实验6

task4

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#define N 10
typedef struct {
    char isbn[20];
    char name[80];
    char author[80];
    double sales_price;
    int sales_count;
}Book;
void output(Book x[], int n);
void sort(Book x[], int n);
double sales_amount(Book x[], int n);
int main() {
    Book x[N] = { {"978-7-5327-6082-4", "门将之死", "罗纳德.伦", 42, 51},
                  {"978-7-308-17047-5", "自由与爱之地：入以色列记", "云也退", 49 , 30},
                  {"978-7-5404-9344-8", "伦敦人", "克莱格泰勒", 68, 27},
                  {"978-7-5447-5246-6", "软件体的生命周期", "特德姜", 35, 90},
                  {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49},
                  {"978-7-5133-5750-0", "主机战争", "布莱克.J.哈里斯", 128, 42},
                  {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44},
                  {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42},
                  {"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},
                  {"978-7-229-14156-1", "源泉", "安.兰德", 84, 59} };
    printf("图书销量排名（按销售册数）：\n");
    sort(x, N);
    output(x, N);
    printf("\n图书销售总额：%.2f\n", sales_amount(x, N));
}
void output(Book x[], int n) {
    printf("ISBN号\t\t    书名\t\t          作者\t\t      售价\t\t    销售册数\n");
    for (int i = 0; i < n; i++) {
        printf("%-20s%-30s%-20s%-20lf%-20d\n", x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count);
    }
}
void sort(Book x[], int n) {
    Book temp;
    for (int i = 0; i < n-1; i++) {
        for (int j = 0; j < n-i-1; j++) {
            if (x[j].sales_count < x[j + 1].sales_count) {
                temp = x[j];
                x[j] = x[j + 1];
                x[j + 1] = temp;
            }
        }
    }

}
double sales_amount(Book x[], int n) {
    double total = 0;
    for (int i = 0; i < n; i++) {
        total += x[i].sales_price * x[i].sales_count;
    }
    return total;

}

task 5

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>

typedef struct {
    int year;
    int month;
    int day;
} Date;

// 函数声明
void input(Date* pd);                   // 输入日期给pd指向的Date变量
int day_of_year(Date d);                // 返回日期d是这一年的第多少天
int compare_dates(Date d1, Date d2);    // 比较两个日期: 
// 如果d1在d2之前，返回-1;
// 如果d1在d2之后，返回1
// 如果d1和d2相同，返回0

void test1() {
    Date d;
    int i;

    printf("输入日期:(以形如2025-12-19这样的形式输入)\n");
    for (i = 0; i < 3; ++i) {
        input(&d);
        printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day, day_of_year(d));
    }
}

void test2() {
    Date Alice_birth, Bob_birth;
    int i;
    int ans;

    printf("输入Alice和Bob出生日期:(以形如2025-12-19这样的形式输入)\n");
    for (i = 0; i < 3; ++i) {
        input(&Alice_birth);
        input(&Bob_birth);
        ans = compare_dates(Alice_birth, Bob_birth);

        if (ans == 0)
            printf("Alice和Bob一样大\n\n");
        else if (ans == -1)
            printf("Alice比Bob大\n\n");
        else
            printf("Alice比Bob小\n\n");
    }
}

int main() {
    printf("测试1: 输入日期, 打印输出这是一年中第多少天\n");
    test1();

    printf("\n测试2: 两个人年龄大小关系\n");
    test2();
}

// 补足函数input实现
// 功能: 输入日期给pd指向的Date变量
void input(Date* pd) {
   
    scanf("%d%d%d", &pd->year, &pd->month, &pd->day);
    pd->day *= -1;
    pd->month *= -1;
   
}

// 补足函数day_of_year实现
// 功能：返回日期d是这一年的第多少天
int day_of_year(Date d) {
    int sum=0;
    for (int i = 1; i < d.month; i++) {
        if (i == 1 || i == 3 || i == 5 || i == 7 || i == 8 || i == 10)
            sum += 31;
        else if (i == 2)
            sum += 28;
        else
            sum += 30;
    }
    sum += d.day;
    if (d.year % 4 == 0 && d.year % 100 != 0 || d.year % 400 == 0)
        sum += 1;
    return sum;
}

// 补足函数compare_dates实现
// 功能：比较两个日期: 
// 如果d1在d2之前，返回-1;
// 如果d1在d2之后，返回1
// 如果d1和d2相同，返回0
int compare_dates(Date d1, Date d2) {
    if (d1.year < d2.year)
        return -1;
    else if (d1.year > d2.year)
        return 1;
    else {
        if (day_of_year(d1) < day_of_year(d2))
            return -1;
        else if (day_of_year(d1) > day_of_year(d2))
            return 1;
        else
            return 0;
    }
}

task6

#include <stdio.h>
#include <string.h>

enum Role { admin, student, teacher };

typedef struct {
    char username[20];  // 用户名
    char password[20];  // 密码
    enum Role type;     // 账户类型
} Account;


// 函数声明
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

int main() {
    Account x[] = { {"A1001", "123456", student},
                    {"A1002", "123abcdef", student},
                    {"A1009", "xyz12121", student},
                    {"X1009", "9213071x", admin},
                    {"C11553", "129dfg32k", teacher},
                    {"X3005", "921kfmg917", student} };
    int n;
    n = sizeof(x) / sizeof(Account);
    output(x, n);

    return 0;
}

// 待补足的函数output()实现
// 功能：遍历输出账户数组x中n个账户信息
//      显示时，密码字段以与原密码相同字段长度的*替代显示
void output(Account x[], int n) {
    for (int i = 0; i < n; i++) {
        int m;
       m= strlen(x[i].password);
       int y = 40 - m;
       printf("%-20s", x[i].username);
       for (int j = 0; j < m; j++)
           printf("*");
       for (int j = 0; j < y; j++)
           printf(" ");
       switch (x[i].type) {
       case admin: {
           printf("admin\n");
           break;
       }
       case teacher: {
           printf("teacher\n");
           break;
       }
       case student: {
           printf("student\n");
           break;
       }
       }
  }
    
}

task7

#include <stdio.h>
#include <string.h>
typedef struct {
char name[20]; // 姓名
char phone[12]; // 手机号
int vip; // 是否为紧急联系人，是取1；否则取0
} Contact;
// 函数声明
void set_vip_contact(Contact x[], int n, char name[]); // 设置紧急联系人
void output(Contact x[], int n); // 输出x中联系人信息
void display(Contact x[], int n); // 按联系人姓名字典序升序显示信息，紧急联系人最先显示
#define N 10
int main() {
Contact list[N] = {{"刘一", "15510846604", 0},
{"陈二", "18038747351", 0},
{"张三", "18853253914", 0},
{"李四", "13230584477", 0},
{"王五", "15547571923", 0},
{"赵六", "18856659351", 0},
{"周七", "17705843215", 0},
{"孙八", "15552933732", 0},
{"吴九", "18077702405", 0},
{"郑十", "18820725036", 0}};
int vip_cnt, i;
char name[20];
printf("显示原始通讯录信息: \n");
output(list, N);
printf("\n输入要设置的紧急联系人个数: ");
scanf("%d", &vip_cnt);
printf("输入%d个紧急联系人姓名:\n", vip_cnt);
for(i = 0; i < vip_cnt; ++i) {
scanf("%s", name);
set_vip_contact(list, N, name);
}
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");
display(list, N);
return 0;
}
void set_vip_contact(Contact x[], int n, char name[]) {
    for (int i = 0; i < n; i++) {
        if (strcmp(x[i].name, name) == 0) {
            x[i].vip = 1;
            break;
        }
    }
}
void display(Contact x[], int n) {
    int i, j;
    Contact temp;

    for (i = 0; i < n - 1; i++) {
        for (j = 0; j < n - 1 - i; j++) {
            int need_swap = 0;

            if (x[j].vip < x[j + 1].vip) {
                need_swap = 1;
            }
            else if (x[j].vip == x[j + 1].vip && strcmp(x[j].name, x[j + 1].name) > 0) {
                need_swap = 1;
            }

            if (need_swap) {
                temp = x[j];
                x[j] = x[j + 1];
                x[j + 1] = temp;
            }
        }
    }

    for (i = 0; i < n; i++) {
        printf("%-10s %-15s", x[i].name, x[i].phone);
        if (x[i].vip) {
            printf(" *");
        }
        printf("\n");
    }
}
void output(Contact x[], int n) {
    int i;
    for (i = 0; i < n; ++i) {
        printf("%s %s", x[i].name, x[i].phone);
        if (x[i].vip)
            printf(" *");
        printf("\n");
    }
}