实验六
任务4:
源代码
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 }
任务5:
源代码
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 }
任务6:
源代码
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 }
任务7:
源代码
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 }