前三個任務代碼均編寫並運行驗證
Task4:
1 #include <stdio.h>
2 #define N 10
3 typedef struct {
4 char isbn[20]; // isbn號
5 char name[80]; // 書名
6 char author[80]; // 作者
7 double sales_price; // 售價
8 int sales_count; // 銷售冊數
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", "門將之死", "羅納德.倫", 42, 51},
18 {"978-7-308-17047-5", "自由與愛之地:入以色列記", "雲也退", 49 , 30},
19 {"978-7-5404-9344-8", "倫敦人", "克萊格泰勒", 68, 27},
20 {"978-7-5447-5246-6", "軟件體的生命週期", "特德姜", 35, 90},
21 {"978-7-5722-5475-8", "芯片簡史", "汪波", 74.9, 49},
22 {"978-7-5133-5750-0", "主機戰爭", "布萊克.J.哈里斯", 128, 42},
23 {"978-7-2011-4617-1", "世界盡頭的咖啡館", "約翰·史崔勒基", 22.5, 44},
24 {"978-7-5133-5109-6", "你好外星人", "英國未來出版集團", 118, 42},
25 {"978-7-1155-0509-5", "無窮的開始:世界進步的本源", "戴維·多伊奇", 37.5,55},
26 {"978-7-229-14156-1", "源泉", "安.蘭德", 84, 59}
27 };
28 printf("圖書銷量排名(按銷售冊數): \n");
29 sort(x, N);
30 output(x, N);
31 printf("\n圖書銷售總額: %.2f\n", sales_amount(x, N));
32 return 0;
33 }
34 void output(Book x[], int n) {
35 printf("ISBN號 書名 作者 售價 銷售冊數\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 // 函數聲明
8 void input(Date *pd); // 輸入日期給pd指向的Date變量
9 int day_of_year(Date d); // 返回日期d是這一年的第多少天
10 int compare_dates(Date d1, Date d2); // 比較兩個日期:
11 // 如果d1在d2之前,返回-1;
12 // 如果d1在d2之後,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("輸入日期:(以形如2025-12-19這樣的形式輸入)\n");
18 for(i = 0; i < 3; ++i) {
19 input(&d);
20 printf("%d-%02d-%02d是這一年中第%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("輸入Alice和Bob出生日期:(以形如2025-12-19這樣的形式輸入)\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一樣大\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("測試1: 輸入日期, 打印輸出這是一年中第多少天\n");
43 test1();
44 printf("\n測試2: 兩個人年齡大小關係\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]; // 密碼
7 enum Role type; // 賬户類型
8 } Account;
9 // 函數聲明
10 void output(Account x[], int n); // 輸出賬户數組x中n個賬户信息,其中,密碼用*替代顯示
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("密碼: ");
30 for (int j = 0; j < len; j++) {
31 printf("*");
32 }
33 printf("\n");
34
35 printf("賬户類型: ");
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]; // 手機號
6 int vip; // 是否為緊急聯繫人,是取1;否則取0
7 } Contact;
8 // 函數聲明
9 void set_vip_contact(Contact x[], int n, char name[]); // 設置緊急聯繫人
10 void output(Contact x[], int n); // 輸出x中聯繫人信息
11 void display(Contact x[], int n); // 按聯繫人姓名字典序升序顯示信息,緊急聯繫人最先顯示
12 #define N 10
13 int main() {
14 Contact list[N] = {{"劉一", "15510846604", 0},
15 {"陳二", "18038747351", 0},
16 {"張三", "18853253914", 0},
17 {"李四", "13230584477", 0},
18 {"王五", "15547571923", 0},
19 {"趙六", "18856659351", 0},
20 {"周七", "17705843215", 0},
21 {"孫八", "15552933732", 0},
22 {"吳九", "18077702405", 0},
23 {"鄭十", "18820725036", 0}};
24 int vip_cnt, i;
25 char name[20];
26 printf("顯示原始通訊錄信息: \n");
27 output(list, N);
28 printf("\n輸入要設置的緊急聯繫人個數: ");
29 scanf("%d", &vip_cnt);
30 printf("輸入%d個緊急聯繫人姓名:\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顯示通訊錄列表:(按姓名字典序升序排列,緊急聯繫人最先顯示)\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 }
本文章為轉載內容,我們尊重原作者對文章享有的著作權。如有內容錯誤或侵權問題,歡迎原作者聯繫我們進行內容更正或刪除文章。