实验6
任务4
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", "门将之死", "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 38 void output(Book x[],int n) 39 { 40 int i; 41 printf("%-20s%-30s%-20s%10s%10s\n", "ISBN号", "书名", "作者", "售价", "销售册数"); 42 for(i=0;i<n;i++) 43 { 44 printf("%-20s%-30s%-20s%10.1lf%10d\n", x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count); 45 } 46 } 47 48 49 void sort(Book x[], int n) 50 { 51 int i,j; 52 Book t; 53 for(i=0;i<n-1;i++) 54 { 55 for(j=0;j<n-1-i;j++) 56 { 57 if(x[j+1].sales_count>x[j].sales_count) 58 { 59 t=x[j+1]; 60 x[j+1]=x[j]; 61 x[j]=t; 62 } 63 } 64 } 65 66 } 67 68 69 70 double sales_amount(Book x[], int n) 71 { 72 double sum=0; 73 int i; 74 for(i=0;i<n;i++) 75 { 76 sum+=x[i].sales_price*x[i].sales_count; 77 } 78 return sum; 79 }
任务5
1 #include <stdio.h> 2 3 typedef struct { 4 int year; 5 int month; 6 int day; 7 } Date; 8 9 10 void input(Date *pd); 11 int day_of_year(Date d); 12 int compare_dates(Date d1, Date d2); 13 14 15 void test1() { 16 Date d; 17 int i; 18 19 printf("输入日期:(以形如2024-12-16这样的形式输入)\n"); 20 for(i = 0; i < 3; ++i) { 21 input(&d); 22 printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day, day_of_year(d)); 23 } 24 } 25 26 void test2() { 27 Date Alice_birth, Bob_birth; 28 int i; 29 int ans; 30 31 printf("输入Alice和Bob出生日期:(以形如2024-12-16这样的形式输入)\n"); 32 for(i = 0; i < 3; ++i) { 33 input(&Alice_birth); 34 input(&Bob_birth); 35 ans = compare_dates(Alice_birth, Bob_birth); 36 37 if(ans == 0) 38 printf("Alice和Bob一样大\n\n"); 39 else if(ans == -1) 40 printf("Alice比Bob大\n\n"); 41 else 42 printf("Alice比Bob小\n\n"); 43 } 44 } 45 46 int main() { 47 printf("测试1: 输入日期, 打印输出这是一年中第多少天\n"); 48 test1(); 49 50 printf("\n测试2: 两个人年龄大小关系\n"); 51 test2(); 52 } 53 54 55 void input(Date *pd) 56 { 57 scanf("%d-%d-%d",&pd->year,&pd->month,&pd->day); 58 59 } 60 61 62 int day_of_year(Date d) 63 { 64 int a=0,i; 65 int month[12]={31,28,31,30,31,30,31,31,30,31,30,31}; 66 if((d.year % 4 == 0 && d.year % 100!= 0) || (d.year % 400 == 0)) 67 month[1]=29; 68 if(d.month==1) 69 a=d.day; 70 else 71 { 72 for(i=0;i<d.month-1;i++) 73 { 74 a+=month[i]; 75 } 76 a+=d.day; 77 } 78 79 return a; 80 81 } 82 83 84 int compare_dates(Date d1, Date d2) 85 { 86 if(d1.year<d2.year) 87 return -1; 88 if(d1.year>d2.year) 89 return 1; 90 else 91 { 92 if(d1.month<d2.month) 93 return -1; 94 if(d1.month>d2.month) 95 return 1; 96 else 97 { 98 if(d1.day<d2.day) 99 return -1; 100 if(d1.day>d2.day) 101 return 1; 102 else 103 return 0; 104 } 105 } 106 }
任务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 void output(Account x[], int n); 14 15 int main() { 16 Account x[] = {{"A1001", "123456", student}, 17 {"A1002", "123abcdef", student}, 18 {"A1009", "xyz12121", student}, 19 {"X1009", "9213071x", admin}, 20 {"C11553", "129dfg32k", teacher}, 21 {"X3005", "921kfmg917", student}}; 22 int n; 23 n = sizeof(x)/sizeof(Account); 24 output(x, n); 25 26 return 0; 27 } 28 29 void output(Account x[], int n) 30 { 31 int i,j; 32 33 for(i=0;i<n;i++) 34 { 35 printf("%-10s",x[i].username); 36 for(j=0;j<strlen(x[i].password);j++) 37 { 38 printf("*"); 39 } 40 printf("\t"); 41 if(x[i].type==0) 42 printf("admin\n"); 43 if(x[i].type==1) 44 printf("student\n"); 45 if(x[i].type==2) 46 printf("teacher\n"); 47 } 48 49 }
任务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 12 void set_vip_contact(Contact x[], int n, char name[]); 13 void output(Contact x[], int n); 14 void display(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 { 53 int i; 54 for(i=0;i<n;i++) 55 { 56 if(strcmp(x[i].name,name)==0) 57 x[i].vip=1; 58 } 59 60 } 61 62 63 void display(Contact x[], int n) 64 { 65 Contact t; 66 int i,j; 67 for (i=0;i<n-1;i++) 68 { 69 for (j=0;j<n-i-1;j++) 70 { 71 if (x[j].vip < x[j + 1].vip) 72 { 73 t=x[j+1]; 74 x[j+1]=x[j]; 75 x[j]=t; 76 } 77 else if (x[j].vip == x[j + 1].vip) 78 { 79 if (strcmp(x[j].name, x[j + 1].name) > 0) 80 { 81 t=x[j+1]; 82 x[j+1]=x[j]; 83 x[j]=t; 84 } 85 } 86 87 } 88 } 89 output(x, n); 90 91 92 93 } 94 95 void output(Contact x[], int n) { 96 int i; 97 98 for(i = 0; i < n; ++i) { 99 printf("%-10s%-15s", x[i].name, x[i].phone); 100 if(x[i].vip) 101 printf("%5s", "*"); 102 printf("\n"); 103 } 104 }
