实验六
task4
1 #include <stdio.h> 2 #include <stdlib.h> 3 #define N 10 4 5 typedef struct { 6 char isbn[20]; 7 char name[80]; 8 char author[80]; 9 double sales_price; 10 int sales_count; 11 } Book; 12 13 void output(Book x[], int n); 14 void sort(Book x[], int n); 15 double sales_amount(Book x[], int n); 16 17 int main() { 18 Book x[N] = {{"978-7-5327-6082-4", "门将之死", "罗纳德.伦", 42, 51}, 19 {"978-7-308-17047-5", "自由与爱之地：入以色列记", "云也退", 49 , 30}, 20 {"978-7-5404-9344-8", "伦敦人", "克莱格泰勒", 68, 27}, 21 {"978-7-5447-5246-6", "软件体的生命周期", "特德姜", 35, 90}, 22 {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49}, 23 {"978-7-5133-5750-0", "主机战争", "布莱克.J.哈里斯", 128, 42}, 24 {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44}, 25 {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42}, 26 {"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}, 27 {"978-7-229-14156-1", "源泉", "安.兰德", 84, 59}}; 28 29 printf("图书销量排名(按销售册数): \n"); 30 sort(x, N); 31 output(x, N); 32 33 printf("\n图书销售总额: %.2f\n", sales_amount(x, N)); 34 35 return 0; 36 } 37 38 void output(Book x[], int n){ 39 printf("图书销售排名（按销售册数）：\n"); 40 printf("%-20s %-25s %-20s %-8s %-s\n","ISBN号","书名","作者","售价","销售册数"); 41 int i; 42 for(i=0;i<n;i++) 43 printf("%-20s %-25s %-20s %-8.1f %-d\n",x[i].isbn,x[i].name,x[i].author,x[i].sales_price,x[i].sales_count); 44 45 } 46 47 void sort(Book x[], int n){ 48 int i,j; 49 Book temp; 50 for(j=0;j<n-1;j++) 51 for(i=0;i<n-j-1;i++) 52 if(x[i]. sales_count>x[i+1]. sales_count) 53 { 54 temp=x[i]; 55 x[i]=x[i+1]; 56 x[i+1]=temp; 57 } 58 } 59 60 double sales_amount(Book x[], int n){ 61 int i; 62 double amount; 63 for(i=0;i<n;i++){ 64 amount+=x[i].sales_price*(x[i].sales_count*(1.0)); 65 } 66 return amount; 67 }
```
task 5
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 14 void test1() { 15 Date d; 16 int i; 17 18 printf("输入日期:(以形如2025-12-19这样的形式输入)\n"); 19 for(i = 0; i < 3; ++i) { 20 input(&d); 21 printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day, day_of_year(d)); 22 } 23 } 24 25 void test2() { 26 Date Alice_birth, Bob_birth; 27 int i; 28 int ans; 29 30 printf("输入Alice和Bob出生日期:(以形如2025-12-19这样的形式输入)\n"); 31 for(i = 0; i < 3; ++i) { 32 input(&Alice_birth); 33 input(&Bob_birth); 34 ans = compare_dates(Alice_birth, Bob_birth); 35 36 if(ans == 0) 37 printf("Alice和Bob一样大\n\n"); 38 else if(ans == -1) 39 printf("Alice比Bob大\n\n"); 40 else 41 printf("Alice比Bob小\n\n"); 42 } 43 } 44 45 int main() { 46 printf("测试1: 输入日期, 打印输出这是一年中第多少天\n"); 47 test1(); 48 49 printf("\n测试2: 两个人年龄大小关系\n"); 50 test2(); 51 } 52 53 void input(Date *pd) { 54 scanf("%d-%d-%d",&(pd->year),&(pd->month),&(pd->day)); 55 } 56 57 int day_of_year(Date d) { 58 int x[11]={31,28,31,30,31,30,31,31,30,31,30,31}; 59 if(d.year%4==0) 60 x[1]=29; 61 int i,days=0; 62 for(i=0;i<(d.month-1);i++) 63 days+=x[i]; 64 days+=d.day; 65 return days; 66 67 } 68 69 int compare_dates(Date d1, Date d2) { 70 if(d1.year<d2.year) 71 return -1; 72 else if(d1.year==d2.year){ 73 if(day_of_year(d1)<day_of_year(d2)) 74 return -1; 75 if(day_of_year(d1)==day_of_year(d2)) 76 return 0; 77 if(day_of_year(d1)>day_of_year(d2)) 78 return 1; 79 } 80 else 81 return 1; 82 }
```
task 6
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 14 void output(Account x[], int n); 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 30 31 void output(Account x[], int n) { 32 int i,j; 33 char *TYPE[]={"admin","student","teacher"}; 34 35 for(i=0;i<n;i++){ 36 int t=strlen(x[i].password); 37 38 printf("%s\t",x[i].username); 39 for(j=0;j<t;j++) 40 printf("*"); 41 printf("\t%s\n",TYPE[x[i].type]); 42 } 43 }
```
task 7
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 void sort(Contact x[], int n); 15 16 17 #define N 10 18 int main() { 19 Contact list[N] = {{"刘一", "15510846604", 0}, 20 {"陈二", "18038747351", 0}, 21 {"张三", "18853253914", 0}, 22 {"李四", "13230584477", 0}, 23 {"王五", "15547571923", 0}, 24 {"赵六", "18856659351", 0}, 25 {"周七", "17705843215", 0}, 26 {"孙八", "15552933732", 0}, 27 {"吴九", "18077702405", 0}, 28 {"郑十", "18820725036", 0}}; 29 int vip_cnt, i; 30 char name[20]; 31 32 printf("显示原始通讯录信息: \n"); 33 output(list, N); 34 35 printf("\n输入要设置的紧急联系人个数: "); 36 scanf("%d", &vip_cnt); 37 38 printf("输入%d个紧急联系人姓名:\n", vip_cnt); 39 for(i = 0; i < vip_cnt; ++i) { 40 scanf("%s", name); 41 set_vip_contact(list, N, name); 42 } 43 44 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"); 45 display(list, N); 46 47 return 0; 48 } 49 50 51 void set_vip_contact(Contact x[], int n, char name[]) { 52 int i,j; 53 for(i=0;i<n;i++){ 54 if(strcmp(x[i].name,name)==0){ 55 x[i].vip=1; 56 return; 57 } 58 } 59 } 60 61 void display(Contact x[], int n){ 62 sort(x,n); 63 int i; 64 for(i=0;i<n;i++){ 65 if(x[i].vip==1) 66 printf("%s\t%s\n",x[i].name,x[i].phone); 67 } 68 for(i=0;i<n;i++){ 69 if(x[i].vip==0) 70 printf("%s\t%s\n",x[i].name,x[i].phone); 71 } 72 } 73 74 void sort(Contact x[], int n) { 75 76 int i,j,t; 77 Contact temp; 78 79 for(i=0;i<n-1;i++){ 80 for(j=0;j<n-i-1;j++){ 81 if(x[j].vip<x[j+1].vip) 82 { 83 temp=x[j]; 84 x[j]=x[j+1]; 85 x[j+1]=temp; 86 } 87 else if(x[j].vip==x[j+1].vip){ 88 if(strcmp(x[j].name,x[j+1].name)>0){ 89 temp=x[j]; 90 x[j]=x[j+1]; 91 x[j+1]=temp; 92 } 93 } 94 } 95 } 96 } 97 void output(Contact x[], int n) { 98 int i; 99 100 for(i = 0; i < n; ++i) { 101 printf("%-10s%-15s", x[i].name, x[i].phone); 102 if(x[i].vip) 103 printf("%5s", "*"); 104 printf("\n"); 105 } 106 }
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