本来前面五题都做完了，写博客时没保存好草稿= =，写了个整合版的程序，实现前五题的关键部分。

1.定义一个叫jojo的结构，存储姓名、替身和力量值，使用动态结构数组初始化二乔、承太郎和乔鲁诺乔巴纳等人的信息。循环地用菜单化的选项提示用户输入，选项1：显示所有人的替身；选项2：按一定比率强化白金之星的力量值，并输出当前所有人力量的平均值；选项3：要求用户输入一系列字符，然后返回相同的字符，其中转换字母大小写，遇到“@”就停止；选项4：退出。如果输入1-4以外的数字，提示用户重新输入，如果输入的不是数字，则提示失败退出程序。

#include
#include
using namespace std;

const double ratio = 0.1;
const int strsize = 100;

struct jojos
{
char name[20];
char stand[20];
double power;
};

int main()
{
jojos *joptr = new jojos[4];

joptr\[0\] =  
{  
    "Joseph",  
    "Hermit Purple",  
    6  
};

joptr\[1\] = { "Jotaro","Star Platinum",9 };

joptr\[2\] =  
{  
    "Giorno Giovanna",  
    "Gold Experience",  
    8  
};

joptr\[3\] = { "me","Repeater",1 };

int choice;  
int flag = 1;  
char store\[strsize\];

char prmt\[\] =  
{  
    "Make your choice:\\n1)show the stand \\t 2)power up \\n3)my stand \\t 4)quit\\n"  
};

while (flag)  
{  
    cout << prmt;  
    if (cin >> choice)  
    {  
        switch (choice)  
        {  
        case 1:  
        {  
            for (int i = 0; i < 4; i++)  
            {  
                cout << joptr\[i\].name << ":" << joptr\[i\].stand << endl;  
            }  
            cout << "\\n";  
            break;  
        }  
        case 2:  
        {  
            cout << "Power up!\\n";  
            joptr\[1\].power = joptr\[1\].power\*(1 + ratio);  
            double sum = 0;  
            for (int i = 0; i < 4; i++)  
            {  
                sum += joptr\[i\].power;  
            }  
            cout << "Jotaro's star platinum:" << joptr\[1\].power << endl;  
            cout << "Power average:" << sum / 4 << endl;  
            cout << "\\n";  
            break;  
        }  
        case 3:  
        {  
            cin.get();  
            cout << "Enter some characters: ";  
            cin.get(store, strsize);  
            for (int i = 0; store\[i\] != '\\0'; i++)  
            {  
                if (isdigit(store\[i\]))  
                    continue;  
                else if (store\[i\] != '@')  
                {  
                    if (isupper(store\[i\]))  
                        store\[i\] = tolower(store\[i\]);  
                    else if (islower(store\[i\]))  
                        store\[i\] = toupper(store\[i\]);  
                    cout << store\[i\];  
                }  
                else if (store\[i\] == '@')  
                {  
                    break;  
                }  
                else {};  
            }  
            cout << "\\n\\n";  
            break;  
        }  
        case 4:  
        {  
            flag = 0;  
            break;  
        }  
        default:  
        {  
            cout << "Please select from 1 to 4.\\n\\n";  
        }  
        }  
    }  
    else  
    {  
        cout << "Bad Input!\\n";  
        flag = 0;  
    }  
}

delete \[\]joptr;  
cout << "Bye.\\n";  
system("pause");  

}

2.编写程序记录一系列捐款者和捐款数目，将所有信息存储在动态结构数组中。首先要求用户输入有几个捐款者，然后循环读取捐款者姓名和捐款数目。读取完数据后，输出所有捐款额超过10000的人的姓名和捐款额，并标记为重要捐款人“Grand Patrons”，然后再输出其它捐款者的姓名，标记为“Patrons”。如果对应标记没有捐款人，则输出“none”。

#include
using namespace std;

struct donator
{
char name[20];
int donation;
};

int main()
{
int counts, counts_temp = 0;

cout << "How many donators: ";  
cin >> counts;

donator \*ptr = new donator\[counts\];

for (int i = 0; i < counts; i++)  
{  
    cout << "#donator " << i + 1 << "#\\n";  
    cout << "name:";  
    cin.get();  
    cin.get(ptr\[i\].name, 20);  
    cin.get();  
    cout << "donation:";  
    cin >> ptr\[i\].donation;  
}

cout << "\*Grand Patrons:";  
for (int i = 0; i < counts; i++)  
{  
    if (ptr\[i\].donation >= 10000)  
    {  
        cout << ptr\[i\].name << " $" << ptr\[i\].donation << "  ";  
        counts\_temp++;  
    }  
    else {};  
}

if (counts\_temp == 0)  
    cout << "none.\\n";  
else  
    counts\_temp = 0;

cout << "\\nPatrons:";  
for (int i = 0; i < counts; i++)  
{  
    if (ptr\[i\].donation < 10000)  
    {  
        cout << ptr\[i\].name<<";";  
        counts\_temp++;  
    }  
    else {};  
}

if (counts\_temp == 0)  
    cout << "none.\\n";  
else {}

delete\[\]ptr;  
system("pause");  

}

3.编写程序，打开一个文件，逐个字符读取该文件，直到文件末尾，然后指出其中包含的字符数。

#include
#include
using namespace std;

const int MAXSIZE = 100;

int main()
{
char store[MAXSIZE] ;
int counts_all = 0, counts_space = 0;

ifstream infile;  
infile.open("test.txt");    //打开文件

infile.get(store, MAXSIZE);        //infile此时可以当cin用

for (int i = 0; store\[i\]!='\\0'; i++)  
{  
    counts\_all++;  
    if (store\[i\] == ' ')  
        counts\_space++;  
    else {}  
}

cout << "Total characters(include spaces):" << counts\_all << endl;  
cout << "Total characters(without spaces):" << counts\_all - counts\_space << endl;

infile.close();

system("pause");  

}

4.修改程序2，这次使用文件存储所有信息，从文件里读取。

文件的格式：

4
Sam Stone
2000
Freida Flass
100500
…

第一行为捐款人数量，从第二行开始，一行写姓名，一行写捐款额。

程序如下。

#include
#include
using namespace std;

struct donator
{
char name[20];
int donation;
};

int main()
{
int counts, counts_temp = 0; int test;

ifstream infile;  
infile.open("test.txt");

cout << "How many donators: ";  
infile >> counts;  
cout << counts << endl;        //每次读入以后输出一下，查看是否正确

donator \*ptr = new donator\[counts\];

//infile会逐行读入，遇到换行符结束，且换行符不会留在缓冲区  
for (int i = 0; i < counts; i++)  
{  
    cout << "#donator " << i + 1 << "#\\n";  
    cout << "name:";  
    infile >> ptr\[i\].name;  
    cout << ptr\[i\].name << endl;  
    cout << "donation:";  
    infile >> ptr\[i\].donation;  
    cout << ptr\[i\].donation << endl;  
}

cout << "\*Grand Patrons:";  
for (int i = 0; i < counts; i++)  
{  
    if (ptr\[i\].donation >= 10000)  
    {  
        cout << ptr\[i\].name << " $" << ptr\[i\].donation << "  ";  
        counts\_temp++;  
    }  
    else {};  
}

if (counts\_temp == 0)  
    cout << "none.\\n";  
else  
    counts\_temp = 0;

cout << "\\nPatrons:";  
for (int i = 0; i < counts; i++)  
{  
    if (ptr\[i\].donation < 10000)  
    {  
        cout << ptr\[i\].name << ";";  
        counts\_temp++;  
    }  
    else {};  
}

if (counts\_temp == 0)  
    cout << "none.\\n";  
else {}

delete\[\]ptr;  
infile.close();  
system("pause");  

}

*经测试，infile对象是逐行读入的，每次遇到换行符（回车）就停止一次输入，并且不会在缓冲区留下换行符，所以不用像原先代码那样用cin.get()来清除。