1.
有一组关于学生成绩的信息，编写函数max，该函数返回值为分数最高的学生的信息（包括学号和分数）。再编写主函数对其进行调用并输出最高分者的信息。假设结构类型定义为：
struct student
{
char *num;
int score;
};
提示：num定义为指针类型，注意向其复制内容之前要先为其申请空间。
yzy’s version:
1 # include 2 #define N 256 3 using namespace std; 4 struct student 5 { 6 char* num; 7 int score; 8 }; 9 struct student max(struct student s[N],int n)10 {11 struct student t = {0,s[0].score };12 t.num = (char*)malloc(sizeof(char));13 for (int i = 0; i < n; i++)14 {15 if (s[i].score > t.score)16 {17 t.score = s[i].score;18 t.num = s[i].num;19 }20 }21 return t;22 }23 int main()24 {25 struct student s[N] = {}, t = {};26 int n,i;27 t.num = (char*)malloc(sizeof(char));28 cout << "输入学生数：" << endl;29 /*char a[N];30 s->num = a;*/31 cin >> n;32 cout << "输入各学生序号及成绩：" << endl;33 for (i = 0; i < n; i++)34 {35 s[i].num = (char*)malloc(sizeof(char));36 cin >> s[i].num >> s[i].score;37 }38 t=max(s,n);39 cout << "最高分者序号为" << *t.num <<"，分值为：" << t.score << endl;40 system("pause");41 return 0;42 }
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good version:
1 #include 2 3 4 5 using namespace std; 6 7 8 9 struct Student { 10 11 char* num; 12 13 int score; 14 15 }; 16 17 18 19 Student max(const Student* students, int count) { 20 21 if (count <= 0) { 22 23 cerr << "Error: Empty student array" << endl; 24 25 exit(EXIT_FAILURE); 26 27 } 28 29 30 31 const Student* maxStudent = &students[0]; 32 33 34 35 for (int i = 1; i < count; ++i) { 36 37 if (students[i].score > maxStudent->score) { 38 39 maxStudent = &students[i]; 40 41 } 42 43 } 44 45 46 47 return *maxStudent; 48 49 } 50 51 52 53 int main() { 54 55 const int MAX_STUDENTS = 100; 56 57 58 59 int numStudents; 60 61 cout << "Enter the number of students: "; 62 63 cin >> numStudents; 64 65 66 67 if (numStudents <= 0 || numStudents > MAX_STUDENTS) { 68 69 cerr << "Invalid number of students" << endl; 70 71 return EXIT_FAILURE; 72 73 } 74 75 76 77 Student* students = new Student[numStudents]; 78 79 80 81 for (int i = 0; i < numStudents; ++i) { 82 83 students[i].num = new char[20]; // Assuming a maximum length of 19 for the student number 84 85 cout << "Enter student " << i + 1 << " information:" << endl; 86 87 cout << "Student Number: "; 88 89 cin >> students[i].num; 90 91 cout << "Score: "; 92 93 cin >> students[i].score; 94 95 } 96 97 98 99 Student highestScoreStudent = max(students, numStudents);100 101 102 103 cout << "Student with the highest score:" << endl;104 105 cout << "Student Number: " << highestScoreStudent.num << endl;106 107 cout << "Score: " << highestScoreStudent.score << endl;108 109 110 111 for (int i = 0; i < numStudents; ++i) {112 113 delete[] students[i].num;114 115 }116 117 118 119 delete[] students;120 121 122 123 system("pause");124 125 126 127 return 0;128 129 }
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2.编写程序，定义一个日期结构变量，计算某日期是本年度的第几天。提示：为简单起见，可定义一个存放12个月中每个月总天数的数组。
yzy’s version:
1 # include 2 #define N 256 3 using namespace std; 4 struct time 5 { 6 int month; 7 int date; 8 }; 9 int change(struct time t,bool run)10 {11 int i,num=0, a[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 }, b[12] = { 31,29,31,30,31,30,31,31,30,31,30,31 };12 if (run == 1)13 for (i = 0; i < t.month-1; i++)14 num += b[i];15 else16 for (i = 0; i < t.month-1; i++)17 num += a[i];18 num += t.date;19 return num;20 }21 int main()22 {23 struct time t;24 bool run;25 int num;26 cout << "今年是不是闰年？（是：1；否：0）" << endl;27 cin >> run;28 cout << "输入一日期（月/日）" << endl;29 cin >> t.month >> t.date;30 num = change(t, run);31 cout << "这是今年的第" << num << "天" << endl;32 system("pause");33 return 0;34 }
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good version:
1 #include 2 3 4 5 using namespace std; 6 7 8 9 struct Date {10 11 int year;12 13 int month;14 15 int day;16 17 };18 19 20 21 int dayOfYear(const Date& date) {22 23 const int daysInMonth[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};24 25 26 27 int totalDays = 0;28 29 30 31 for (int i = 1; i < date.month; ++i) {32 33 totalDays += daysInMonth[i];34 35 }36 37 38 39 totalDays += date.day;40 41 42 43 // Adjust for leap year44 45 if ((date.year % 4 == 0 && date.year % 100 != 0) || (date.year % 400 == 0 && date.month > 2)) {46 47 totalDays++;48 49 }50 51 52 53 return totalDays;54 55 }56 57 58 59 int main() {60 61 Date inputDate;62 63 64 65 cout << "Enter date (year month day): ";66 67 cin >> inputDate.year >> inputDate.month >> inputDate.day;68 69 70 71 int result = dayOfYear(inputDate);72 73 74 75 cout << "The day of the year for the given date is: " << result << endl;76 77 78 79 system("pause");80 81 82 83 return 0;84 85 }
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3.使用结构数组输入10本书的名称和单价，调用函数按照书名的字母顺序序进行排序，在主函数输出排序结果。
yzy’s version:
1 #include 2 #define N 5 3 using namespace std; 4 struct book 5 { 6 char name[100]; 7 int price; 8 }; 9 struct book *sort(struct book b[])10 {11 int i, j;12 struct book a;13 for(i=0;i<N-1;i++)14 for (j = 0; j < N - 1 - i; j++)15 {16 //for(int z=0; b[j].name[z] != '\0' && b[j + 1].name[z] != '\0';z++)17 if (strcmp(b[j].name, b[j+1].name) > 0)18 {19 a = b[j];20 b[j] = b[j + 1];21 b[j + 1] = a;22 }23 }24 struct book* p = b;25 return p;26 }27 int main()28 {29 struct book b[N],*p;30 int i;31 for (i = 0; i < N; i++)32 cin >> b[i].name >> b[i].price;33 p = sort(b);34 cout << "排序后结果：" << endl;35 for (i = 0; i < N; i++)36 cout << p[i].name << ' ' << p[i].price << endl;37 system("pause");38 return 0;39 }
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good version:
1 #include 2 3 #include 4 5 #include 6 7 8 9 using namespace std;10 11 12 13 struct Book {14 15 char title[100];16 17 double price;18 19 };20 21 22 23 bool compareBooks(const Book& book1, const Book& book2) {24 25 return strcmp(book1.title, book2.title) < 0;26 27 }28 29 30 31 int main() {32 33 const int NUM_BOOKS = 10;34 35 Book books[NUM_BOOKS];36 37 38 39 for (int i = 0; i < NUM_BOOKS; ++i) {40 41 cout << "Enter details for book " << i + 1 << ":" << endl;42 43 cout << "Title: ";44 45 cin.getline(books[i].title, sizeof(books[i].title));46 47 cout << "Price: ";48 49 cin >> books[i].price;50 51 cin.ignore(); // Ignore the newline character left in the input buffer52 53 }54 55 56 57 sort(books, books + NUM_BOOKS, compareBooks);58 59 60 61 cout << "Books sorted by title:" << endl;62 63 for (int i = 0; i < NUM_BOOKS; ++i) {64 65 cout << "Title: " << books[i].title << ", Price: " << books[i].price << endl;66 67 }68 69 70 71 system("pause");72 73 74 75 return 0;76 77 }
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