前三个任务代码均编写并运行验证
Task4:
1 #include <stdio.h> 2 #define N 10 3 typedef struct { 4 char isbn[20]; // isbn号 5 char name[80]; // 书名 6 char author[80]; // 作者 7 double sales_price; // 售价 8 int sales_count; // 销售册数 9 } Book; 10 11 void output(Book x[], int n); 12 void sort(Book x[], int n); 13 double sales_amount(Book x[], int n); 14 15 int main() { 16 Book x[N] = { 17 {"978-7-5327-6082-4", "门将之死", "罗纳德.伦", 42, 51}, 18 {"978-7-308-17047-5", "自由与爱之地：入以色列记", "云也退", 49 , 30}, 19 {"978-7-5404-9344-8", "伦敦人", "克莱格泰勒", 68, 27}, 20 {"978-7-5447-5246-6", "软件体的生命周期", "特德姜", 35, 90}, 21 {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49}, 22 {"978-7-5133-5750-0", "主机战争", "布莱克.J.哈里斯", 128, 42}, 23 {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44}, 24 {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42}, 25 {"978-7-1155-0509-5", "无穷的开始:世界进步的本源", "戴维·多伊奇", 37.5,55}, 26 {"978-7-229-14156-1", "源泉", "安.兰德", 84, 59} 27 }; 28 printf("图书销量排名(按销售册数): \n"); 29 sort(x, N); 30 output(x, N); 31 printf("\n图书销售总额: %.2f\n", sales_amount(x, N)); 32 return 0; 33 } 34 void output(Book x[], int n) { 35 printf("ISBN号 书名 作者 售价 销售册数\n"); 36 printf("------------------------------------------------------------------------------------\n"); 37 38 for (int i = 0; i < n; i++) { 39 printf("%-20s %-30s %-15s %-8.2f %-8d\n", 40 x[i].isbn, 41 x[i].name, 42 x[i].author, 43 x[i].sales_price, 44 x[i].sales_count); 45 } 46 } 47 void sort(Book x[], int n) { 48 for (int i = 0; i < n - 1; i++) { 49 for (int j = 0; j < n - i - 1; j++) { 50 if (x[j].sales_count < x[j + 1].sales_count) { 51 Book temp = x[j]; 52 x[j] = x[j + 1]; 53 x[j + 1] = temp; 54 } 55 } 56 } 57 } 58 double sales_amount(Book x[], int n) { 59 double total = 0.0; 60 for (int i = 0; i < n; i++) { 61 total += x[i].sales_price * x[i].sales_count; 62 } 63 return total; 64 }
Task5:
1 #include <stdio.h> 2 typedef struct { 3 int year; 4 int month; 5 int day; 6 } Date; 7 // 函数声明 8 void input(Date *pd); // 输入日期给pd指向的Date变量 9 int day_of_year(Date d); // 返回日期d是这一年的第多少天 10 int compare_dates(Date d1, Date d2); // 比较两个日期: 11 // 如果d1在d2之前，返回-1; 12 // 如果d1在d2之后，返回1 13 // 如果d1和d2相同，返回0 14 void test1() { 15 Date d; 16 int i; 17 printf("输入日期:(以形如2025-12-19这样的形式输入)\n"); 18 for(i = 0; i < 3; ++i) { 19 input(&d); 20 printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day, 21 day_of_year(d)); 22 } 23 } 24 void test2() { 25 Date Alice_birth, Bob_birth; 26 int i; 27 int ans; 28 printf("输入Alice和Bob出生日期:(以形如2025-12-19这样的形式输入)\n"); 29 for(i = 0; i < 3; ++i) { 30 input(&Alice_birth); 31 input(&Bob_birth); 32 ans = compare_dates(Alice_birth, Bob_birth); 33 if(ans == 0) 34 printf("Alice和Bob一样大\n\n"); 35 else if(ans == -1) 36 printf("Alice比Bob大\n\n"); 37 else 38 printf("Alice比Bob小\n\n"); 39 } 40 } 41 int main() { 42 printf("测试1: 输入日期, 打印输出这是一年中第多少天\n"); 43 test1(); 44 printf("\n测试2: 两个人年龄大小关系\n"); 45 test2(); 46 } 47 int is_leap_year(int year) { 48 return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0); 49 } 50 void input(Date *pd){ 51 scanf("%d-%d-%d",&pd->year,&pd->month,&pd->day); 52 } 53 int day_of_year(Date d) { 54 int days_in_month[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; 55 int total_days = 0; 56 57 if (is_leap_year(d.year)) { 58 days_in_month[1] = 29; 59 } 60 61 for (int i = 0; i < d.month - 1; i++) { 62 total_days += days_in_month[i]; 63 } 64 total_days += d.day; 65 return total_days; 66 } 67 int compare_dates(Date d1, Date d2) { 68 if (d1.year < d2.year) return -1; 69 if (d1.year > d2.year) return 1; 70 71 if (d1.month < d2.month) return -1; 72 if (d1.month > d2.month) return 1; 73 74 if (d1.day < d2.day) return -1; 75 if (d1.day > d2.day) return 1; 76 77 return 0; 78 }
Task6:
1 #include <stdio.h> 2 #include <string.h> 3 enum Role {admin, student, teacher}; 4 typedef struct { 5 char username[20]; // 用户名 6 char password[20]; // 密码 7 enum Role type; // 账户类型 8 } Account; 9 // 函数声明 10 void output(Account x[], int n); // 输出账户数组x中n个账户信息，其中，密码用*替代显示 11 int main() { 12 Account x[] = {{"A1001", "123456", student}, 13 {"A1002", "123abcdef", student}, 14 {"A1009", "xyz12121", student}, 15 {"X1009", "9213071x", admin}, 16 {"C11553", "129dfg32k", teacher}, 17 {"X3005", "921kfmg917", student}}; 18 int n; 19 n = sizeof(x)/sizeof(Account); 20 output(x, n); 21 return 0; 22 } 23 void output(Account x[], int n) { 24 for (int i = 0; i < n; i++) { 25 int len = strlen(x[i].password); 26 27 printf("用户名: %s\n", x[i].username); 28 29 printf("密码: "); 30 for (int j = 0; j < len; j++) { 31 printf("*"); 32 } 33 printf("\n"); 34 35 printf("账户类型: "); 36 switch (x[i].type) { 37 case admin: 38 printf("admin\n"); 39 break; 40 case student: 41 printf("student\n"); 42 break; 43 case teacher: 44 printf("teacher\n"); 45 break; 46 } 47 48 printf("------------------------\n"); 49 } 50 }
Task7:
1 #include <stdio.h> 2 #include <string.h> 3 typedef struct { 4 char name[20]; // 姓名 5 char phone[12]; // 手机号 6 int vip; // 是否为紧急联系人，是取1；否则取0 7 } Contact; 8 // 函数声明 9 void set_vip_contact(Contact x[], int n, char name[]); // 设置紧急联系人 10 void output(Contact x[], int n); // 输出x中联系人信息 11 void display(Contact x[], int n); // 按联系人姓名字典序升序显示信息，紧急联系人最先显示 12 #define N 10 13 int main() { 14 Contact list[N] = {{"刘一", "15510846604", 0}, 15 {"陈二", "18038747351", 0}, 16 {"张三", "18853253914", 0}, 17 {"李四", "13230584477", 0}, 18 {"王五", "15547571923", 0}, 19 {"赵六", "18856659351", 0}, 20 {"周七", "17705843215", 0}, 21 {"孙八", "15552933732", 0}, 22 {"吴九", "18077702405", 0}, 23 {"郑十", "18820725036", 0}}; 24 int vip_cnt, i; 25 char name[20]; 26 printf("显示原始通讯录信息: \n"); 27 output(list, N); 28 printf("\n输入要设置的紧急联系人个数: "); 29 scanf("%d", &vip_cnt); 30 printf("输入%d个紧急联系人姓名:\n", vip_cnt); 31 for(i = 0; i < vip_cnt; ++i) { 32 scanf("%s", name); 33 set_vip_contact(list, N, name); 34 } 35 printf("\n显示通讯录列表:(按姓名字典序升序排列,紧急联系人最先显示)\n"); 36 display(list, N); 37 return 0; 38 } 39 void set_vip_contact(Contact x[], int n, char name[]) { 40 for (int i = 0; i < n; i++) { 41 if (strcmp(x[i].name, name) == 0) { 42 x[i].vip = 1; 43 break; 44 } 45 } 46 } 47 void display(Contact x[], int n) { 48 int i, j; 49 Contact temp; 50 51 for (i = 0; i < n - 1; i++) { 52 for (j = 0; j < n - 1 - i; j++) { 53 int need_swap = 0; 54 55 if (x[j].vip < x[j + 1].vip) { 56 need_swap = 1; 57 } 58 else if (x[j].vip == x[j + 1].vip && strcmp(x[j].name, x[j + 1].name) > 0) { 59 need_swap = 1; 60 } 61 62 if (need_swap) { 63 temp = x[j]; 64 x[j] = x[j + 1]; 65 x[j + 1] = temp; 66 } 67 } 68 } 69 70 for (i = 0; i < n; i++) { 71 printf("%-10s %-15s", x[i].name, x[i].phone); 72 if (x[i].vip) { 73 printf(" *"); 74 } 75 printf("\n"); 76 } 77 } 78 void output(Contact x[], int n) { 79 int i; 80 for (i = 0; i < n; ++i) { 81 printf("%s %s", x[i].name, x[i].phone); 82 if (x[i].vip) 83 printf(" *"); 84 printf("\n"); 85 } 86 }
