实验六

task 4：

1.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] = {
18         {"978-7-5327-6082-4", "门将之死", "罗纳德.伦", 42, 51},
19         {"978-7-308-17047-5", "自由与爱之地：入以色列记", "云也退", 49, 30},
20         {"978-7-5404-9344-8", "伦敦人", "克莱格泰勒", 68, 27},
21         {"978-7-5447-5246-6", "软件体的生命周期", "特德姜", 35, 90},
22         {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49},
23         {"978-7-5133-5750-0", "主机战争", "布莱克.J.哈里斯", 128, 42},
24         {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44},
25         {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42},
26         {"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},
27         {"978-7-229-14156-1", "源泉", "安.兰德", 84, 59}
28     };
29 
30     printf("图书销量排名(按销售册数)：\n");
31     sort(x, N);
32     output(x, N);
33     printf("\n图书销售总额：%.2f\n", sales_amount(x, N));
34     return 0;
35 }
36 
37 void output(Book x[], int n)
38 {
39     printf("%-22s %-30s %-15s %-8s %s\n", "ISBN号", "书名", "作者", "售价", "销售册数");
40     for (int i = 0; i < n; i++)
41     {
42         printf("%-22s %-30s %-15s %-8.1f %d\n",
43             x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count);
44     }
45 }
46 
47 void sort(Book x[], int n)
48 {
49     int i, j;
50     Book temp;
51     for (i = 0; i < n - 1; i++)
52     {
53         for (j = 0; j < n - 1 - i; j++)
54         {
55             if (x[j].sales_count < x[j + 1].sales_count)
56             {
57                 temp = x[j];
58                 x[j] = x[j + 1];
59                 x[j + 1] = temp;
60             }
61         }
62     }
63 }
64 
65 double sales_amount(Book x[], int n)
66 {
67     double sum = 0;
68     for (int i = 0; i < n; i++)
69     {
70         sum += x[i].sales_price * x[i].sales_count;
71     }
72     return sum;
73 }
View Code

2.运行效果截图：

4

 

task 5：

1.源代码：

 1 #include <stdio.h>
 2 typedef struct {
 3     int year;
 4     int month;
 5     int day;
 6 } Date;
 7 
 8 void input(Date* pd);
 9 int day_of_year(Date d);
10 int compare_dates(Date d1, Date d2);
11 
12 void test1() {
13     Date d;
14     int i;
15     printf("输入日期:(以形如2025-12-19这样的形式输入)\n");
16     for (i = 0; i < 3; ++i) {
17         input(&d);
18         printf("%d-%02d-%02d是这一年中第%d天\n", d.year, d.month, d.day, day_of_year(d));
19     }
20 }
21 
22 void test2() {
23     Date Alice_birth, Bob_birth;
24     int i;
25     int ans;
26     printf("输入Alice和Bob出生日期:(以形如2025-12-19这样的形式输入)\n");
27     for (i = 0; i < 3; ++i) {
28         input(&Alice_birth);
29         input(&Bob_birth);
30         ans = compare_dates(Alice_birth, Bob_birth);
31         if (ans == 0)
32             printf("Alice和Bob一样大\n");
33         else if (ans == -1)
34             printf("Alice比Bob大\n");
35         else
36             printf("Alice比Bob小\n");
37     }
38 }
39 
40 int main() {
41     printf("测试1: 输入日期，打印输出这是一年中第多少天\n");
42     test1();
43     printf("\n测试2: 两个人年龄大小关系\n");
44     test2();
45 }
46 
47 void input(Date* pd) {
48     scanf_s("%d-%d-%d", &pd->year, &pd->month, &pd->day);
49 }
50 
51 int day_of_year(Date d) {
52     int mon[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
53     int sum = d.day;
54     for (int i = 0; i < d.month - 1; i++) {
55         sum += mon[i];
56     }
57     if ((d.year % 4 == 0 && d.year % 100 != 0) || (d.year % 400 == 0)) {
58         if (d.month > 2) sum++;
59     }
60     return sum;
61 }
62 
63 int compare_dates(Date d1, Date d2) {
64     if (d1.year > d2.year) return 1;
65     if (d1.year < d2.year) return -1;
66     if (d1.month > d2.month) return 1;
67     if (d1.month < d2.month) return -1;
68     if (d1.day > d2.day) return 1;
69     if (d1.day < d2.day) return -1;
70     return 0;
71 }
View Code

2.运行效果截图：

5

 

task 6：

1.源代码：

 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include <stdio.h>
 3 #include <string.h>
 4 
 5 enum Role { admin, student, teacher };
 6 
 7 typedef struct {
 8     char username[20];
 9     char password[20];
10     enum Role type;
11 } Account;
12 
13 void output(Account x[], int n);
14 
15 int main() {
16     Account x[] = {
17         {"A1001", "123456", student},
18         {"A1002", "123abcdef", student},
19         {"A1009", "xyz12121", student},
20         {"X1009", "9213071x", admin},
21         {"C11553", "129dfg32k", teacher},
22         {"x3005", "921kfmg917", student}
23     };
24     int n;
25     n = sizeof(x) / sizeof(Account);
26     output(x, n);
27     return 0;
28 }
29 
30 void output(Account x[], int n) {
31     for (int i = 0; i < n; i++) {
32         char pwd_mask[20] = { 0 };
33         int len = strlen(x[i].password);
34         for (int j = 0; j < len; j++) {
35             pwd_mask[j] = '*';
36         }
37         char role_str[20];
38         if (x[i].type == admin) strcpy(role_str, "admin");
39         else if (x[i].type == student) strcpy(role_str, "student");
40         else strcpy(role_str, "teacher");
41         printf("用户名：%s  密码：%s  角色：%s\n", x[i].username, pwd_mask, role_str);
42     }
43 }
View Code

2.运行效果截图：

6

 

task 7：

1.源代码：

 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include <stdio.h>
 3 #include <string.h>
 4 
 5 typedef struct {
 6     char name[20];
 7     char phone[12];
 8     int vip;
 9 } Contact;
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 #define N 10
16 int main() {
17     Contact list[N] = {
18         {"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     };
29     int vip_cnt, i;
30     char name[20];
31 
32     printf("显示原始通讯录信息：\n");
33     output(list, N);
34 
35     printf("\n输入要设置的紧急联系人个数：");
36     scanf("%d", &vip_cnt);
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 output(Contact x[], int n) {
59     int i;
60     for (i = 0; i < n; ++i) {
61         printf("%-10s%-15s", x[i].name, x[i].phone);
62         if (x[i].vip)
63             printf("%5s", "*");
64         printf("\n");
65     }
66 }
67 
68 void display(Contact x[], int n) {
69     Contact temp[N];
70     int vip_num = 0;
71     for (int i = 0; i < n; i++) {
72         if (x[i].vip == 1) {
73             temp[vip_num++] = x[i];
74         }
75     }
76     for (int i = 0; i < n; i++) {
77         if (x[i].vip == 0) {
78             temp[vip_num++] = x[i];
79         }
80     }
81     for (int i = 0; i < n - 1; i++) {
82         for (int j = 0; j < n - 1 - i; j++) {
83             if (temp[j].vip == 0 && temp[j + 1].vip == 1) {
84                 Contact t = temp[j];
85                 temp[j] = temp[j + 1];
86                 temp[j + 1] = t;
87             }
88             else if (temp[j].vip == temp[j + 1].vip && strcmp(temp[j].name, temp[j + 1].name) > 0) {
89                 Contact t = temp[j];
90                 temp[j] = temp[j + 1];
91                 temp[j + 1] = t;
92             }
93         }
94     }
95     output(temp, n);
96 }
View Code

2.运行效果截图：

7

 

posted @ 2026-06-16 21:30  槊  阅读(9)  评论(0)    收藏  举报