/*******************************************************************************************************************
Author : Cui mingyang
Blog : cx_12586
Time : 2017/10/28
From : C++ Primer Plus第五版第七章編程練習 第1題
Problem : 編寫一個程序，不斷要求用戶輸入兩個數，直到其中一個爲0。對於每兩個數，程序將使用一個函數來計算它們的
調和平均數，並將結果返回給main()，而後者將報告結果。調和平均數指的是倒數平均值的倒數，計算公式如下：
調和平均數 = 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第五版第七章編程練習 第2題
Problem : 編寫一個程序，要求用戶輸入最多10個高爾夫成績，並將其存儲在一個數組中。程序允許用戶提早結束輸入，並在
一行上顯示所有成績，然後報告平均成績。請使用3個數組處理函數來分別進行輸入、顯示和計算平均成績。請使用3個數組
處理函數來分別
*******************************************************************************************************************/
#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第五版第七章編程練習 第3題
Problem : 下面是一個結構聲明：
struct box
{
char maker[40];
float height;
float width;
float length;
float volume;
};
a.編寫一個函數，按值傳遞box結構，並顯示每個成員的值
b.編寫一個函數，傳遞box結構的地址，並將volume成員設置爲其他三維長度的乘積。
c.編寫一個使用這兩個函數的簡單程序。
*******************************************************************************************************************/
#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第五版第七章編程練習 第4題
Problem : 許多州的彩票發行機構都使用如程序清單7.4所示的簡單彩票玩法的變體。在這些玩法中，玩家從一組被稱爲域號碼
(field number)的號碼中選擇幾個。例如，可以從域號碼1~47中選擇5個號碼：還可以從第二個區間（如1~27）選擇一個號碼
（稱爲特選號碼）。要贏得頭獎，必須正確猜中所有的號碼。中頭獎的機率是選中所有域號碼的機率與選中特選號碼機率的乘積。
例如，在這個例子中，中頭獎的機率是從47個號碼中正確選取5個號碼的機率與從27個號碼中選擇1個號碼的機率的成績。請修改
程序清單7.4，以計算中得這種彩票頭獎的機率。
*******************************************************************************************************************/
#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第五版第七章編程練習 第5題
Problem : 定義一個遞歸函數，接受一個整型參數，並返回該參數的階乘。前面講過，3的階乘寫作3！，等於3 * 2！，以此類推：
而0！被定義爲1.通用的計算公式是，如果n大於零 ， 則n! = n * (n - 1)!。在程序中對該函數進行測試，程序使用循環讓用戶
輸入不同的值，程序將報告這些值的階乘。
*******************************************************************************************************************/
#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第五版第七章編程練習 第6題
Problem : 編寫一個程序，它使用下列函數：
Fill_array()將一個double數組的名稱和長度作爲參數。它提示用戶輸入double值，並將這些值存儲到數組中。當數組被填滿或
用戶輸入了非數字時，輸入將停止，並返回實際輸入了多少個數字。
Show_array()將一個double數組的名稱和長度作爲參數，並顯示該數組的內容。
Reverse-array()將一個double數組的名稱和長度作爲參數，並將存儲在數組中的值的順序反轉。
程序將使用這些函數來填充數組，然後顯示數組；反轉數組，然後顯示數組；反轉數組中除第一個和最後一個元素之外的所有元素，
然後顯示數組
*******************************************************************************************************************/
#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第五版第七章編程練習 第7題
Problem : 修改程序清單7.7中的3個數組處理函數，使之使用兩個指針參數來表示區間。file_array()函數不返回實際讀取了多少個
數字，而是返回一個指針，該指針指向最後被填充的位置：其他的函數可以將該指針作爲第二個參數，以標識數據結尾。
*******************************************************************************************************************/
#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第五版第七章編程練習 第8題
Problem : 這個練習讓您編寫處理數組和結構的函數。下面是程序的框架，請提供其中描述的函數，以完成該程序。
#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第五版第七章編程練習 第9題
Problem: 設計一個名爲calculate()的函數，它接受兩個double值和一個指向函數的指針,而被指向的函數接受兩個double參數，
並返回一個double值。calculate()函數的類型也爲double，並返回被指向的函數使用calculate()的兩個double參數計算得到的值。
例如，架設add()函數的定義如下：
double add(double x, double y)
{
return x + y;
}
則下述代碼中的函數調用將導致calculate()把2.5和10.4傳遞給add()函數，並返回add()的返回值(12.9):
double q = calculate(2.5, 10.4, add);
請編寫一個程序，它調用上述兩個函數和至少另一個與add()類似的函數。該程序使用循環來讓用戶成對地輸入數字。對於每對數字，
程序都使用calculate()來調用add()和至少一個其他的函數。如果讀者愛冒險，可以嘗試創建一個指針數組，其中的指針指向add()
樣式的函數，並編寫一個循環，使用這些指針連續讓calculate()調用這些函數。提示：下面是聲明這樣指針數組的方式，其中包含
三個指針：
double (*pf[3])(double, double);
可以採用數組初始化語法，並將函數名作爲地址來初始化這樣的數組。
*******************************************************************************************************************/
#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;
}
