task4.
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] = {{"978-7-5327-6082-4", "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 void output(Book x[], int n)
37 {
38 int i;
39 printf("ISBN号 书名 作者 售价 销售册数\n");
40 for (i = 0; i < n; i++)
41 {
42 printf("%-15s %-25s %-20s %-5g %-5d\n", x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count);
43 }
44 }
45 void sort(Book x[], int n)
46 {
47 int i,j;
48 for ( i = 0; i < n - 1; i++)
49 {
50 for (j = 0; j < n-1-i; j++)
51 {
52 if (x[j].sales_count < x[j+1].sales_count)
53 {
54 Book temp = x[j];
55 x[j] = x[j+1];
56 x[j+1] = temp;
57 }
58 }
59 }
60 }
61 double sales_amount(Book x[], int n)
62 {
63 double sum = 0;
64 int i;
65 for ( i = 0; i < n; i++)
66 {
67 sum += x[i].sales_price * x[i].sales_count;
68 }
69 return sum;
70 }
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] = {{"978-7-5327-6082-4", "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 void output(Book x[], int n)
37 {
38 int i;
39 printf("ISBN号 书名 作者 售价 销售册数\n");
40 for (i = 0; i < n; i++)
41 {
42 printf("%-15s %-25s %-20s %-5g %-5d\n", x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count);
43 }
44 }
45 void sort(Book x[], int n)
46 {
47 int i,j;
48 for ( i = 0; i < n - 1; i++)
49 {
50 for (j = 0; j < n-1-i; j++)
51 {
52 if (x[j].sales_count < x[j+1].sales_count)
53 {
54 Book temp = x[j];
55 x[j] = x[j+1];
56 x[j+1] = temp;
57 }
58 }
59 }
60 }
61 double sales_amount(Book x[], int n)
62 {
63 double sum = 0;
64 int i;
65 for ( i = 0; i < n; i++)
66 {
67 sum += x[i].sales_price * x[i].sales_count;
68 }
69 return sum;
70 }
task5.
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2024-12-16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, day_of_year(d));
25 }
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2024-12-16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
39 if(ans == 0)
40 printf("Alice和Bob一样大\n\n");
41 else if(ans == -1)
42 printf("Alice比Bob大\n\n");
43 else
44 printf("Alice比Bob小\n\n");
45 }
46 }
47
48 int main() {
49 printf("测试1: 输入日期, 打印输出这是一年中第多少天\n");
50 test1();
51
52 printf("\n测试2: 两个人年龄大小关系\n");
53 test2();
54 }
55
56 void input(Date *pd)
57 {
58 scanf("%d-%d-%d",&(*pd).year,&(*pd).month,&(*pd).day);
59 }
60 int day_of_year(Date d)
61 {
62 int days=0,i;
63 int days_month[]={31,28,31,30,31,30,31,31,30,31,30,31};
64 if(d.year%4==0&&d.year%100!=0||d.year%400==0)
65 days_month[1]+=1;
66 for(i=0;i<d.month-1;++i)
67 days+=days_month[i];
68 days+=d.day;
69
70 return days;
71 }
72
73 int compare_dates(Date d1, Date d2)
74 {
75 if(d1.year<d2.year)
76 return -1;
77 else if(d1.year>d2.year)
78 return 1;
79 else
80 {
81 if(d1.month<d2.month)
82 return -1;
83 else if(d1.month>d2.month)
84 return 1;
85 else
86 {
87 if(d1.day<d2.day)
88 return -1;
89 else if(d1.day>d2.day)
90 return 1;
91 else
92 return 0;
93 }
94 }
95
96 }
task6.
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
13 // 函数声明
14 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*替代显示
15
16 int main() {
17 Account x[] = {{"A1001", "123456", student},
18 {"A1002", "123abcdef", student},
19 {"A1009", "xyz12121", student},
20 {"X1009", "9213071x", admin},
21 {"C11553", "129dfg32k", teacher},
22 {"X3005", "921kfmg917", student}};
23 int n;
24 n = sizeof(x)/sizeof(Account);
25 output(x, n);
26
27 return 0;
28 }
29 void output(Account x[], int n)
30 {
31 int i,j,sum;
32 char word[20];
33
34 for(i=0;i<n;++i)
35 {
36 printf("%s\t",x[i].username);
37 sum=strlen(x[i].password);
38 for(j=0;j<sum;++j)
39 word[j]='*';
40 word[j]='\0';
41 printf("%-20s",word);
42 switch(x[i].type)
43 {
44 case 0:printf("admin\n");break;
45 case 1:printf("student\n");break;
46 case 2:printf("teacher\n");break;
47 default:printf("somethint must be wrong");break;
48 }
49 }
50
51 }
task7.
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 void set_vip_contact(Contact x[], int n, char name[]);
11 void output(Contact x[], int n);
12 void display(Contact x[], int n);
13
14
15 #define N 10
16 int main() {
17 Contact list[N] = {{"刘一", "15510846604", 0},
18 {"陈二", "18038747351", 0},
19 {"张三", "18853253914", 0},
20 {"李四", "13230584477", 0},
21 {"王五", "15547571923", 0},
22 {"赵六", "18856659351", 0},
23 {"周七", "17705843215", 0},
24 {"孙八", "15552933732", 0},
25 {"吴九", "18077702405", 0},
26 {"郑十", "18820725036", 0}};
27 int vip_cnt, i;
28 char name[20];
29
30 printf("显示原始通讯录信息: \n");
31 output(list, N);
32
33 printf("\n输入要设置的紧急联系人个数: ");
34 scanf("%d", &vip_cnt);
35
36 printf("输入%d个紧急联系人姓名:\n", vip_cnt);
37 for(i = 0; i < vip_cnt; ++i) {
38 scanf("%s", name);
39 set_vip_contact(list, N, name);
40 }
41
42 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");
43 display(list, N);
44
45 return 0;
46 }
47
48 void set_vip_contact(Contact x[], int n, char name[]) {
49 int i;
50 for(i=0;i<n;i++){
51 if(strcmp(x[i].name,name)==0){
52 x[i].vip=1;
53 }
54 }
55 }
56
57 void display(Contact x[], int n) {
58 Contact temp;
59 int i,j;
60 for(i=0;i<n-1;i++){
61 for(j=0;j<n-i-1;j++){
62 if (x[j].vip < x[j + 1].vip) {
63 temp = x[j];
64 x[j] = x[j + 1];
65 x[j + 1] = temp;
66 } else if (x[j].vip == x[j + 1].vip && strcmp(x[j].name, x[j + 1].name) > 0) {
67 temp = x[j];
68 x[j] = x[j + 1];
69 x[j + 1] = temp;
70 }
71 }
72 }
73 output(x,n);
74 }
75
76 void output(Contact x[], int n) {
77 int i;
78
79 for(i = 0; i < n; ++i) {
80 printf("%-10s%-15s", x[i].name, x[i].phone);
81 if(x[i].vip)
82 printf("%5s", "*");
83 printf("\n");
84 }
85 }
标签：int,void,sales,978,Book,实验,printf From： https://www.cnblogs.com/zm060204/p/18610338