task4
1 #include <stdio.h>
2 #include <string.h>
3 #define N 10
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 void output(Book x[], int n);
12 void sort(Book x[], int n);
13 double sales_amount(Book x[], int n);
14 int main() {
15 Book x[N] = { {"978-7-229-14156-1", "源泉", "安.兰德", 84, 59},
16 {"978-7-5133-5261-1", "李白来到旧金山", "谭夏阳", 48, 16},
17 {"978-7-5617-4347-8", "陌生人日记", "周怡芳", 72.6, 27},
18 {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49},
19 {"978-7-5046-9568-0", "数据化决策", "道格拉斯·W·哈伯德", 49,
20 42},
21 {"978-7-5133-4388-6", "美好时代的背后", "凯瑟琳.布", 34.5, 39},
22 {"978-7-1155-0509-5", "无穷的开始:世界进步的本源", "戴维·多伊奇",
23 37.5, 55},
24 {"978-7-5321-5691-7", "何为良好生活", "陈嘉映", 29.5 , 31},
25 {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118,
26 42},
27 {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5,
28 44} };
29 printf("图书销量排名: \n");
30 sort(x, N);
31 output(x, N);
32 printf("\n图书销售总额: %.2f\n", sales_amount(x, N));
33 return 0;
34 }
35
36 void output(Book x[], int n) {
37 printf("ISBN号\t\t\t书名\t\t\t\t作者\t\t\t售价\t销售册数\n");
38 for (int i = 0; i < n; i++) {
39 printf("%-20s\t%-30s\t%-20s\t%g\t%d\n", x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count);
40 }
41 }
42
43 void sort(Book x[], int n) {
44 for (int i = 0; i < n; i++) {
45 for (int j = i; j < n; j++) {
46 if (x[i].sales_count <= x[j].sales_count) {
47 int t = x[i].sales_count;
48 x[i].sales_count = x[j].sales_count;
49 x[j].sales_count = t;
50 t = x[i].sales_price;
51 x[i].sales_price = x[j].sales_price;
52 x[j].sales_price = t;
53 char tt[100];
54 strcpy(tt, x[i].isbn);
55 strcpy(x[i].isbn, x[j].isbn);
56 strcpy(x[j].isbn, tt);
57 strcpy(tt, x[i].name);
58 strcpy(x[i].name, x[j].name);
59 strcpy(x[j].name, tt);
60 strcpy(tt, x[i].author);
61 strcpy(x[i].author, x[j].author);
62 strcpy(x[j].author, tt);
63 }
64 }
65 }
66 }
67
68 double sales_amount(Book x[], int n) {
69 double sum = 0.0;
70 for (int i = 0; i < n; i++) {
71 sum += x[i].sales_price * x[i].sales_count * 1.0;
72 }
73 return sum;
74 }
task5
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
11 void test1() {
12 Date d;
13 int i;
14 printf("输入日期:(以形如2023-12-11这样的形式输入)\n");
15 for (i = 0; i < 3; ++i) {
16 input(&d);
17 printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day,
18 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2023-12-11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\n");
33 else if (ans == -1)
34 printf("Alice比Bob大\n\n");
35 else
36 printf("Alice比Bob小\n\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("%d-%d-%d", &(*pd).year, &(*pd).month, &(*pd).day);
49 }
50
51 int day_of_year(Date d) {
52 int m[13] = { 0, 31,28,31,30,31,30,31,31,30,31,30,31 };
53 if ((d.year % 4 == 0 && d.year % 400 == 0) || d.year % 4 == 0) {
54 m[2] = 29;
55 }
56 int sum = 0;
57 for (int i = 1; i <= d.month - 1; i++) {
58 sum += m[i];
59 }
60 sum += d.day;
61 return sum;
62 }
63
64 int compare_dates(Date d1, Date d2) {
65 int ans;
66 if (d1.year < d2.year) {
67 ans = -1;
68 return ans;
69 }
70 else if (d1.year > d2.year) {
71 ans = 1;
72 return ans;
73 }
74 else {
75 if (d1.month < d2.month) {
76 ans = -1;
77 return ans;
78 }
79 else if (d1.month > d2.month) {
80 ans = 1;
81 return ans;
82 }
83 else {
84 if (d1.day < d2.day) {
85 ans = -1;
86 return ans;
87 }
88 else if (d1.day > d2.day) {
89 ans = 1;
90 return ans;
91 }
92 else {
93 ans = 0;
94 return ans;
95 }
96 }
97 }
98 }
task6
1 #include <stdio.h>
2 #include <string.h>
3 enum Role { admin, student, teacher };
4 typedef struct {
5 char username[20];
6 char password[20];
7 enum Role type;
8 } Account;
9
10 void output(Account x[], int n);
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
24 void output(Account x[], int n) {
25 for (int i = 0; i < n; i++) {
26 printf("%s\t", x[i].username);
27 int s = strlen(x[i].password);
28 for (int i = 0; i < s; i++) {
29 printf("*");
30 }
31 if (s < 7) {
32 printf("\t\t");
33 }
34 else {
35 printf("\t");
36 }
37 switch (x[i].type) {
38 case admin:
39 printf("admin");
40 break;
41 case student:
42 printf("student");
43 break;
44 case teacher:
45 printf("teacher");
46 }
47 printf("\n");
48 }
49 }
task7
1 #include <stdio.h>
2 #include <string.h>
3 typedef struct {
4 char name[20];
5 char phone[12];
6 int vip;
7 } Contact;
8
9 void set_vip_contact(Contact x[], int n, char name[]);
10 void output(Contact x[], int n);
11 void display(Contact x[], int n);
12
13 #define N 10
14 int main() {
15 Contact list[N] = { {"刘一", "15510846604", 0},
16 {"陈二", "18038747351", 0},
17 {"张三", "18853253914", 0},
18 {"李四", "13230584477", 0},
19 {"王五", "15547571923", 0},
20 {"赵六", "18856659351", 0},
21 {"周七", "17705843215", 0},
22 {"孙八", "15552933732", 0},
23 {"吴九", "18077702405", 0},
24 {"郑十", "18820725036", 0} };
25 int vip_cnt, i;
26 char name[20];
27 printf("显示原始通讯录信息: \n");
28 output(list, N);
29 printf("\n输入要设置的紧急联系人个数: ");
30 scanf("%d", &vip_cnt);
31 printf("输入%d个紧急联系人姓名:\n", vip_cnt);
32 for (i = 0; i < vip_cnt; ++i) {
33 scanf("%s", name);
34 set_vip_contact(list, N, name);
35 }
36 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");
37 display(list, N);
38 return 0;
39 }
40
41 void set_vip_contact(Contact x[], int n, char name[]) {
42 for (int i = 0; i < n; i++) {
43 if (strcmp(x[i].name, name) == 0) {
44 x[i].vip = 1;
45 }
46 }
47 }
48
49 void display(Contact x[], int n) {
50
51 for (int i = n - 1; i >= 0; i--) {
52 for (int j = i; j >= 0; j--) {
53 if (x[i].vip == 1 && x[j].vip == 0) {
54 char nn[100];
55 strcpy(nn, x[i].name);
56 strcpy(x[i].name, x[j].name);
57 strcpy(x[j].name, nn);
58 strcpy(nn, x[i].phone);
59 strcpy(x[i].phone, x[j].phone);
60 strcpy(x[j].phone, nn);
61 int ss = x[i].vip;
62 x[i].vip = x[j].vip;
63 x[j].vip = ss;
64 }
65 }
66 }
67
68 for (int i = 0; i < n; i++) {
69 for (int j = i; j < n; j++) {
70 if (strcmp(x[i].name, x[j].name) > 0 && x[i].vip != 1 && x[j].vip != 1) {
71 char nn[100];
72 strcpy(nn, x[j].name);
73 strcpy(x[j].name, x[i].name);
74 strcpy(x[i].name, nn);
75 strcpy(nn, x[j].phone);
76 strcpy(x[j].phone, x[i].phone);
77 strcpy(x[i].phone, nn);
78 }
79 }
80 }
81
82 for (int i = 0; i < n; i++) {
83 for (int j = i; j < n; j++) {
84 if (x[i].vip == 1 && x[j].vip == 1) {
85 if (strcmp(x[i].name, x[j].name) > 0){
86 char nn[100];
87 strcpy(nn, x[j].name);
88 strcpy(x[j].name, x[i].name);
89 strcpy(x[i].name, nn);
90 strcpy(nn, x[j].phone);
91 strcpy(x[j].phone, x[i].phone);
92 strcpy(x[i].phone, nn);
93 }
94 }
95 }
96 }
97
98 output(x, n);
99 }
100 void output(Contact x[], int n) {
101 int i;
102 for (i = 0; i < n; ++i) {
103 printf("%-10s%-15s", x[i].name, x[i].phone);
104 if (x[i].vip)
105 printf("%5s", "*");
106 printf("\n");
107 }
108 }
标签：name,int,void,vip,strcpy,实验,printf From： https://www.cnblogs.com/VitaminC114514/p/17894376.html