实验六

任务4:

 源代码

复制代码
 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]={{"978-7-5327-6082-4","门之将死","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 
37 void output(Book x[],int n){
38     int i;
39     
40     printf("ISBN号              书名                          作者                售价        销售册数\n");
41     for(i = 0;i < n;i++)
42         printf("%s   %-27s   %-17s   %-10.2f   %2d\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     int i,j;
47     Book t;
48     
49     for(i = 0;i < n-1;i++)
50         for(j = 0;j < n-1-i;j++)
51             if(x[j].sales_count<x[j+1].sales_count){
52                 t=x[j];
53                 x[j]=x[j+1];
54                 x[j+1]=t;
55             }
56 }
57 
58 double sales_amount(Book x[],int n){
59     int i;
60     double amount=0.0;
61     
62     for(i = 0;i < n;i++)
63         amount+=x[i].sales_price*x[i].sales_count;
64     
65     return amount;
66 }
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任务5:

源代码

复制代码
 1 #include <stdio.h>
 2 
 3 typedef struct {
 4     int year;
 5     int month;
 6     int day;
 7 } Date;
 8 
 9 void input(Date *pb);
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2024-12-16这样的形式输入)\n");
18     for(i = 0;i < 3;++i){
19         input(&d);
20         printf("%d-%02d-%02d是这一年中的第%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2024--12-16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的第多少天\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 *pb){
53     scanf("%d-%d-%d",&pb->year,&pb->month,&pb->day);
54 }
55 
56 int day_of_year(Date d){
57     int i,t,x=0;
58     int days[12]={31,28,31,30,31,30,31,31,30,31,30,31};
59     
60     if((d.year%4==0&&d.year%100!=0)||d.year%400==0)
61        t=1;
62     else
63        t=0;
64        
65     if(t)
66        days[1]=29;
67        
68     for(i = 0;i < d.month-1;i++)
69         x+=days[i];
70         
71     x+=d.day;
72     
73     return x;
74 }
75 
76 int compare_dates(Date d1,Date d2){
77     if(d1.year<d2.year)
78        return -1;
79     else if(d1.year>d2.year)
80        return 1;
81     else{
82         if(d1.month<d2.month)
83            return -1;
84         else if(d1.month>d2.month)
85            return 1;
86         else{
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 }
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任务6:

源代码

复制代码
 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 void output(Account x[],int n);
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     n = sizeof(x)/sizeof(Account);
23     output(x,n);
24     
25     return 0;
26 }
27 
28 void output(Account x[],int n){
29     int i,j,t;
30     
31     for(i = 0;i < n;i++){
32         printf("%-20s",x[i].username);
33         
34         t=strlen(x[i].password);
35         for(j = 0;j < t;j++)
36             printf("*");
37         for(j = 0;j < 20-t;j++)
38             printf(" ");    
39         
40         switch(x[i].type) {
41             case 0:printf("admin\n");break;
42             case 1:printf("student\n");break;
43             case 2:printf("teacher\n");break;
44         }
45     }
46 }
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 任务7:

源代码

复制代码
  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 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 #define N 10
 15 int main(){
 16     Contact list[N]={{"刘一","15510846604",0},
 17                      {"陈二","18038747351",0},
 18                      {"张三","18853253914",0},
 19                      {"李四","13230584477",0},
 20                      {"王五","15547571923",0},
 21                      {"赵六","18856659351",0},
 22                      {"周七","17705843215",0},
 23                      {"孙八","15552933732",0},
 24                      {"吴九","18077702405",0},
 25                      {"郑十","18820725036",0}};
 26     int vip_cnt,i;
 27     char name[20];
 28     
 29     printf("显示原始通讯录信息:\n");
 30     output(list,N);
 31     
 32     printf("\n输入要设置的紧急联系人个数:");
 33     scanf("%d",&vip_cnt);
 34     printf("输入%d个紧急联系人姓名:\n",vip_cnt);
 35     for(i = 0;i < vip_cnt; ++i){
 36         scanf("%s",name);
 37         set_vip_contact(list,N,name);
 38     }
 39     
 40     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");
 41     display(list,N);
 42     
 43     return 0;
 44 }
 45 
 46 void set_vip_contact(Contact x[],int n,char name[]){
 47     int i;
 48     
 49     for(i = 0;i < n;i++)
 50         if(strcmp(x[i].name,name)==0){
 51             x[i].vip=1;
 52             break;
 53         }
 54 }
 55 
 56 void display(Contact x[],int n){
 57     Contact vip[10],fvip[10];
 58     int vipcount=0,fvipcount=0;
 59     int i,j;
 60     
 61     for(i = 0;i < n;i++){
 62         if(x[i].vip)
 63            vip[vipcount++]=x[i];
 64         else
 65            fvip[fvipcount++]=x[i];
 66     }
 67     
 68     Contact y[20];
 69     int ycount=0;
 70     
 71     for(i = 0;i < vipcount;i++)
 72         y[ycount++]=vip[i];
 73     for(i = 0;i < fvipcount;i++)
 74         y[ycount++]=fvip[i];
 75     
 76     Contact t;
 77     
 78     for(i = 0;i < vipcount-1;i++)
 79         for(j = 0;j < vipcount-1-i;j++)
 80             if(strcmp(y[j].name,y[j+1].name)>0){
 81                 t=y[j];
 82                 y[j]=y[j+1];
 83                 y[j+1]=t;
 84             }
 85     for(i = vipcount;i < ycount-1;i++)
 86         for(j = vipcount;j < ycount-1-i;j++)
 87             if(strcmp(y[j].name,y[j+1].name)>0){
 88                 t=y[j];
 89                 y[j]=y[j+1];
 90                 y[j+1]=t;
 91             }
 92     
 93     for(i = 0;i < ycount;i++){
 94         printf("%-10s%-15s",y[i].name,y[i].phone);
 95         if(y[i].vip)
 96            printf("%5s","*");
 97         printf("\n");
 98     }
 99 }
100 
101 void output(Contact x[],int n){
102     int i;
103     
104     for(i = 0;i < n;++i){
105         printf("%-10s%-15s",x[i].name,x[i].phone);
106         if(x[i].vip)
107            printf("%5s","*");
108         printf("\n");
109     }
110 }
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posted @   hydrangea1123  阅读(5)  评论(0编辑  收藏  举报
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