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<iostream> #include<cctype> 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<iostream> 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<iostream> #include<fstream> 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<iostream> #include<fstream> 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。