout << double(sum_line / cyclelimit) << endl;
			avg_wait_time = double(line_wait / served);
			cout << "  average wait time: " << avg_wait_time << " minutes\n";
		}
		else
			cout << "No customers!\n";
	} while ((perhour++) && ((avg_wait_time < 0.9) || (avg_wait_time > 1.1)));

	cout << "When perhour = " << perhour << ", the average wait time will be about 1 minute.\n";
	cout << "Done!\n";
	return 0;
}

bool newcustomer(double x) {
	return (std::rand() * x / RAND_MAX < 1);
}

6. Heather银行想知道，如果再开设一台ATM，情况将如何。请对模拟进行修改，以包含两个队列。假设当第一台ATM前的排队人数少于第二台ATM时，客户将排在第一队，否则排在第二队。然后再找出要使平均等候时间为1分钟，每小时到达的客户数应该为多少（注意，这是一个非线性问题，即将ATM数量加倍，并不能保证每小时处理的客户数量也加倍，并确保客户等候的时间少于1分钟）？

// 12-6
#include <iostream> 
#include <cstdlib>    // rand(), srand(), RAND_MAX
#include <ctime>      // time()
#include "queue.h"
const int MIN_PER_HR = 60;

bool newcustomer(double x);   // 判断是否有新用户到来

int main() {
	using std::cin;
	using std::cout;
	using std::endl;
	using std::ios_base;
	std::srand(std::time(0));  // 随机初始化rand()

	cout << "Case Study: Bank of Heather Automatic Teller\n";
	cout << "Enter maximum size of queue: ";
	int qs;
	cin >> qs;
	Queue line1(qs);
	Queue line2(qs);

	cout << "Enter the number of simulation hours (no less than 100): ";
	int hours;
	cin >> hours;
	// 循环每1分钟运行一次
	long cyclelimit = MIN_PER_HR * hours;

	//cout << "Enter the average number of customers per hour: ";
	//double perhour;
	//cin >> perhour;
	//double min_per_cust;
	//min_per_cust = MIN_PER_HR / perhour;

	Item temp;
	long turnaways = 0;           // 因满队列而拒绝接收的人数
	long customers = 0;           // 加入队列的数量
	long served = 0;              // 仿真期间服务的人数
	long sum_line = 0;            // 累计队列长度
	int wait_time1 = 0;           // 自动存取机1的空闲时间
	int wait_time2 = 0;           // 自动存取机2的空闲时间
	int line1_size = 0;           // 队列1当前长度
	int line2_size = 0;           // 队列2当前长度
	long line_wait = 0;           // 累计在队列中的时间
	double avg_wait_time = 0.0;   // 顾客平均等待时间

	// 每小时使用人数初始值设为15，之后进行+1枚举测试
	double perhour = 15;
	double min_per_cust;

	// 仿真开始
	do {
		min_per_cust = MIN_PER_HR / perhour;
		turnaways = 0;           // 因满队列而拒绝接收的人数
		customers = 0;           // 加入队列的数量
		served = 0;              // 仿真期间服务的人数
		sum_line = 0;            // 累计队列长度
		wait_time1 = 0;            // 自动存取机1的空闲时间
		wait_time2 = 0;
		line1_size = 0;
		line2_size = 0;
		line_wait = 0;           // 累计在队列中的时间
		avg_wait_time = 0.0;   // 顾客平均等待时间

	    // 清空队列
		while (!line1.isempty())
			line1.dequeue(temp);
		while (!line2.isempty())
			line2.dequeue(temp);

		for (int cycle = 0; cycle < cyclelimit; cycle++) {
			if (newcustomer(min_per_cust))   // 有新顾客到来
			{
				if (line1.isfull() && line2.isfull())
					turnaways++;
				else if(line1_size > line2_size){
					customers++;
					temp.set(cycle);          // cycle = time of arrival
					line2.enqueue(temp);
					line2_size++;
				}
				else {
					customers++;
					temp.set(cycle);
					line1.enqueue(temp);
					line1_size++;
				}
			}
			if (wait_time1 <= 0 && !line1.isempty()) {
				line1.dequeue(temp);  // 下一个客户
				line1_size--;
				wait_time1 = temp.ptime();
				line_wait += cycle - temp.when();
				served++;
			}
			if (wait_time2 <= 0 && !line2.isempty()) {
				line2.dequeue(temp);  // 下一个客户
				line2_size--;
				wait_time2 = temp.ptime();
				line_wait += cycle - temp.when();
				served++;
			}
			if (wait_time1 > 0)
				wait_time1--;
			if (wait_time2 > 0)
				wait_time2--;
			sum_line += line1.queuecount();
			sum_line += line2.queuecount();
		}

		// 报告结果
		if (customers > 0) {
			cout << "custormers accepted: " << customers << endl;
			cout << "   customers served: " << served << endl;
			cout << "          turnaways: " << turnaways << endl;
			cout << " average queue size: ";
			cout.precision(2);
			cout.setf(ios_base::fixed, ios_base::floatfield);
			cout << double(sum_line / cyclelimit) << endl;
			avg_wait_time = double(line_wait / served);
			cout << "  average wait time: " << avg_wait_time << " minutes\n";
		}
		else
			cout << "No customers!\n";
	} while ((perhour++) && ((avg_wait_time < 0.9) || (avg_wait_time > 1.1)));

	cout << "When perhour = " << perhour << ", the average wait time will be about 1 minute.\n";
	cout << "Done!\n";
	return 0;
}

bool newcustomer(double x) {
	return (std::rand() * x / RAND_MAX < 1);
}