实验6

task1

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define N 3
 4 typedef struct student{
 5     int id;
 6     char name[20];
 7     char subject[20];
 8     double perf;
 9     double mid;
10     double final;
11     double total;
12     char level[10];
13 }stu;
14 void input(stu[],int);
15 void output(stu[],int);
16 void calc(stu[],int);
17 int fail(stu[],stu[],int);
18 void sort(stu[],int);
19 int main(){
20     stu st[N],fst[N];
21     int k;
22     printf("录入学生成绩信息\n");
23     input(st,N);
24     printf("\n成绩处理...\n");
25     calc(st,N);
26     k=fail(st,fst,N);
27     sort(st,N);
28     printf("\n学生成绩排名情况:\n");
29     output(st,N);
30     printf("不及格学生信息:\n");
31     output(fst,k);
32     return 0;
33 }
34 void input(stu s[],int n){
35     int i;
36     for(i=0;i<n;i++){
37         scanf("%d %s %s %lf %lf %lf",
38         &s[i].id,s[i].name,s[i].subject,&s[i].perf,
39         &s[i].mid,&s[i].final);
40     }
41 }
42 void output(stu s[],int n){
43     int i;
44     printf("----------------\n");
45     printf("学号   姓名     科目   平时   期中   期末   总评   等级\n");
46     for(i=0;i<n;i++){
47         printf("%d   %-6s   %-4s   %-4.0f   %-4.0f   %-4.0f   %-4.1f   %s\n",
48         s[i].id,s[i].name,s[i].subject,s[i].perf,
49         s[i].mid,s[i].final,s[i].total,s[i].level);
50     }
51 }
52 void calc(stu s[],int n){
53     int i;
54     for(i=0;i<n;i++){
55         s[i].total=s[i].perf*0.2+s[i].mid*0.2+s[i].final*0.6;
56         if(s[i].total>=90){
57             strcpy(s[i].level,""); 
58         }
59         else if(s[i].total>=80){
60             strcpy(s[i].level,"");
61         }
62         else if(s[i].total>=70){
63             strcpy(s[i].level,"");
64         }
65         else if(s[i].total>=60){
66             strcpy(s[i].level,"及格");
67         }
68         else{
69             strcpy(s[i].level,"不及格"); 
70         }
71     }
72 }
73 int fail(stu s[],stu t[],int n){
74     int i,cnt=0;
75     for(i=0;i<n;++i){
76         if(s[i].total<60){
77             t[cnt++]=s[i];
78         }
79     }
80     return cnt;
81 }
82 void sort(stu s[],int n){
83     int i,j;
84     stu t;
85     for(i=0;i<n;++i){
86         for(j=0;j<n-1-i;++j){
87             if(s[j].total<s[j+1].total){
88                 t=s[j];
89                 s[j]=s[j+1];
90                 s[j+1]=t;    
91             }
92         }
93     }
94 }
View Code

屏幕截图 2025-12-19 134333

task2

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define N 10
 4 #define M 80
 5 typedef struct {
 6     char name[M];
 7     char author[M];
 8 }book;
 9 int main(){
10     book x[N] = { {"《一九八四》", "乔治.奥威尔"},
11                   {"《美丽新世界》", "赫胥黎"},
12                   {"《昨日的世界》", "斯蒂芬.茨威格"}, 
13                   {"《万历十五年》", "黄仁宇"},
14                   {"《一只特立独行的猪》", "王小波"},
15                   {"《百年孤独》", "马尔克斯"},
16                   {"《查令十字街84号》", "海莲.汉芙"},
17                   {"《只是孩子》", "帕蒂.史密斯"}, 
18                   {"《刀锋》", "毛姆"},
19                   {"《沉默的大多数》", "王小波"} };
20     book *ptr;
21     int i;
22     char author[M];
23     int found;
24     printf("-------------------所有图书信息-------------------\n");
25     for(ptr=x;ptr<x+N;ptr++){
26         printf("%-30s%-30s\n",ptr->name,ptr->author);
27     } 
28     printf("\n-------------------按作者查询图书-------------------\n");
29     printf("输入作者名: ");
30     gets(author);
31     found=0;
32     for(ptr=x;ptr<x+N;ptr++){
33         if(strcmp(ptr->author,author)==0){
34             found=1;
35             printf("%-30s%-30s\n",ptr->name,ptr->author);
36         }
37     }
38     if(!found){
39         printf("暂未收录该作者书籍\n"); 
40     }
41     return 0;
42 }
task2

屏幕截图 2025-12-19 141720

屏幕截图 2025-12-19 141657

task3-1

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #define N 80
 4 typedef struct filminfo{
 5     char name[N];
 6     char director[N];
 7     char region[N];
 8     int year; 
 9     struct filminfo *next;
10 }film;
11 void output(film *head);
12 film *insert(film *head,int n); 
13 int main(){
14     int n;
15     film *head;
16     head=NULL;
17     printf("输入影片数目: ");
18     scanf("%d",&n);
19     head=insert(head,n);
20     printf("\n所有影片信息如下: \n");
21     output(head);
22     return 0;
23 } 
24 film *insert(film *head,int n){
25     int i;
26     film *p;
27     for(i=1;i<=n;++i){
28         p=(film *)malloc(sizeof(film));
29         printf("请输入第%d部影片信息: ", i);
30         scanf("%s %s %s %d",p->name,p->director,p->region,&p->year);
31         p->next=head;
32         head=p;
33     }
34     return head;
35 }
36 void output(film *head){
37     film *p;
38     p=head;
39     while(p!=NULL){
40         printf("%-20s %-20s %-20s %d\n",p->name ,p->director ,p->region ,p->year );
41         p=p->next;
42     }
43 }
task3-1

屏幕截图 2025-12-19 161829

task3-2

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #define N 80
 4 typedef struct FilmInfo {
 5     char name[N];
 6     char director[N];
 7     char region[N];
 8     int year;
 9     struct FilmInfo *next;
10 } Film;
11 void output(Film *head);  
12 Film *insert(Film *head, int n);  
13 int main() {
14     int n;          
15     Film *head;     
16     Film *p;  
17     p = (Film *)malloc(sizeof(Film));
18     p->next = NULL;
19     head = p;   
20     printf("输入影片数目: ");
21     scanf("%d", &n);
22     head = insert(head, n);
23     printf("\n所有影片信息如下: \n");
24     output(head);
25     return 0;
26 }
27 Film *insert(Film *head, int n) {
28     int i;
29     Film *p;
30     for(i = 1; i <= n; ++i) {
31         p = (Film *)malloc(sizeof(Film));
32         printf("请输入第%d部影片信息: ", i);
33         scanf("%s %s %s %d", p->name, p->director, p->region, &p->year);
34         p->next = head->next;
35         head->next = p;
36     }
37     return head;
38 }
39 void output(Film *head) {
40     Film *p;
41     p = head->next;
42     while(p != NULL) {
43         printf("%-20s %-20s %-20s %d\n", p->name, p->director, p->region, p->year);
44         p = p -> next;
45     }
46 }
task3-2

屏幕截图 2025-12-19 162706

task4

 1 #include<stdio.h>
 2 #define N 10
 3 typedef struct{
 4     char isbn[20];
 5     char name[80];
 6     char author[80];
 7     double saleprice;
 8     int salecount;
 9     double saleamout;
10 }book;
11 void output(book x[],int n);
12 void sort(book x[],int n);
13 double saleamount(book x[],int n);
14 int main(){
15     book x[N] = {{"978-7-5327-6082-4", "门将之死", "罗纳德.伦", 42, 51},
16                   {"978-7-308-17047-5", "自由与爱之地:入以色列记", "云也退", 49 , 30},
17                   {"978-7-5404-9344-8", "伦敦人", "克莱格泰勒", 68, 27},
18                   {"978-7-5447-5246-6", "软件体的生命周期", "特德姜", 35, 90}, 
19                   {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49},
20                   {"978-7-5133-5750-0", "主机战争", "布莱克.J.哈里斯", 128, 42},
21                   {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44},
22                   {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42},
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                   {"978-7-229-14156-1", "源泉", "安.兰德", 84, 59}};
25     printf("图书销量排名(按销售册数): \n");
26     sort(x,N);
27     output(x,N); 
28     printf("\n图书销售总额: %.2f\n", saleamount(x, N));
29     return 0;
30 }
31 void sort(book x[],int n){
32     int i,j;
33     book t;
34     for(i=0;i<n;++i){
35         for(j=0;j<n-1-i;++j){
36             if(x[j].salecount<x[j+1].salecount){
37             t=x[j];
38             x[j]=x[j+1];
39             x[j+1]=t;
40             }
41         }
42     }
43 }
44 void output(book x[],int n){
45     int i;
46     printf("ISBN号            书名                      作者                  售价         销售册数\n");
47     for(i=0;i<n;++i){
48         printf("%-10s %-25s %-20s  %.1lf %10d\n",x[i].isbn,x[i].name,x[i].author,
49                                               x[i].saleprice,x[i].salecount);
50     }
51 }
52 double saleamount(book x[],int n){
53     int i;
54     double sum=0.0;
55     for(i=0;i<n;++i){
56         sum+=x[i].saleprice*x[i].salecount;
57     }
58     return sum;
59 }
task4

屏幕截图 2025-12-19 170232

task5

 1 #include<stdio.h>
 2 typedef struct{
 3     int year;
 4     int month;
 5     int day;
 6 }date;
 7 void input(date *pd);
 8 int dateofyear(date d);
 9 int comdates(date d1,date d2);
10 void test1() {
11     date d;
12     int i;
13 
14     printf("输入日期:(以形如2025-12-19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, dayofyear(d));
18     }
19 }
20 void test2() {
21     date Alice_birth, Bob_birth;
22     int i;
23     int ans;
24 
25     printf("输入Alice和Bob出生日期:(以形如2025-12-19这样的形式输入)\n");
26     for(i = 0; i < 3; ++i) {
27         input(&Alice_birth);
28         input(&Bob_birth);
29         ans = comdates(Alice_birth, Bob_birth);
30         
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 int main(){
40     printf("测试1: 输入日期, 打印输出这是一年中第多少天\n");
41     test1();
42     printf("\n测试2: 两个人年龄大小关系\n");
43     test2();
44     return 0;
45 } 
46 void input(date *pd){
47         scanf("%d-%d-%d",&pd->year,&pd->month,&pd->day);
48 }
49 int dayofyear(date d){
50     int total=0,i;
51     int a[]={31,28,31,30,31,30,31,31,30,31,30,31};
52     if(d.year%400==0||(d.year%4==0&&d.year%100==0)){
53         a[1]=29;
54     }
55     for(i=0;i<d.month-1;++i){
56         total+=a[i];
57     }
58     total+=d.day;
59     return total;
60 }
61 int comdates(date d1,date d2){
62     int ans;
63     if(d1.year<d2.year){
64         ans=-1;
65     }
66     else if(d1.year>d2.year){
67         ans=1;
68     } 
69     else {
70         if(d1.month<d2.month){
71             ans=-1;
72         }
73         else if(d1.month>d2.month){
74             ans=1;
75         }
76         else{
77             if(d1.day<d2.day){
78                 ans=-1;
79             }
80             else if(d1.day>d2.day){
81                 ans=1;
82             }
83             else{
84                 ans=0;
85             }
86         }
87     }
88 return ans;
89 }
task5

屏幕截图 2025-12-21 154130

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 void output(account x[],int n);
10 int main(){
11     int n;
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     n=sizeof(x)/sizeof(account);
19     output(x,n);
20     return 0;
21 }
22 void output(account x[],int n){
23     int i,j;
24     for(i=0;i<n;++i){
25         printf("%-10s",x[i].username);
26         int len=strlen(x[i].password);
27         for(j=0;j<n;++j){
28             printf("*");
29         }
30         printf("     ");
31         switch(x[i].type){
32         case admin:printf("admin\n");break;
33         case student:printf("student\n");break;
34         case teacher:printf("teacher\n");break;
35         default:printf("wrong");
36         }
37         printf("\n");
38     }
39 }
task6

屏幕截图 2025-12-21 162856

task7

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define N 10
 4 typedef struct{
 5     char name[20];
 6     char phone[12];
 7     int vip;
 8 }contact;
 9 void setvipcontact(contact x[],int n,char name[]);
10 void output(contact x[],int n);
11 void display(contact x[],int n);
12 int main(){
13     int cnt,i;
14     char name[20];
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     printf("显示原始通讯录信息: \n"); 
26     output(list, N);
27     printf("\n输入要设置的紧急联系人个数: ");
28     scanf("%d", &cnt);
29     printf("输入%d个紧急联系人姓名:\n",cnt);
30     for(i = 0; i < cnt; ++i) {
31         scanf("%s", name);
32         setvipcontact(list, N, name);
33     }
34     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");
35     display(list, N);
36     return 0;
37 }
38 void setvipcontact(contact x[],int n,char name[]){
39     int i;
40     for(i=0;i<n;++i){
41         if(strcmp(x[i].name,name)==0){
42             x[i].vip=1;
43         }
44     }
45 }
46 void display(contact x[],int n){
47     int i,j;
48     contact temp;
49     for(i=0;i<n;++i){
50         for(j=0;j<n-1-i;++j){
51             if(strcmp(x[j].name,x[j+1].name)>0){
52                 temp=x[j];
53                 x[j]=x[j+1];
54                 x[j+1]=temp;
55             }
56         }
57     }
58     for(i=0;i<n;++i){
59         for(j=0;j<n-1-i;++j){
60             if(x[j].vip<x[j+1].vip){
61                 temp=x[j];
62                 x[j]=x[j+1];
63                 x[j+1]=temp;
64             }
65         }
66     }
67     output(x,n);
68 }
69 void output(contact x[], int n) {
70     int i;
71     for(i = 0; i < n; ++i) {
72         printf("%-10s%-15s", x[i].name, x[i].phone);
73         if(x[i].vip)
74             printf("%5s", "*");
75         printf("\n");
76     }
77 }
task7

屏幕截图 2025-12-21 190937

posted @ 2025-12-21 19:11  ZL425  阅读(0)  评论(0)    收藏  举报