实验六
实验4
1 #include <stdio.h> 2 #include <string.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] = { 19 {"978-7-5327-6082-4", "门将之死", "罗纳德.伦", 42, 51}, 20 {"978-7-308-17047-5", "自由与爱之地:入以色列记", "云也退", 49, 30}, 21 {"978-7-5404-9344-8", "伦敦人", "克莱格泰勒", 68, 27}, 22 {"978-7-5447-5246-6", "软件体的生命周期", "特德姜", 35, 90}, 23 {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49}, 24 {"978-7-5133-5750-0", "主机战争", "布莱克.J.哈里斯", 128, 42}, 25 {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44}, 26 {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42}, 27 {"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}, 28 {"978-7-229-14156-1", "源泉", "安·兰德", 84, 59} 29 }; 30 31 printf("图书销量排名(按销售册数):\n"); 32 sort(x, N); 33 output(x, N); 34 printf("\n图书销售总额:%.2f\n", sales_amount(x, N)); 35 36 return 0; 37 } 38 39 void output(Book x[], int n) { 40 printf("ISBN号\t\t\t书名\t\t\t\t\t作者\t\t\t售价\t销售册数\n"); 41 int i; 42 for (i = 0; i < n; i++) { 43 printf("%s\t%s\t%s\t%.1f\t%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 (i = 0; i < n - 1; i++) { 51 for (j = 0; j < n - 1 - i; j++) { 52 if (x[j].sales_count < x[j+1].sales_count) { 53 temp = x[j]; 54 x[j] = x[j+1]; 55 x[j+1] = temp; 56 } 57 } 58 } 59 } 60 61 double sales_amount(Book x[], int n) { 62 double total = 0.0; 63 int i; 64 for (i = 0; i < n; i++) { 65 total += x[i].sales_price * x[i].sales_count; 66 } 67 return total; 68 }

实验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 void test1() { 14 Date d; 15 int i; 16 printf("输入日期:(以形如2025-12-19这样的形式输入)\n"); 17 for(i = 0; i < 3; ++i) { 18 input(&d); 19 printf("%d-%02d-%02d是这一年中第%d天\n", d.year, d.month, d.day, day_of_year(d)); 20 } 21 } 22 23 void test2() { 24 Date Alice_birth, Bob_birth; 25 int i; 26 int ans; 27 printf("输入Alice和Bob出生日期:(以形如2025-12-19这样的形式输入)\n"); 28 for(i = 0; i < 3; ++i) { 29 input(&Alice_birth); 30 input(&Bob_birth); 31 ans = compare_dates(Alice_birth, Bob_birth); 32 if(ans == 0) 33 printf("Alice和Bob一样大\n"); 34 else if(ans == -1) 35 printf("Alice比Bob大\n"); 36 else 37 printf("Alice比Bob小\n"); 38 } 39 } 40 41 int main() { 42 printf("测试1: 输入日期,打印输出这是一年中第多少天\n"); 43 test1(); 44 printf("\n测试2: 两个人年龄大小关系\n"); 45 test2(); 46 return 0; 47 } 48 49 void input(Date *pd) { 50 scanf("%d-%d-%d", &pd->year, &pd->month, &pd->day); 51 } 52 53 int day_of_year(Date d) { 54 int days[] = {31,28,31,30,31,30,31,31,30,31,30,31}; 55 int total = 0; 56 int i; 57 if((d.year % 4 == 0 && d.year % 100 != 0) || (d.year % 400 == 0)) { 58 days[1] = 29; 59 } 60 for(i = 0; i < d.month - 1; i++) { 61 total += days[i]; 62 } 63 total += d.day; 64 return total; 65 } 66 67 int compare_dates(Date d1, Date d2) { 68 if(d1.year < d2.year) { 69 return -1; 70 } else if(d1.year > d2.year) { 71 return 1; 72 } else { 73 int day1 = day_of_year(d1); 74 int day2 = day_of_year(d2); 75 if(day1 < day2) { 76 return -1; 77 } else if(day1 > day2) { 78 return 1; 79 } else { 80 return 0; 81 } 82 } 83 }

实验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 void output(Account x[], int n); 13 14 int main() { 15 Account x[] = { 16 {"A1001", "123456", student}, 17 {"A1002", "123abcdef", student}, 18 {"A1009", "xyz12121", student}, 19 {"X1009", "9213071x", admin}, 20 {"C11553", "129dfg32k", teacher}, 21 {"X3005", "921kfmg917", student} 22 }; 23 int n; 24 n = sizeof(x)/sizeof(Account); 25 output(x, n); 26 return 0; 27 } 28 29 void output(Account x[], int n) { 30 int i, j; 31 char *role_name; 32 for(i = 0; i < n; i++) { 33 printf("%s\t\t", x[i].username); 34 for(j = 0; j < strlen(x[i].password); j++) { 35 printf("*"); 36 } 37 printf("\t\t"); 38 switch(x[i].type) { 39 case admin: role_name = "admin"; break; 40 case student: role_name = "student"; break; 41 case teacher: role_name = "teacher"; break; 42 default: role_name = "unknown"; 43 } 44 printf("%s\n", role_name); 45 } 46 }

实验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 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 16 int main() { 17 Contact list[N] = { 18 {"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 }; 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 void set_vip_contact(Contact x[], int n, char name[]) { 51 int i; 52 for(i = 0; i < n; i++) { 53 if(strcmp(x[i].name, name) == 0) { 54 x[i].vip = 1; 55 break; 56 } 57 } 58 } 59 60 void output(Contact x[], int n) { 61 int i; 62 for(i = 0; i < n; ++i) { 63 printf("%-10s%-15s", x[i].name, x[i].phone); 64 if(x[i].vip) { 65 printf("%5s", "*"); 66 } 67 printf("\n"); 68 } 69 } 70 71 void display(Contact x[], int n) { 72 Contact temp, sorted[N]; 73 int i, j, k = 0; 74 75 for(i = 0; i < n; i++) { 76 sorted[i] = x[i]; 77 } 78 79 for(i = 0; i < n - 1; i++) { 80 for(j = 0; j < n - 1 - i; j++) { 81 if(sorted[j].vip < sorted[j+1].vip) { 82 temp = sorted[j]; 83 sorted[j] = sorted[j+1]; 84 sorted[j+1] = temp; 85 } else if(sorted[j].vip == sorted[j+1].vip) { 86 if(strcmp(sorted[j].name, sorted[j+1].name) > 0) { 87 temp = sorted[j]; 88 sorted[j] = sorted[j+1]; 89 sorted[j+1] = temp; 90 } 91 } 92 } 93 } 94 95 output(sorted, n); 96 }

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