实验任务4
代码task4.c
1 #include <stdio.h> 2 #define N 10 3 4 typedef struct { 5 char isbn[20]; // isbn号 6 char name[80]; // 书名 7 char author[80]; // 作者 8 double sales_price; // 售价 9 int sales_count; // 销售册数 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] = {{"978-7-229-14156-1", "源泉", "安.兰德", 84, 59}, 18 {"978-7-5133-5261-1", "李白来到旧金山", "谭夏阳", 48, 16}, 19 {"978-7-5617-4347-8", "陌生人日记", "周怡芳", 72.6, 27}, 20 {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49}, 21 {"978-7-5046-9568-0", "数据化决策", "道格拉斯·W·哈伯德", 49, 42}, 22 {"978-7-5133-4388-6", "美好时代的背后", "凯瑟琳.布", 34.5, 39}, 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 {"978-7-5321-5691-7", "何为良好生活", "陈嘉映", 29.5 , 31}, 25 {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42}, 26 {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44}}; 27 28 printf("图书销量排名: \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 printf("isbn号\t\t\t书名\t\t\t\t作者\t\t\t售价\t销售册数\n"); 40 for(i = 0; i < n; i++){ 41 printf("%-20s\t%-30s\t%-20s\t%g\t%d\n", x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count); 42 } 43 } 44 45 void sort(Book x[], int n){ 46 Book t; 47 int i, j; 48 for(i = 0; i < n-1; i++){ 49 for(j = 0; j < n-i-1; j++){ 50 if(x[j].sales_count < x[j+1].sales_count){ 51 t = x[j]; 52 x[j] = x[j+1]; 53 x[j+1] = t; 54 } 55 } 56 } 57 } 58 59 double sales_amount(Book x[], int n){ 60 double avg; 61 int s = 0, i; 62 for(i = 0; i < n; i++){ 63 s += x[i].sales_price; 64 } 65 avg = s / n; 66 return avg; 67 }
运行结果
实验任务5
代码task5.c
1 #include <stdio.h> 2 3 typedef struct { 4 int year; 5 int month; 6 int day; 7 } Date; 8 9 // 函数声明 10 void input(Date *pd); // 输入日期给pd指向的Date变量 11 int day_of_year(Date d); // 返回日期d是这一年的第多少天 12 int compare_dates(Date d1, Date d2); // 比较两个日期: 13 // 如果d1在d2之前，返回-1; 14 // 如果d1在d2之后，返回1 15 // 如果d1和d2相同，返回0 16 17 void test1() { 18 Date d; 19 int i; 20 21 printf("输入日期:(以形如2023-12-11这样的形式输入)\n"); 22 for(i = 0; i < 3; ++i) { 23 input(&d); 24 printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day, day_of_year(d)); 25 } 26 } 27 28 void test2() { 29 Date Alice_birth, Bob_birth; 30 int i; 31 int ans; 32 33 printf("输入Alice和Bob出生日期:(以形如2023-12-11这样的形式输入)\n"); 34 for(i = 0; i < 3; ++i) { 35 input(&Alice_birth); 36 input(&Bob_birth); 37 ans = compare_dates(Alice_birth, Bob_birth); 38 39 if(ans == 0) 40 printf("Alice和Bob一样大\n\n"); 41 else if(ans == -1) 42 printf("Alice比Bob大\n\n"); 43 else 44 printf("Alice比Bob小\n\n"); 45 } 46 } 47 48 int main() { 49 printf("测试1: 输入日期, 打印输出这是一年中第多少天\n"); 50 test1(); 51 52 printf("\n测试2: 两个人年龄大小关系\n"); 53 test2(); 54 } 55 56 void input(Date *pd) { 57 scanf("%d-%d-%d", &(*pd).year, &(*pd).month, &(*pd).day); 58 } 59 60 int day_of_year(Date d) { 61 int m[13] = { 0, 31,28,31,30,31,30,31,31,30,31,30,31 }; 62 if ((d.year % 4 == 0 && d.year % 400 == 0) || d.year % 4 == 0) { 63 m[2] = 29; 64 } 65 int sum = 0, i; 66 for ( i = 1; i <= d.month - 1; i++) { 67 sum += m[i]; 68 } 69 sum += d.day; 70 return sum; 71 } 72 73 int compare_dates(Date d1, Date d2) { 74 if(d1.year < d2.year) 75 return -1; 76 else if(d1.year > d2.year) 77 return 1; 78 else 79 { 80 if(day_of_year(d1) < day_of_year(d2)) 81 return -1; 82 else if(day_of_year(d1) == day_of_year(d2)) 83 return 0; 84 else 85 return 1; 86 } 87 }
运行结果
实验任务6
代码task6.c
1 #include <stdio.h> 2 #include <string.h> 3 4 enum Role {admin, student, teacher}; 5 6 typedef struct { 7 char username[20]; // 用户名 8 char password[20]; // 密码 9 enum Role type; // 账户类型 10 } Account; 11 12 13 // 函数声明 14 void output(Account x[], int n); // 输出账户数组x中n个账户信息，其中，密码用*替代显示 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 // 待补足的函数output()实现 31 // 功能：遍历输出账户数组x中n个账户信息 32 // 显示时，密码字段以与原密码相同字段长度的*替代显示 33 void output(Account x[], int n) { 34 int i; 35 char k[10][10] = {"admin","student","teacher"}; 36 for(i = 0; i < n; i++) 37 { 38 int j = 0; 39 while(x[i].password[j] != '\0') 40 { 41 x[i].password[j] = '*'; 42 j++; 43 } 44 printf("%s\t\t%-9s\t\t%s\n", x[i].username, x[i].password, k[x[i].type]); 45 } 46 }
运行结果
