task4
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] = {{"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
32 printf("\n图书销售总额: %.2f\n", sales_amount(x, N));
33
34 return 0;
35 }
36
37 void output(Book x[], int n){
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 printf("%-20s %-30s %-20s %.1lf %d\n",x[i].isbn,x[i].name,x[i].author,x[i].sales_price,x[i].sales_count);
42 }
43 }
44
45 void sort(Book x[], int n){
46 int i,j;
47 Book t;
48 for(i=0;i<n-1;i++){
49 for(j=0;j<n-i-1;j++){
50 if(x[j+1].sales_count>x[j].sales_count){
51 t=x[j];
52 x[j]=x[j+1];
53 x[j+1]=t;
54 }
55 }
56 }
57
58 }
59
60 double sales_amount(Book x[], int n){
61 double count=0;
62 int i;
63 for(i=0;i<n;i++){
64 count+=x[i].sales_count*x[i].sales_price;
65 }
66 return count;
67 }
结果
task5
代码
1 #include <stdio.h>
2
3 typedef struct {
4 int year;
5 int month;
6 int day;
7 } Date;
8
9 void input(Date *pd);
10 int day_of_year(Date d);
11 int compare_dates(Date d1, Date d2);
12
13 void test1() {
14 Date d;
15 int i;
16
17 printf("输入日期:(以形如2023-12-11这样的形式输入)\n");
18 for(i = 0; i < 3; ++i) {
19 input(&d);
20 printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day, day_of_year(d));
21 }
22 }
23
24 void test2() {
25 Date Alice_birth, Bob_birth;
26 int i;
27 int ans;
28
29 printf("输入Alice和Bob出生日期:(以形如2023-12-11这样的形式输入)\n");
30 for(i = 0; i < 3; ++i) {
31 input(&Alice_birth);
32 input(&Bob_birth);
33 ans = compare_dates(Alice_birth, Bob_birth);
34
35 if(ans == 0)
36 printf("Alice和Bob一样大\n\n");
37 else if(ans == -1)
38 printf("Alice比Bob大\n\n");
39 else
40 printf("Alice比Bob小\n\n");
41 }
42 }
43
44 int main() {
45 printf("测试1: 输入日期, 打印输出这是一年中第多少天\n");
46 test1();
47
48 printf("\n测试2: 两个人年龄大小关系\n");
49 test2();
50 }
51
52 void input(Date *pd) {
53 scanf("%d-%d-%d",&pd->year,&pd->month,&pd->day);
54
55 }
56
57 int day_of_year(Date d) {
58 int count=0,n=0;
59
60 if(d.month==1){
61 n=1;
62 }else if(d.month==2){
63 n=-1;
64 }else if(d.month==5){
65 n=1;
66 }else if(d.month==7){
67 n=2;
68 }else if(d.month==8||d.month==9){
69 n=3;
70 }else if(d.month==10||d.month==11){
71 n=4;
72 }else if(d.month==12){
73 n=5;
74 }
75
76 if((d.year%4==0&d.year%100!=0)||(d.year%400==0)){
77 count=(d.month-1)*30+d.day+n;
78 }else{
79 count=(d.month-1)*30+d.day+n-1;
80 }
81 return count;
82 }
83
84 int compare_dates(Date d1, Date d2) {
85 int i=0;
86 if(d1.year==d2.year){
87 if(d1.month==d2.month){
88 if(d1.day==d2.day){
89 i=0;
90 }else if(d1.day>d2.day){
91 i=1;
92 }else {
93 i=-1;
94 }
95 }else if(d1.month>d2.month){
96 i=1;
97 }else {
98 i=-1;
99 }
100 }else if(d1.year>d2.year){
101 i=1;
102 }else {
103 i=-1;
104 }
105
106 return i;
107 }
结果
task6
代码
1 #include <stdio.h>
2 #include <string.h>
3
4 enum Role {admin, student, teacher};
5
6 typedef struct {
7 char username[20];
8 char password[20];
9 enum Role type;
10 } Account;
11
12
13 void output(Account x[], int n);
14
15 int main() {
16 Account x[] = {{"A1001", "123456", student},
17 {"A1002", "123abcdef", student},
18 {"A1009", "xyz12121", student},
19 {"X1009", "9213071x", admin},
20 {"C11553", "129dfg32k", teacher},
21 {"X3005", "921kfmg917", student}};
22 int n;
23 n = sizeof(x)/sizeof(Account);
24 output(x, n);
25
26 return 0;
27 }
28
29
30 void output(Account x[], int n) {
31 int i,j;
32
33 for(i=0;i<n;i++){
34 printf("%-20s",x[i].username);
35 int count=0;
36 count=strlen(x[i].password);
37 while(count!=0){
38 printf("*");
39 count--;
40 }
41
42 j=20;
43 while(j>strlen(x[i].password)){
44 printf(" ");
45 j--;
46 }
47
48 switch(x[i].type){
49 case admin:printf("admin\n");break;
50 case student:printf("student\n");break;
51 case teacher:printf("teacher\n");break;
52 }
53
54 }
55 }
结果
task7
代码
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
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
16 #define N 10
17 int main() {
18 Contact list[N] = {{"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 int vip_cnt, i;
29 char name[20];
30
31 printf("显示原始通讯录信息: \n");
32 output(list, N);
33
34 printf("\n输入要设置的紧急联系人个数: ");
35 scanf("%d", &vip_cnt);
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
49 void set_vip_contact(Contact x[], int n, char name[]) {
50 int i;
51 for(i=0;i<n;i++){
52 if(strcmp(x[i].name,name)==0){
53 x[i].vip=1;
54 }
55 }
56
57 }
58
59
60 void display(Contact x[], int n) {
61 Contact t;
62 int i,j;
63 for(i=0;i<n-1;i++){
64 for(j=0;j<n-i-1;j++){
65 if(x[j].vip==0&&x[j+1].vip==1){
66 t=x[j];
67 x[j]=x[j+1];
68 x[j+1]=t;
69 }
70 }
71 }
72
73 int count=0;
74 for(i=0;i<n;i++){
75 if(x[i].vip==1){
76 count++;
77 }
78 }
79 for(i=0;i<count-1;i++){
80 for(j=0;j<count-i-1;j++){
81 if(strcmp(x[j].name,x[j+1].name)>0){
82 t=x[j];
83 x[j]=x[j+1];
84 x[j+1]=t;
85 }
86 }
87 }
88 for(i=count;i<n-1;i++){
89 for(j=count;j<n-i-1;j++){
90 if(strcmp(x[j].name,x[j+1].name)>0){
91 t=x[j];
92 x[j]=x[j+1];
93 x[j+1]=t;
94 }
95 }
96 }
97
98 for(i = 0; i < n; ++i) {
99 printf("%-10s%-15s", x[i].name, x[i].phone);
100 if(x[i].vip)
101 printf("%5s", "*");
102 printf("\n");
103 }
104 }
105
106 void output(Contact x[], int n) {
107 int i;
108
109 for(i = 0; i < n; ++i) {
110 printf("%-10s%-15s", x[i].name, x[i].phone);
111 if(x[i].vip)
112 printf("%5s", "*");
113 printf("\n");
114 }
115 }
结果
标签：int,void,else,实验,printf,month,day From： https://www.cnblogs.com/cy-t520/p/17894971.html