实验6
实验任务四
源代码：
1 #include <stdio.h> 2 #define N 10 3 4 typedef struct { 5 char isbn[20]; 6 char name[80]; 7 char author[80]; 8 double sales_price; 9 int sales_count; 10 } Book; 11 12 void output(Book x[], int n); 13 void sort(Book x[], int n); 14 double sales_amount(Book x[], int n); 15 16 int main() { 17 Book x[N] = {{"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 32 printf("\n图书销售总额: %.2f\n", sales_amount(x, N)); 33 34 return 0; 35 } 36 37 void output(Book x[], int n){ 38 printf("ISBN编号 书名 作者 售价 销售册数\n"); 39 for (int i = 0; i < n; i++) { 40 printf("%-20s %-30s %-20s %-4.0f %-10d\n", 41 x[i].isbn, x[i].name, x[i].author, 42 x[i].sales_price, x[i].sales_count); 43 } 44 } 45 46 47 void sort(Book x[], int n){ 48 int i, j; 49 Book t; 50 for(i = 0; i < n-1; i++) 51 for(j = 0; j < n-1-i; j++) 52 if(x[j].sales_count < x[j+1].sales_count) { 53 t = x[j]; 54 x[j] = x[j+1]; 55 x[j+1] = t; 56 } 57 } 58 59 double sales_amount(Book x[], int n){ 60 double total = 0; 61 for (int i = 0; i < n; i++) { 62 total += x[i].sales_price * x[i].sales_count; 63 } 64 return total; 65 }
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实验任务5：
源代码：
1 #include <stdio.h> 2 3 typedef struct { 4 int year; 5 int month; 6 int day; 7 } Date; 8 9 10 void input(Date *pd); 11 int day_of_year(Date d); 12 int compare_dates(Date d1, Date d2); 13 14 void test1() { 15 Date d; 16 int i; 17 18 printf("输入日期:(以形如2024-12-16这样的形式输入)\n"); 19 for(i = 0; i < 3; ++i) { 20 input(&d); 21 printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day, day_of_year(d)); 22 } 23 } 24 25 void test2() { 26 Date Alice_birth, Bob_birth; 27 int i; 28 int ans; 29 30 printf("输入Alice和Bob出生日期:(以形如2024-12-16这样的形式输入)\n"); 31 for(i = 0; i < 3; ++i) { 32 input(&Alice_birth); 33 input(&Bob_birth); 34 ans = compare_dates(Alice_birth, Bob_birth); 35 36 if(ans == 0) 37 printf("Alice和Bob一样大\n\n"); 38 else if(ans == -1) 39 printf("Alice比Bob大\n\n"); 40 else 41 printf("Alice比Bob小\n\n"); 42 } 43 } 44 45 int main() { 46 printf("测试1: 输入日期, 打印输出这是一年中第多少天\n"); 47 test1(); 48 49 printf("\n测试2: 两个人年龄大小关系\n"); 50 test2(); 51 } 52 53 54 void input(Date *pd) { 55 printf("请输入日期 (年 月 日): "); 56 57 scanf("%d %d %d", &pd->year, &pd->month, &pd->day); 58 59 } 60 61 int day_of_year(Date d) { 62 int days_in_month[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; 63 64 if ((d.year % 4 == 0 && d.year % 100 != 0) || (d.year % 400 == 0)) { 65 days_in_month[1] = 29; 66 } 67 68 int day_of_year = 0; 69 for (int i = 0; i < d.month - 1; i++) { 70 day_of_year += days_in_month[i]; 71 } 72 day_of_year += d.day; 73 74 return day_of_year; 75 } 76 77 int compare_dates(Date d1, Date d2) { 78 if (d1.year < d2.year) 79 return -1; 80 if (d1.year > d2.year) 81 return 1; 82 83 if (d1.month < d2.month) 84 return -1; 85 if (d1.month > d2.month) 86 return 1; 87 88 if (d1.day < d2.day) 89 return -1; 90 if (d1.day > d2.day) 91 return 1; 92 93 return 0; 94 }
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实验任务6
源代码：
1 #include <stdio.h> 2 #include <string.h> 3 4 enum Role {admin, student, teacher}; 5 6 typedef struct { 7 char username[20]; 8 char password[20]; 9 enum Role type; 10 } Account; 11 12 13 void output(Account x[], int n); 14 15 int main() { 16 Account x[] = {{"A1001", "123456", student}, 17 {"A1002", "123abcdef", student}, 18 {"A1009", "xyz12121", student}, 19 {"X1009", "9213071x", admin}, 20 {"C11553", "129dfg32k", teacher}, 21 {"X3005", "921kfmg917", student}}; 22 int n; 23 n = sizeof(x)/sizeof(Account); 24 output(x, n); 25 26 return 0; 27 } 28 29 void output(Account x[], int n) { 30 31 for (int i = 0; i < n; ++i) { 32 printf("用户名: %-20s ", x[i].username); 33 34 for (int j = 0; j < strlen(x[i].password); ++j) { 35 printf("*"); 36 } 37 38 switch (x[i].type) { 39 case admin: printf(" admin"); break; 40 case student: printf(" student"); break; 41 case teacher: printf(" teacher"); break; 42 default: printf(" unknown"); break; 43 } 44 printf("\n"); 45 } 46 }
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实验任务7：
源代码：
1 #include <stdio.h> 2 #include <string.h> 3 4 typedef struct { 5 char name[20]; 6 char phone[12]; 7 int vip; 8 } Contact; 9 10 11 void set_vip_contact(Contact x[], int n, char name[]); 12 void output(Contact x[], int n); 13 void display(Contact x[], int n); 14 15 16 #define N 10 17 int main() { 18 Contact list[N] = {{"刘一", "15510846604", 0}, 19 {"陈二", "18038747351", 0}, 20 {"张三", "18853253914", 0}, 21 {"李四", "13230584477", 0}, 22 {"王五", "15547571923", 0}, 23 {"赵六", "18856659351", 0}, 24 {"周七", "17705843215", 0}, 25 {"孙八", "15552933732", 0}, 26 {"吴九", "18077702405", 0}, 27 {"郑十", "18820725036", 0}}; 28 int vip_cnt, i; 29 char name[20]; 30 31 printf("显示原始通讯录信息: \n"); 32 output(list, N); 33 34 printf("\n输入要设置的紧急联系人个数: "); 35 scanf("%d", &vip_cnt); 36 37 printf("输入%d个紧急联系人姓名:\n", vip_cnt); 38 for(i = 0; i < vip_cnt; ++i) { 39 scanf("%s", name); 40 set_vip_contact(list, N, name); 41 } 42 43 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"); 44 display(list, N); 45 46 return 0; 47 } 48 49 void set_vip_contact(Contact x[], int n, char name[]) { 50 for (int i = 0; i < n; ++i) { 51 if (strcmp(x[i].name, name) == 0) { 52 x[i].vip = 1; 53 break; 54 } 55 } 56 } 57 58 void display(Contact x[], int n) { 59 for (int i = 0; i < n-1; ++i) { 60 for (int j = i + 1; j < n; ++j) { 61 if (x[i].vip < x[j].vip) { 62 Contact temp = x[i]; 63 x[i] = x[j]; 64 x[j] = temp; 65 } 66 else if (x[i].vip == x[j].vip && strcmp(x[i].name, x[j].name) > 0) { 67 Contact temp = x[i]; 68 x[i] = x[j]; 69 x[j] = temp; 70 } 71 } 72 } 73 74 output(x, n); 75 } 76 77 void output(Contact x[], int n) { 78 int i; 79 80 for(i = 0; i < n; ++i) { 81 printf("%-10s%-15s", x[i].name, x[i].phone); 82 if(x[i].vip) 83 printf("%5s", "*"); 84 printf("\n"); 85 } 86 }
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