task4 code
1 #include <stdio.h>
2 #define N 10
3
4 typedef struct {
5 char isbn[20]; // isbn号
6 char name[80]; // 书名
7 char author[80]; // 作者
8 double sales_price; // 售价
9 int sales_count; // 销售册数
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] = { {"978-7-229-14156-1", "源泉", "安.兰德", 84, 59},
18 {"978-7-5133-5261-1", "李白来到旧金山", "谭夏阳", 48, 16},
19 {"978-7-5617-4347-8", "陌生人日记", "周怡芳", 72.6, 27},
20 {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49},
21 {"978-7-5046-9568-0", "数据化决策", "道格拉斯·W·哈伯德", 49,42},
22 {"978-7-5133-4388-6", "美好时代的背后", "凯瑟琳.布", 34.5, 39},
23 {"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},
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", 118,42},
26 {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5,44} };
27
28 printf("图书销量排名: \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
33 return 0;
34 }
35
36 void sort(Book x[], int n) {
37 int i, j;
38 Book temp;
39 Book* ptr = x;
40
41 for (i = 0; i < n - 1; i++)
42 for (j = 0; j < n - 1 - i; j++) {
43 if (x[j].sales_count < x[j + 1].sales_count) {
44 temp = x[j];
45 x[j] = x[j+1];
46 x[j + 1] = temp;
47
48 }
49 }
50 }
51
52 void output(Book x[], int n) {
53 Book* ptr;
54 int i;
55
56 printf("%-25s%-25s%-25s%-25s%-25s\n", "ISBN号", "书名", "作者", "售价", "销售册数");
57 for (ptr = x; ptr < x + n; ptr++)
58 printf("%-25s%-25s%-25s%-25.1lf%-25d\n", ptr->isbn, ptr->name, ptr->author, ptr->sales_price,
59 ptr->sales_count);
60 }
61
62 double sales_amount(Book x[], int n) {
63 double total = 0;
64 int i;
65
66 for (i = 0; i < n; i++)
67 total += (x[i].sales_price * x[i].sales_count);
68
69 }
View Code
task4 result
task5 code
1 #include <stdio.h>
2
3 typedef struct {
4 int year;
5 int month;
6 int day;
7 } Date;
8
9 // 函数声明
10 void input(Date* pd); // 输入日期给pd指向的Date变量
11 int day_of_year(Date d); // 返回日期d是这一年的第多少天
12 int compare_dates(Date d1, Date d2); // 比较两个日期:
13 // 如果d1在d2之前，返回-1;
14 // 如果d1在d2之后，返回1
15 // 如果d1和d2相同，返回0
16
17 void test1() {
18 Date d;
19 int i;
20
21 printf("输入日期:(以形如2023-12-11这样的形式输入)\n");
22 for (i = 0; i < 3; ++i) {
23 input(&d);
24 printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day,
25 day_of_year(d));
26 }
27 }
28 void test2() {
29 Date Alice_birth, Bob_birth;
30 int i;
31 int ans;
32
33 printf("输入Alice和Bob出生日期:(以形如2023-12-11这样的形式输入)\n");
34 for (i = 0; i < 3; ++i) {
35 input(&Alice_birth);
36 input(&Bob_birth);
37 ans = compare_dates(Alice_birth, Bob_birth);
38 if (ans == 0)
39 printf("Alice和Bob一样大\n\n");
40 else if (ans == -1)
41 printf("Alice比Bob大\n\n");
42 else
43 printf("Alice比Bob小\n\n");
44 }
45 }
46 int main() {
47 printf("测试1: 输入日期, 打印输出这是一年中第多少天\n");
48 test1();
49 printf("\n测试2: 两个人年龄大小关系\n");
50 test2();
51 }
52 // 补足函数input实现
53 // 功能: 输入日期给pd指向的Date变量
54 void input(Date* pd) {
55 int t1, t2;
56 scanf("%d", &pd->year);
57 scanf("%d", &t1);
58 pd->month = -t1;
59 scanf("%d", &t2);
60 pd->day = -t2;
61 }
62 // 补足函数day_of_year实现 31
63 // 功能：返回日期d是这一年的第多少天
64 int day_of_year(Date d) {
65 int leap = 0;
66 int sum = 0;
67
68 if (d.year % 4 == 0 && d.year % 100 != 0 || d.year % 400 == 0)
69 leap = 1;
70
71 if (leap) {
72 switch (d.month) {
73 case 1:sum = 0; break;
74 case 2:sum = 31; break;
75 case 3:sum = 60; break;
76 case 4:sum = 91; break;
77 case 5:sum = 121; break;
78 case 6:sum = 152; break;
79 case 7:sum = 182; break;
80 case 8:sum = 213; break;
81 case 9:sum = 244; break;
82 case 10:sum = 274; break;
83 case 11:sum = 305; break;
84 case 12:sum = 334; break;
85 }
86 sum += d.day;
87 }
88 else {
89 switch (d.month) {
90 case 1:sum = 0; break;
91 case 2:sum = 31; break;
92 case 3:sum = 59; break;
93 case 4:sum = 90; break;
94 case 5:sum = 120; break;
95 case 6:sum = 151; break;
96 case 7:sum = 181; break;
97 case 8:sum = 212; break;
98 case 9:sum = 243; break;
99 case 10:sum = 273; break;
100 case 11:sum = 304; break;
101 case 12:sum = 334; break;
102 }
103 sum += d.day;
104 }
105
106 return sum;
107 }
108 // 补足函数compare_dates实现
109 // 功能：比较两个日期:
110 // 如果d1在d2之前，返回-1;
111 // 如果d1在d2之后，返回1
112 // 如果d1和d2相同，返回0
113 int compare_dates(Date d1, Date d2) {
114 if (d1.year < d2.year)
115 return -1;
116 else if (d1.year > d2.year)
117 return 1;
118 else {
119 if (day_of_year(d1) < day_of_year(d2))
120 return -1;
121 else if (day_of_year(d1) > day_of_year(d2))
122 return 1;
123 else
124 return 0;
125 }
126
127 }
View Code
task5 result
task6 code
1 #include <stdio.h>
2 #include <string.h>
3
4 enum Role { admin, student, teacher };
5
6 typedef struct {
7 char username[20]; // 用户名
8 char password[20]; // 密码
9 enum Role type; // 账户类型
10 } Account;
11
12 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
13
14 int main() {
15 Account x[] = { {"A1001", "123456", student},
16 {"A1002", "123abcdef", student},
17 {"A1009", "xyz12121", student},
18 {"X1009", "9213071x", admin},
19 {"C11553", "129dfg32k", teacher},
20 {"X3005", "921kfmg917", student} };
21 int n;
22
23 n = sizeof(x) / sizeof(Account);
24 output(x, n);
25
26 return 0;
27 }
28 // 功能：遍历输出账户数组x中n个账户信息
29 // 显示时，密码字段以与原密码相同字段长度的*替代显示
30 void output(Account x[], int n) {
31 int i, j, k;
32 char t[20] = {'\0'};
33
34 for (i = 0; i < n; i++) {
35 k = strlen(x[i].password);
36 for (j = 0; j < k; j++)
37 t[j] = '*';
38 t[k] = '\0';
39
40 printf("%-9s %-9s ", x[i].username, t);
41 switch (x[i].type) {
42 case 0: printf(" admin"); break;
43 case 1: printf(" student"); break;
44 case 2: printf(" teacher"); break;
45 }
46 printf("\n");
47
48 for (j = 0; j < 20; j++)
49 t[j] = '\0';
50 }
51 }
View Code
task6 result
标签：case,int,sum,printf,break,实验,year From： https://www.cnblogs.com/nuist0415/p/17894960.html