(2011.07.31)自学《C++ Primer Plus》时做过的书上的习题！

        那时候做的题目，可能做出的质量不高，不过可以费尽了心思了，每次做题目都很有收获，而且，每当做出了一道题，都会相当地有成就感，做题目即能巩固问题，又要增加经验，发现问题，真是提升的好方法，所以也很喜欢做题，只是有时候会一道题花费很多时间，不过做出来了就后就会觉得是值得的。
        五个月左右，零零散散的时间，自学完了这个C++ Primer Plus ，下面是做过的题目，自己常看看可以温故而知新呢！

// 编程练习_02_06.cpp 
/*******************第02章（开始学C++），第06题，
编写一个程序，要求用户输入小时数和分钟数，在main()函数中，将这两个值传递给一个void函数，后者以下面这样的格式显示这两个值。
Enter the number of hours: 9
Enter the number of minutes: 28
Time: 9:28
******************************************************************************************************************************/

#include <iostream>									// a preprocesor directive
using namespace std;								// make definitions visible
int main()											// function header
{													// start of function body
	int a;											// declare an integar variable a
	int b;											// declare an integar variable b
	cout << "Enter the number of hours: ";			// message
	cin >> a;										// input the value of the variable a
	cout << "Enter the number of minutes: ";		// message
	cin >> b;										// input the value of the variable a 
	void display(int&, int&);						// function prototype for display()
	display(a, b);									// call the display()
	return 0;										// terminate main()
}													// end of function body

void display(int& x, int& y)						// define the display() function
{
	cout << "Time: " 
		<< x
		<< ":"
		<< y
		<<endl;
}

// 编程练习 04.01_display message.cpp

/***************************************************************
1.编写一个C++程序，如下述输出范例所示的那样请求并显示信息：
	What is your first name? Betty Sue 
	What is your last name? Yew
	What letter grade do you deserve? B
	What is your age? 22
	Name: Yew, Betty Sue
	Grade: C
	注意，该程序应该接受的名字包含多个单词。另外，程序将向下调整成绩，即向上调一个字母。
	假设用户请求A、B或C,所以不必担心D和F之间的空档。
********************************************************************************************/

#include <iostream>							// a preprocessor directive
#include <string>							// for enter the name
using namespace std;

int main()
{
// define
	string fname;							// define the first name
	string lname;							// define the last name
	char grade;								// define the grade
	int age;								// define the age

// calculate and cout
	cout << "What is your first name? ";
	getline(cin, fname);

	cout << "What is your last name? ";
	getline(cin, lname);

	cout << "What letter grade do you deserve? ";
	cin >> grade;
	getline(cin, *(new string));		//下面getline一次,忽略1行
	// cout << endl;

	cout << "What is your age? ";
	cin >> age;
	// cout << endl;

// cout the last message
	cout << "Name: " << lname << "," << fname <<endl;
	cout << "Grade: " << (char(grade+1)) << endl;

	return 0;
}

// 编程练习 04.01_display message_2.cpp

/***************************************************************
1.编写一个C++程序，如下述输出范例所示的那样请求并显示信息：
	What is your first name? Betty Sue 
	What is your last name? Yew
	What letter grade do you deserve? B
	What is your age? 22
	Name: Yew, Betty Sue
	Grade: C
	注意，该程序应该接受的名字包含多个单词。另外，程序将向下调整成绩，即向上调一个字母。
	假设用户请求A、B或C,所以不必担心D和F之间的空档。
********************************************************************************************/

#include <iostream>							// a preprocessor directive
#include <string>							// for enter the name
using namespace std;

struct message
{
	char *FirstName;
	char *LastName;
	char Grade;
	int age;
};
/*有错，连对象都还没有定义
int main()
{
	string temp;
	cout << "What is your first name? ";
	cin >> temp;
	message.FirstName = new char [strlen(temp) + 1];
	strcpy(message.FirstName, temp);
	cout << "What is your last name? ";
	cin >> temp;
	message.LastName = new char [strlen(temp) + 1];
	strcpy(message.LastName, temp);
	cout << "What letter grade do you deserve? ";
	cin.get();
	cin.get(message.Grade);
	cout << "What is your age? ";
	cin >> message.age;
	cout << "Name: " << message.*LastName << ", " << message.*FirstName;
	cout << endl << "Grade: " << ((char)(message.Grade  + 1)) << endl;
	cin.get();
	return 0;
}
***/
/*************************************************************************
打算参考书上的 
char *ps;
ps = new char[strlen(animal) + 1];
strcpy(ps, animal);
谁知道呢？挺多错误的。 
 G:\Temp\编程练习_04.01_display message_2.cpp:43: error: `LastName' undeclared (first use this function)
G:\Temp\编程练习_04.01_display message_2.cpp:43: error: (Each undeclared identifier is reported only once for each function it appears in.)
G:\Temp\编程练习_04.01_display message_2.cpp:43: error: expected primary-expression before '.*' token
************************************************************************/ 
int main()
{
	string temp;
	cout << "What is your first name? ";
	cin >> temp;
	message msg;
	msg.FirstName = new char [strlen(temp.c_str()) + 1];  // string 类型获取长度方法　strlen(string.c_str())
	strcpy(msg.FirstName, temp.c_str());
	cout << "What is your last name? ";
	cin >> temp;
	msg.LastName = new char [strlen(temp.c_str()) + 1];
	strcpy(msg.LastName, temp.c_str());
	cout << "What letter grade do you deserve? ";
	cin.get();
	cin.get(msg.Grade);
	cout << "What is your age? ";
	cin >> msg.age;
	cout << "Name: " << msg.LastName << ", " << msg.FirstName;
	cout << endl << "Grade: " << ((char)(msg.Grade  + 1)) << endl;
	cin.get();
	return 0;
} 

// 编程练习_04.09_nestedloopdisplaychar.cpp -- display linechar use nested loop

/**************************** **  题目要求  ** *******************************
9.编写一个使用嵌套循环的程序，要求用户输入一个值，指出要显示多少行。
　然后，程序将显示相应的行数的星号，
　其中第一行包括一个星号，第二行包括两个星号，依此类推。
　第一行包含的字符数等于用户指定的行数，在星号不够的情况下，在星号前面加上句点。
　该程序的运行情况如下：
  Enter number of rows: 5
  ....*
  ...**
  ..***
  .****
  *****
******************************************************************************/ 
 
 #include <iostream>
 using namespace std;
 
 int main()
 {
     cout << "Enter number of rows: ";
     int n;
	 cin >> n;
     while (n <= 0)
     	{
     	cout << "Please Enter again and the number must be equal or greater than zero: ";
     	cin >> n;
		}

	char point = '.';
	char asterisk = '*';
	int i;					// define two variable for circulate 
	int j;
	int m = n;				// mark the biggest number
	for (; n; --n)
	{
	 	for (i = n-1; i; --i)
	 		cout << point;
		for (j = 0; j < (m - n + 1); ++j)
			cout << asterisk;
		cout << endl;
	}
     return 0;
 }
 

// 很奇怪，在GCC里面运行正常，然后放在VC++6.0就不正常了。
// VC++ 2010 运行正常。


/************************************************************************************
网友修改后
--------------------Configuration: test - Win32 Debug--------------------
Compiling...
test.cpp
F:\Temp\test.cpp(26) : error C2248: 'brand' : cannot access private member declared in class 'CandyBar'
        F:\Temp\test.cpp(17) : see declaration of 'brand'
F:\Temp\test.cpp(30) : error C2248: 'weight' : cannot access private member declared in class 'CandyBar'
        F:\Temp\test.cpp(18) : see declaration of 'weight'
F:\Temp\test.cpp(34) : error C2248: 'calorie' : cannot access private member declared in class 'CandyBar'
        F:\Temp\test.cpp(19) : see declaration of 'calorie'
F:\Temp\test.cpp(41) : error C2248: 'brand' : cannot access private member declared in class 'CandyBar'
        F:\Temp\test.cpp(17) : see declaration of 'brand'
F:\Temp\test.cpp(42) : error C2248: 'weight' : cannot access private member declared in class 'CandyBar'
        F:\Temp\test.cpp(18) : see declaration of 'weight'
F:\Temp\test.cpp(43) : error C2248: 'calorie' : cannot access private member declared in class 'CandyBar'
        F:\Temp\test.cpp(19) : see declaration of 'calorie'
F:\Temp\test.cpp(118) : error C2593: 'operator >>' is ambiguous
F:\Temp\test.cpp(132) : error C2593: 'operator >>' is ambiguous
F:\Temp\test.cpp(145) : error C2593: 'operator <<' is ambiguous
Error executing cl.exe.

test.obj - 9 error(s), 0 warning(s)

************************************************************************************/


// 编程练习_04.9_CandyBarmain.cpp
/***************************************
9.结构 CandyBar 包含3个成员，
第一个成员存储了糖块的品牌；
第二个成员存储糖块的重量（可有小数）；
第三个成员存储了糖块的卡路里含量（整数）。
请编写一个程序，声明这个类。
使用new动态分配数组。
*****************************************/


#include <iostream>
#include <string>
using namespace std;

const int NuLL = 0;

// Candy Bar.h -- class include three data members

class CandyBar
{
 public:
  friend istream&operator >> (istream&, CandyBar&);  // operator << redefine 
  friend ostream&operator << (ostream&, CandyBar&);  // operator >> redefine
friend CandyBar* update();//---------------------------------------------------------------------------------
friend CandyBar* display();//因为要在类外访问私有,没有添加友元
 private:
  string brand;
  double weight;
  int calorie;
  CandyBar* next;    // to create a linked list
};

istream& operator >> (istream& input, CandyBar& cinCandyBar)
{
 cout << "Please input the brand of the sugar:";
 input >> cinCandyBar.brand ;
 cout << endl;
 
 cout << "Please input the weight of the sugar:";
 input >> cinCandyBar.weight;
 cout << endl;

 cout << "Please input how many calorie in this sugar:";
 input >> cinCandyBar.calorie;
 cout << endl;
 return input;//-------------------------------需要返回流,为了可以后续同类操作  cout>>a>>1>>"d";
}

ostream& operator << (ostream& output, CandyBar& coutCandyBar)
{
 output << coutCandyBar.brand << "\t\t" 
  << coutCandyBar.weight << "\t"
  << coutCandyBar.calorie << endl;
 return output;//----------------------------------------------------------同上
}

// CandyBarmain.cpp -- the main function of CandyBar
//---------------------------------------------------------------------------------------------------------------------
//---------------------------------------------------后面的函数要用,放main里访问不到
 CandyBar* head = NuLL;
 CandyBar* pa;
 CandyBar* pb;

int main()
{
 // CandyBar* pc;
 // CandyBar* temp;

 int choice();
 choice();
 return 0;
}

int choice()
{
 // the Candy Bar function
 CandyBar* update();
 CandyBar* display();

 // circulate choice
 while(1)
 {
  // display the message
  cout << "Welcome to our Candy Bar. " << endl
   << " 1.display the message of Candy Bar."<<endl
   << " 2.update the message of Candy Bar."<<endl
   << " 3.exit the system."<<endl
   << "Now, please choice a function you want to come true and enter the function number: ";

  //start realize the function
  char ch;
  switch (ch=cin.get()){      //      -----------------------------switch()  缺个大括号 {
case '1': 
 cout << "\n";//-------------------------------这之后的好几个\n写成/n了
 display();
 break;
case '2':
 cout << "\n";
 update();
 break;
case '3':
 cout << "\n Success to exit the system! ";
 return 0;
default : 
 cin.clear();//---------------------------------写错  clean
 while(ch!='\n')
  cin.get();
 cout << "\nplease enter a true number";
 break;
  }//-------------------------------------------------------同上,缺}
 }
 return 0;
}

CandyBar* update()
{

 // make sure update number
 cout << "How many message you want to update? ";
 int n;
 cin >> n;
 cout << endl;

 if (head == NuLL) // if it is first input
 { 
  pb = new CandyBar;
  head = pb;
  cin >> *head;
  pa = head;
  --n;
 }
 else    // search the address of last linked list 
  for (pa = head; pa->next != NuLL; pa = pa -> next);


 // input the data
 for (int i = 0; i<n; i++)
 {
  pb = new CandyBar;
  pa->next = pb;
  pa = pb;
  cin >> *pa;
 }

 pa->next = NuLL;
 return (head);
}

CandyBar* display()
{
 cout << " \n Brand\t\tWeight\tcalorie\n" ;
 for (pb = head; pb ; pb = pb->next)    //pb->next = NuLL   =号是赋值 ==是判断
                                                                  //这里循环后才执行  pb = pb->next,
                                                                  //所以判断pb就行,
  cout << *pb;                               //需要传的是引用,不是指针,先解指针
 return 0;
} 

// 编程练习_05.03_计算投资价值
/*********************  第3题  *****************************************
3.Daphne 以10%的单利投资了100美元。也就是说，每一年的利润都是投资额的10%，
即每年10元美元：
利息＝0.10*原始存款
而Cleo以5%的复利投资了100美元。
也就是说，利息是当前存款（包括获得的利息）的5%,:
利息＝0.05*当前存款
Cleo在第一年投资100美元的盈利是5%---得到了105美元。
下一年的盈利是105美元的5%即5.25美元，依此类推。
请编写一个程序，计算多少年后，Cleo的投资价值才能超过Daphne的投资价值，
并显示此时两个人的投资价值。
**************************************************************************/

#include <iostream>
using namespace std;
int main()
{
 	const double depositd = 100;			// d的本金 
 	double depositc = 100;					// c的本金 
 	const double interestrd = 0.10;	
 	const double interestrc = 0.05;
 	double interestd;						// d的利息 
 	double interestc;						// c的利息 
 	double sumd = 0;
	double &sumc = depositc;
 	int year;
	
	//先计算出第一年的 
	sumd += depositd;
	interestd = interestrd * depositd;
	
 	interestc = interestrc * depositc;
	 sumc += interestc;
 	
	 for (year = 2; sumc > sumd; ++year)
	 {
	 	 sumd += interestd;
	 	 interestc = interestrc * depositc;
	 	 sumc += interestc;
	  }
	  	  	  
	  cout << "Invest of Cleo: " << sumc << endl;
	  cout << "Invest of Daphne: " << sumd << endl;
	  cout << "After year: " << year << endl;
	  
	  while(2);
	  return 0;
}														  

// 编程练习_05.04_用循环显示数据.cpp
/******************************  第4题  **************************************
4.假设要销售 C++ For Fools 一书。请编写一个程序，输入全年中每个月的销售量
（图书数量，而不是销售额）。程序通过循环，使用初始化为月份字符串的char* 数组
（或string对象数组）逐月进行提示，并将输入的数据储存在一个int数组中。
然后，程序计算数组中各元素的总数，并报告这一年的销售情况。
*****************************************************************************/
#include <iostream>
#include <string>
using namespace std;
 	
int main()
{
 	const int monnum = 12;
 	int sale [monnum];
// string smonth[]是用C++中的string 定义的字符串数组
// 如果你想用C语言中的指针(呵呵，同样也是C++中的，只是在C++中少用了，大多都用string代替了)，
// 你就把 string smonth[]= 用 char *scmonth[]= 代替
 	string pointout[monnum] =
// char *scmonth[] =
	 {
	  	   "Please enter the distribution records of January: ",
		   "Please enter the distribution records of February: ",
		   "Please enter the distribution records of March: ",
		   "Please enter the distribution records of April: ",
		   "Please enter the distribution records of May: ",
		   "Please enter the distribution records of June: ",
		   "Please enter the distribution records of July: ",
		   "Please enter the distribution records of August: ",
		   "Please enter the distribution records of September: ",
		   "Please enter the distribution records of October: ",
		   "Please enter the distribution records of November: ",
		   "Please enter the distribution records of December: "
		    };
    int i = 0;
	int sum = 0; 
 	for (; i < 12; ++i)
 	 {
	 	cout << pointout[i];
 // 你用char *scmonth[] 就把 cout << smonth[i];用cout << scmonth[i];代替
 //  cout << scmonth[i];
		 cin >> sale[i];
		 cout << endl;
	 	sum += sale[i];
     }
     
     cout << "\nTotal: " << sum << endl;
   
     return 0;
}
    
	 	

// 编程练习_05.06_设计car结构.cpp 简单的动态链表 
/*********************************  第6题 *************************************
6.设计一个名为car的结构，用它存储下述有关汽车的信息：生产商（存储在字符数组成
string 对象中的字符串）、生产年份（整数）。编写一个程序，向用户询问有多少辆汽车。
随后，程序使用new来创建一个由相应数量的car 结构数组。接下来，程序提示用户输入每
辆辐的生产商（可能由多个单词组成）和年份信息。请注意，这需要特别小心，因为它将交
替读取数值和字符串（参见第四章）。最后，程序将显示每个结构的内容。该程序的运行情
况如下：
How many cars do you wish to catalog? 2
Car #1:
Please enter the make: Hudson Hornet
Peease enter the year made: 1952
Car #2:
Please enter the make: Kaiser
Please enter the year made: 1951
Here is your collection:
1952 Hudson Hornet
1951 Kaiser
*******************************************************************************/

#include <iostream>
using namespace std;
int n;

class car 
{
 	  public:
	  		 string name;
	  		 int year;
	  		 car* next;
};

int main()
{
 	car* creat();
	car* display(car*);
	display(creat());
    while(1);
	return 0;
}	

car* creat()
{
 	 car *head;
	 car *pa;
	 car *pb;
	 cout << "How many cars do you wish to catalog?";
	 cin >> n;
	 while (!cin )
	 	{
	 	cout << "Enter Error, please enter again:";
	 	cin.clear();
	 	while(cin.get() != '\n')
			continue;
		cin >> n;
		}
     
     int i;
     head = new car;
     pa = head;
     for (i = 1; i <= n; ++i)
     	 {
		  	  pb = pa;
		  	  cout << "Car #" << i << ":" << endl;
		  	  cout << "Please enter the make: ";
			  cin >> pa -> name; 
			  cout << "Please enter the year made: ";
			  cin >> pa -> year;
			  pa = new car; 
			  pb -> next = pa;
          }
     pa = 0;
     pb -> next = 0;
  	 return head;
}

car* display(car *head)
{
 	 cout << "Here is your collection: " << endl;
 	 car *pa;
 	 for (pa = head; pa != 0; pa = pa -> next)
	  cout <<  pa -> year << "  " << pa -> name << endl;
     return 0;
}

// 编程练习_05.7_计算读取单词.cpp
/*******************　第07题　************************************************
7.编写一个程序，它使用一个 char 数组和循环来每次读取一个单词，
直到用户输入 done 为止。随后，该程序指出用户输入了多少个单词（不包括done在内).
下面是该程序的运行情况：
Enter words (to stop, type the word done):
anteaster birthday category dumpster
anvyfiagle geometry done for sure
You entered a total of 7 words.
您应在程序中包含头文件 cstring，并使用函数 strcmp() 来进行测试.
******************************************************************************/ 

#include <iostream>
#include <string>
using namespace std;

int main()
{
 	cout << "Enter words (to stop, type the word done): " << endl;
 	const int ch_ar_si = 100;
 	char word[ch_ar_si];
 	int n = 0;
 	while(cin >> word && strcmp("done", word))
 	++n;

    cout << "You enterd a total of " << n << "words.";
    
    while(2);
    return 0;
}

// 编程练习_06.06_动态链表应用.cpp
/***************************　　第六题  　*********************************** 
6.编写一个程序，记录捐助给”维护合法权利团体“的钱。
该程序要求用户输入捐献者数目，然后要求用户输入每一个捐献者的姓名和款项。
这些信息被储存在一个动态分配的结构数组中。每个结构有两个成员：
用来存储姓名的字符数组（或string）和用来存储款项的double成员。
读取所有的数据后，程序将显示所有捐款超过10000的捐款者的姓名及其捐款数额。
该列表前应包含一个标题，指出下面的捐款者是重要的捐款人（Grand Patrons).
然后，程序将列出其他的捐款者，该列表要以Patrons开头。
如果某种类别没有捐款者，则程序将打印单词"none"。
该程序只显示这两种类别，而不进行排序。
*****************************************************************************/ 
 
#include <iostream>
#include <string>
using namespace std;

struct jq
{
 	string name;
 	double money;
 	jq *next;
};

int main()
{
// 	enter donors numbers
 	int n;
 	int t;
	cout << "Please enter the number of donors：";
	do
	{
    t = 0;
	cin >> n;
    if (!cin)
       {
	   cout << "Enter error, Please enter again: ";
       cin.clear();
       while(cin.get() != '\n');
       t++;
	   }
 	}while(t);

// prototype
    jq* creat(int n);
    jq* print(jq *head, int n);
    
// function call
	jq* temp = creat(n);
    print(temp, n);
    
// The end
    while(2);
    return 0;
}

jq* creat(int n)
{
// define three pointer to make the list
 	jq* head = new jq;			// 链表结点 
 	jq* pa = head;
 	jq* pb = pa;
    
// enter data
    int i;
    for (i = 1; i <= n; ++i)
    {
	   	pb -> next = pa;
		pb = pa;
	    
		cout << "Please enter number " << i << " donors money: ";
	    int t;
		do
	    {
	  	 t = 0;
		 cin >> pa -> money;
         if (!cin)
          {
	      cout << "Enter error, Please enter again: ";
          cin.clear();
          while(cin.get() != '\n');
    	   }
		}while(t);
		
		cout << "Please enter number " << i << " name: ";
		while(cin.get() != '\n');			// 消回车 
		getline(cin, pa -> name);			// 输字符串 
		if (pa -> name == "")				// string 不读取 \n 
		    pa -> name = "none";			// 若不输入名则自动为none 
	    cout << endl;
	    
		pa = new jq;
     }
     pa = 0;
     pb -> next = 0;
     
    return head;
}

jq* print(jq* head, int n)
{
   jq* pa;

// screen 1
   cout << "Grand Patrons List: " << endl;
   for (pa = head; pa != 0; pa = pa -> next)
   	   if(pa -> money >= 10000)
   			 cout << pa -> name << "\t\t" << pa -> money << endl;

// screen 2
   cout << endl << "Other Patrons List: " << endl;
   for (pa = head; pa != 0; pa = pa -> next)
   	   if(pa -> money < 10000)
   			 cout << pa -> name << "\t\t" << pa -> money << endl;	
				 
cout << "Finish!";
return 0;
}

// 编程练习_07.02_处理数组的函数.cpp
/**********************************************************************
2.编写一个程序，要求用户输入最多10个高尔夫成绩，并将其存储在一个数组中。
　程序允许用户提早结束输入，并在一行上显示所有成绩，然后报告平均成绩。
　请使用３个数组处理函数来分别进行输入、显示和计算平均成绩。
***********************************************************************/

#include <iostream>

 int const GolfArSize = 10;
 int TrueSize = 0;

int main()
{
	using namespace std;
	cout << "Now, begin to remember the score." << endl;
	void SaveGolf(int* Golf);
	void ShowGolf (const int * Golf);
	void AverageGolf  (const int* Golf);
	int Golf[GolfArSize];
	SaveGolf(Golf);
	ShowGolf(Golf);
	AverageGolf(Golf);
	return 0;
}

void SaveGolf(int* Golf)
{
	using namespace std;
	int i = 0;
	cout << "If you want to end the programme,Please enter a letter." << endl;
	for (i; i < GolfArSize; i++)
	{
		cout << "Please enter " << i+1 << " golf score: ";
		if (cin >> Golf[i])
		{
		}
		else
		{
			TrueSize = i;
			break;
		}
		TrueSize = i;
	}
	cout << "Enter End" << endl;
}

void ShowGolf (const int * Golf)
{
	using namespace std;
	int i= 0;
	cout << endl;
	cout << "Now show the score in next line: " << endl;
	for (i; i < TrueSize; i++)
	{
		cout << Golf[i] << '\t';
	}
	cout << endl;
}

void AverageGolf (const int* Golf)
{
	using namespace std;
	int i = 0;
	int sum = 0;
	for (i; i < TrueSize; ++i)
	{
		sum += Golf[i];
	}
	cout << "Average Score is : " << sum / TrueSize << endl;
}


// 本次编程时遇到了可提早结束输入的问题，后来想到多定义一个全局变量解决问题。
// 另外，这里还学习注意用const访问数组。用三种函数对数组实现功能。

// 编程练习_07.08_处理数组和结构的函数.cpp

/******************************************************************
8.这个练习让您编写处理数组和结构的函数。下面是程序的框架，
请提供其中描述的函数，以完成该程序。

#include <iostream>
using namespace std;

const int SLEN = 30;
struct student {
	char fullname[SLEN];
	char hobby[SLEN];
	int ooplevel;
	};
// getinfo() has two arguments: a pointer to the first element of 
// an array of student structures and an int representing the 
// number of elements of the array. The function solicits and
// stores data about students. It terminates input upon filling
// the array or upon encoutering a blank line for the student
// name. The function returns the actual namber of array elements
// filled.
int getinfo(student pa[], int n);

// display1()takes a student structure as an argument
// and displays its contents
void display1(student st);

// display2()takes the address of student structure as an 
// argument and displays the structure's contents
void display2(const student *ps);

// display3()takes the address of the first element of an array
// of student structures and the number of array elements as

// arrguments and displays the contents of the structures 
void display3(const student pa[], int n);

int main()
{
	cout << "Enter class size:";
	int class_size;
	cin >> class_size;
	while(cin.get() != '\n')
		continue;

	student * ptr_stu = new student [class_size);
	int entered = getinfo(ptr_stu, class_size);
	for (int i = 0; i < entered; i++)
	{
		display1(ptr_stu[i]);
		display2(&ptr_stu[i]);
	}
		display3(ptr_stu, entered);
		delete[] ptr_stu;
		cout << "Done\n";
		return 0;
}

*****************************************************************/

#include <iostream>
using namespace std;

const int SLEN = 30;
struct student 
	{
	char fullname[SLEN];
	char hobby[SLEN];
	int ooplevel;
	};
// getinfo() has two arguments: a pointer to the first element of 
// an array of student structures and an int representing the 
// number of elements of the array. The function solicits and
// stores data about students. It terminates input upon filling
// the array or upon encoutering a blank line for the student
// name. The function returns the actual namber of array elements
// filled.
int getinfo(student pa[], int n);

// display1()takes a student structure as an argument
// and displays its contents
void display1(student st);

// display2()takes the address of student structure as an 
// argument and displays the structure's contents
void display2(const student *ps);

// display3()takes the address of the first element of an array
// of student structures and the number of array elements as

// arrguments and displays the contents of the structures 
void display3(const student pa[], int n);

int main()
{
	cout << "Enter class size:";
	int class_size;
	cin >> class_size;
	while(cin.get() != '\n')
		continue;

	student * ptr_stu = new student [class_size];
	int entered = getinfo(ptr_stu, class_size);
	for (int i = 0; i < entered; i++)
	{
		display1(ptr_stu[i]);
		display2(&ptr_stu[i]);
	}
		display3(ptr_stu, entered);
		delete[] ptr_stu;
		cout << "Done\n";
		cin.get();
		cin.get();
		cin.get();
		return 0;
}


// getinfo() has two arguments: a pointer to the first element of 
// an array of student structures and an int representing the 
// number of elements of the array. The function solicits and
// stores data about students. It terminates input upon filling
// the array or upon encoutering a blank line for the student
// name. The function returns the actual namber of array elements
// filled.
int getinfo(student pa[], int n)
{
    int i;
	for (i = 0; i < n ; ++i)
	{
		cout << "Please enter this student's fullname: ";
        
		int j;
		cin.get(pa[i].fullname[0]);
 		if (pa[i].fullname[0] == '\n')
 		{
 			break;
         }

		for (j = 1; j < SLEN && (pa[i].fullname[j-1] != '\n'); ++j)
		{
			 cin.get(pa[i].fullname[j]);
			 pa[i].fullname[j+1] = '\0';
		}
		
		cout << "Please enter this student's hobby: ";
		cin >> pa[i].hobby;
		cout << "Please enter this student's ooplevel: ";
		cin >> pa[i].ooplevel;
		cin.get();

	}
	return i;
}

// display1()takes a student structure as an argument
// and displays its contents
void display1(student st)
{
	cout << st.fullname << '\t' << st.hobby << '\t' 
		<< st.ooplevel << endl;
}

// display2()takes the address of student structure as an 
// argument and displays the structure's contents
void display2(const student *ps)
{
	cout << ps -> fullname << '\t' 
		<< ps -> hobby << '\t' 
		<< ps -> ooplevel << endl;
}


// display3()takes the address of the first element of an array
// of student structures and the number of array elements as

// arrguments and displays the contents of the structures 
void display3(const student pa[], int n)
{
	for (int i = 0; i < n; ++i)
	{
		if (pa[i].fullname == "")
		{
			break;
		}
		cout << pa[i].fullname << '\t' ;
		cout << pa[i].hobby << '\t' ;
		cout << pa[i].ooplevel << endl;
	}
}

 

附：2011.10.08

刚刚，非常感谢一位网友，指出了这里面一道题目的错误，然后，还帮发给我了正确的答案，后来，我仔细看了一下，加上了注释，知道了错在哪里了，嘿嘿，非常感谢。

下面是源代码：

 

// 编程练习_05.03_计算投资价值
/*********************  第3题  *****************************************
  3.Daphne 以10%的单利投资了100美元。也就是说，每一年的利润都是投资额的10%，
  即每年10元美元：
  利息＝0.10*原始存款
  而Cleo以5%的复利投资了100美元。
  也就是说，利息是当前存款（包括获得的利息）的5%,:
  利息＝0.05*当前存款
  Cleo在第一年投资100美元的盈利是5%---得到了105美元。
  下一年的盈利是105美元的5%即5.25美元，依此类推。
  请编写一个程序，计算多少年后，Cleo的投资价值才能超过Daphne的投资价值，
  并显示此时两个人的投资价值。
 **************************************************************************/

// g++ -O3 -o benifit.exe -lstdc++ benifit.cpp

#include <iostream>
using namespace std;
int main()
{
  const double depositd = 100;   // d的本金 
  double depositc = 100;     // c的本金 
  const double interestrd = 0.10; 
  const double interestrc = 0.05;
  double interestd;      // d的利息 
  double interestc;      // c的利息 
  double sumd = 0;
  double &sumc = depositc;
  int year;

  /* 
 
  //先计算出第一年的 
  sumd += depositd;      // d的第一年本金加利息
  interestd = interestrd * depositd; // d的固定利息

  interestc = interestrc * depositc; // c的固定利息
  sumc += interestc;     // c的第一年的本金加利息

  for (year = 2; sumc > sumd; ++year) // 因为计算了第一年，所以从第一年开始算，当c的总额大于d的时候，结束循环
  {
    sumd += interestd;
 // **********************************************************************************************
    interestc = interestrc * depositc;  // 求出c的利息，利率* c的本金 // <- 就是错在这里了
 // **********************************************************************************************
    sumc += interestc;
  }
  */
  // 
  sumd = depositd;       // d的第一年本金加利息
  interestd = interestrd * depositd;  // d的固定利息

  year = 0;
  while(true) {        // 从第一年开始计算，不断循环，符合条件则用break跳出循环
    year ++;

    sumd += interestd;      // d可以直接进行利息累加
    sumc += sumc * interestrc;    // c的计算方法是上一年的乘以利息再加上本金

    cout << year << " year later, Daphne got " << sumd <<
         " and Cleo got " << sumc << endl;
    if(sumc > sumd) {
      break;
    }
  }

  cout << "Invest of Cleo: " << sumc << endl;
  cout << "Invest of Daphne: " << sumd << endl;
  cout << "After year: " << year << endl;

 


  system("pause");
  return 0;
}