C++ Primer Plus第五版 第七章 编程练习答案

/*******************************************************************************************************************  Author : Cui mingyang Blog : cx_12586 Time : 2017/10/28 From : C++ Primer Plus第五版第七章编程练习 第1题  Problem : 编写一个程序，不断要求用户输入两个数，直到其中一个为0。对于每两个数，程序将使用一个函数来计算它们的 调和平均数，并将结果返回给main()，而后者将报告结果。调和平均数指的是倒数平均值的倒数，计算公式如下： 调和平均数 = 2.0 * x * y / (x + y) *******************************************************************************************************************/ #include<iostream>using namespace std;double harmonic_mean(double x,double y);int main(){cout << "Enter two numbers (0 represent end): ";double x,y;while(cin >> x >> y && x*y!=0){cout << "The harmonic mean of " << x << " and " << y << " is " << harmonic_mean(x,y) << endl;cout << "Enter two numbers (0 represent end): ";}cout << "Bye! " <<endl;system("pause");return 0;}double harmonic_mean(double x,double y){return 2.0*x*y/(x+y);}

/*******************************************************************************************************************  Author : Cui mingyang Blog : cx_12586 Time : 2017/10/28 From : C++ Primer Plus第五版第七章编程练习 第2题  Problem : 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。请使用3个数组 处理函数来分别  *******************************************************************************************************************/ #include<iostream>using namespace std;const int SIZE = 10;int input_scords(double scords[], int size);void show(const double scords[],int size);double mean(const double scords[],int size);int main(){cout << "Enter " << SIZE << " scords: ";double scords [10]={0};int num;num= input_scords(scords, SIZE);show(scords,num);cout << "The mean scords is " << mean(scords,num) <<endl;system("pause");return 0;}int input_scords(double scords[], int SIZE){double temp;int i=0;for(i=0;i<SIZE;i++){cin>>temp;if(!cin)         //bad input{cin.clear();while(cin.get()!='\n')continue;cout<<"Bad input:input process terminated.\n";break;}else if(temp<0)break;scords[i]=temp;}return i;}void show(const double scords[],int size){for (int i=0; i<size ;i++){cout << scords[i] << "  ";}cout << endl;}double mean(const double scords[],int size){double sum=0;for (int i=0; i<size ;i++)sum+=scords[i];double mean_scords=sum/size;return mean_scords;}

/*******************************************************************************************************************  Author : Cui mingyang Blog : cx_12586 Time : 2017/10/28 From : C++ Primer Plus第五版第七章编程练习 第3题  Problem : 下面是一个结构声明： struct box { char maker[40]; float height; float width; float length; float volume; }; a.编写一个函数，按值传递box结构，并显示每个成员的值 b.编写一个函数，传递box结构的地址，并将volume成员设置为其他三维长度的乘积。 c.编写一个使用这两个函数的简单程序。 *******************************************************************************************************************/ #include<iostream>using namespace std;struct box{char maker[40];float height;float width;float length;float volum;};void show_box (const box box1);void box_fun(box *p);int main(){box box1;cout << "Please enter the name of maker: ";cin.getline(box1.maker,40);cout << "Enter the height, width and length: ";cin >> box1.height >> box1.width >> box1.length ;box_fun(&box1);show_box(box1);system("pause");return 0;}void box_fun(box *p){p->volum = p->height*p->length*p->width;}void show_box (const box box1){cout << "The maker of the box is " << box1.maker << endl;cout << "The height, width and length of the box are " << box1.height << ", "<< box1.width << ", "<< box1.length << endl; cout << "The volum of the box is " << box1.volum <<endl;}

/*******************************************************************************************************************  Author : Cui mingyang Blog : cx_12586 Time : 2017/10/28 From : C++ Primer Plus第五版第七章编程练习 第4题  Problem : 许多州的彩票发行机构都使用如程序清单7.4所示的简单彩票玩法的变体。在这些玩法中，玩家从一组被称为域号码 (field number)的号码中选择几个。例如，可以从域号码1~47中选择5个号码：还可以从第二个区间（如1~27）选择一个号码 （称为特选号码）。要赢得头奖，必须正确猜中所有的号码。中头奖的几率是选中所有域号码的几率与选中特选号码几率的乘积。 例如，在这个例子中，中头奖的几率是从47个号码中正确选取5个号码的几率与从27个号码中选择1个号码的几率的成绩。请修改 程序清单7.4，以计算中得这种彩票头奖的几率。 *******************************************************************************************************************/ #include <iostream>using namespace std;long double probability(unsigned numbers1, unsigned picks1, unsigned numbers2, unsigned picks2);int main(){double total1, choices1, total2, choice2;cout << "Enter the total number of choices on the game card and\n""the number of picks allowed of field number and Selected number:\n";while ((cin >> choices1 >> total1>> choice2 >> total2) && choices1 <= total1 && choice2 <= total2){cout << "You have one chance in ";cout << probability(total1, choices1, total2, choice2);    cout << " of winning.\n";cout << "Next two numbers (q to quit): ";}cout << "bye\n";system("pause");return 0;}long double probability(unsigned numbers1, unsigned picks1, unsigned numbers2, unsigned picks2){long double result = 1.0; long double n;unsigned p;for (n = numbers1, p = picks1; p > 0; n--, p--)result = result * n / p ; for (n = numbers2, p = picks2; p > 0; n--, p--)result = result * n / p ; return result;}

/*******************************************************************************************************************  Author : Cui mingyang Blog : cx_12586 Time : 2017/10/28 From : C++ Primer Plus第五版第七章编程练习 第5题  Problem : 定义一个递归函数，接受一个整型参数，并返回该参数的阶乘。前面讲过，3的阶乘写作3！，等于3 * 2！，以此类推： 而0！被定义为1.通用的计算公式是，如果n大于零 ， 则n! = n * (n - 1)!。在程序中对该函数进行测试，程序使用循环让用户 输入不同的值，程序将报告这些值的阶乘。 *******************************************************************************************************************/ #include<iostream>using namespace std;long long times(int num);int main(){cout << "Enter an integer (negative represent the end): ";int num;while(cin >> num){if (num < 0)break;cout << "The factorial of " << num << " is " << times(num) << endl;cout << "Enter another integer (negative represent the end): ";}system("pause");return 0;}long long times(int num){long long ans;if (num == 1 || num == 0)return 1;ans = num*times(num-1);return ans;}

/*******************************************************************************************************************  Author : Cui mingyang Blog : cx_12586 Time : 2017/10/28 From : C++ Primer Plus第五版第七章编程练习 第6题  Problem : 编写一个程序，它使用下列函数： Fill_array()将一个double数组的名称和长度作为参数。它提示用户输入double值，并将这些值存储到数组中。当数组被填满或 用户输入了非数字时，输入将停止，并返回实际输入了多少个数字。 Show_array()将一个double数组的名称和长度作为参数，并显示该数组的内容。 Reverse-array()将一个double数组的名称和长度作为参数，并将存储在数组中的值的顺序反转。 程序将使用这些函数来填充数组，然后显示数组；反转数组，然后显示数组；反转数组中除第一个和最后一个元素之外的所有元素， 然后显示数组 *******************************************************************************************************************/ #include<iostream>using namespace std;const int SIZE =10;int Fill_array(double arr[],int SIZE);void Show_array(double arr[],int size);void Reverse_array(double arr[],int size);int main(){double arr[SIZE];int length;length=Fill_array(arr,SIZE);Show_array(arr,length);Reverse_array(arr,length);Show_array(arr,length);Reverse_array(arr+1,length-2);Show_array(arr,length);system("pause");return 0;}int Fill_array(double arr[],int SIZE){cout << "Enter double numbers(less than " << SIZE << ", q to quit):  \n";int i=0;double temp;while(cin >> temp){if (i<SIZE){arr[i]=temp;i++;}else break;}return i;}void Show_array(double arr[],int size){cout << "The array is ";for (int i=0;i<size;i++){cout << arr[i] <<"  ";}cout<<endl;}void Reverse_array(double arr[],int size){double temp;for (int i=0;i<size/2;i++){temp=arr[i];arr[i]=arr[size-1-i];arr[size-1-i]=temp;}}

/*******************************************************************************************************************  Author : Cui mingyang Blog : cx_12586 Time : 2017/10/28 From : C++ Primer Plus第五版第七章编程练习 第7题  Problem : 修改程序清单7.7中的3个数组处理函数，使之使用两个指针参数来表示区间。file_array()函数不返回实际读取了多少个 数字，而是返回一个指针，该指针指向最后被填充的位置：其他的函数可以将该指针作为第二个参数，以标识数据结尾。 *******************************************************************************************************************/ #include <iostream>  using namespace std;  const int SIZE=10;double *fill_array(double *a);  void show_array(double *a , double *b);  void revalue(double r , double *a , double *b);  int main()  {  double a[SIZE];  double *e = fill_array(a);  show_array(a , e);  revalue(0.5 , a , e);  show_array(a , e);  system("pause");return 0;  }  double *fill_array(double *a)  {  cout << "Enter double numbers(less than " << SIZE << ", q to quit):  \n";int i = 0;  while(cin >> a[i++])  if(i == 5)  break;  return &a[i];  }  void show_array(double *a , double *b)  {  while(a != b){  cout << *a << " ";  ++ a;  }  cout << endl;  }  void revalue(double r , double *a , double *b)  {  while(a != b){  (*a) *= r;  ++ a;  }  }  

/*******************************************************************************************************************  Author : Cui mingyang Blog : cx_12586 Time : 2017/10/28 From : C++ Primer Plus第五版第七章编程练习 第8题  Problem :  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 argumnets: a pointer to the first element of //an array of student structures and an int representing the //number of elemnets of the array. The function solicits and //stores data about students. It terminates input upon filling //the array or upon encountering a blank line for the student //nmae. The function returns the actual number of array elemnets //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 struture as an //argument and displays the stucture's contents void display2(const student * ps); //display3() takes the address of the first elemnet of an array //of student structures and the number of array elemnets as //arguments 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 argumnets: a pointer to the first element of  //an array of student structures and an int representing the  //number of elemnets of the array. The function solicits and  //stores data about students. It terminates input upon filling  //the array or upon encountering a blank line for the student  //nmae. The function returns the actual number of array elemnets  //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 struture as an  //argument and displays the stucture's contents  void display2(const student * ps);  //display3() takes the address of the first elemnet of an array  //of student structures and the number of array elemnets as  //arguments 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;  }  int getinfo(student pa[],int n)    {    int count=0;    for(int i=0;i<n;i++)    {    cout<<"Please enter the fullname:";    cin>>pa[i].fullname;    cout<<"\nPlease enter the hobby:";    cin>>pa[i].hobby;    cout<<"\nPlease enter the ooplevel:";    cin>>pa[i].ooplevel;    count++;    }    cout<<"\nEnter end!";    return count;    }    void display1(student st)   {    cout<<"\ndisplay1:FullName:"<<st.fullname<<"\nhobby:"<<st.hobby    <<"\nooplevel:"<<st.ooplevel<<endl;    }    void display2(const student *ps)  {    cout<<"\ndispaly2:FullName:"<<ps->fullname<<"\nhobby:"<<ps->hobby    <<"\nooplevel:"<<ps->ooplevel<<endl;    }    void display3(const student pa[],int n)    {    cout<<"\ndispaly3:"<<endl;    for(int i=0;i<n;i++)    cout<<i<<"::FullName:"<<pa[i].fullname<<"\nhobby:"<<pa[i].hobby    <<"\nooplevel:"<<pa[i].ooplevel<<endl;    }    

/*******************************************************************************************************************  Author : Cui mingyang Blog : cx_12586 Time : 2017/10/28 From : C++ Primer Plus第五版第七章编程练习 第9题Problem: 设计一个名为calculate()的函数，它接受两个double值和一个指向函数的指针,而被指向的函数接受两个double参数，并返回一个double值。calculate()函数的类型也为double，并返回被指向的函数使用calculate()的两个double参数计算得到的值。例如，架设add()函数的定义如下：double add(double x, double y){return x + y;}则下述代码中的函数调用将导致calculate()把2.5和10.4传递给add()函数，并返回add()的返回值(12.9):double q = calculate(2.5, 10.4, add);请编写一个程序，它调用上述两个函数和至少另一个与add()类似的函数。该程序使用循环来让用户成对地输入数字。对于每对数字，程序都使用calculate()来调用add()和至少一个其他的函数。如果读者爱冒险，可以尝试创建一个指针数组，其中的指针指向add()样式的函数，并编写一个循环，使用这些指针连续让calculate()调用这些函数。提示：下面是声明这样指针数组的方式，其中包含三个指针：double (*pf[3])(double, double);可以采用数组初始化语法，并将函数名作为地址来初始化这样的数组。*******************************************************************************************************************/#include <iostream>  using namespace std;double calculate(double a, double b, double (*pf)(double a, double b));  double add(double a, double b);  double max(double a, double b);  double min(double a, double b);  int main()  {  double a, b;  double (*pf[3])(double a, double b);  pf[0] = add;  pf[1] = max;  pf[2] = min;cout << "Enter two double numbers(q two quit): ";while(cin >> a >> b)  {cout << "The sum of " << a << " and " << b << " is " << (*pf[0])(a, b) <<endl;cout << "The max of " << a << " and " << b << " is " << (*pf[1])(a, b) <<endl;  cout << "The min of " << a << " and " << b << " is " << (*pf[2])(a, b) <<endl;  cout << "Enter another two double numbers(q two quit): ";}system("pause");return 0;  }  double calculate(double a, double b, double (*pf)(double a, double b))  {  return (*pf)(a, b);  }  double add(double a, double b)  {  return a + b;  }  double max(double a, double b)  {  return a > b ? a : b;  }  double min(double a, double b)  {  return a > b? b : a;  }