实验6

task4

#include <stdio.h>
#define N 10
typedef struct {
    char isbn[20];  // isbn号
    char name[80];  // 书名
    char author[80];  // 作者
    double sales_price;  // 售价
    int  sales_count;  // 销售册数
} 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));

    return 0;
}

void output(Book x[], int n) {
    
    printf("图书销量排名(按销售册数):\n");
    printf("%-20s %-25s %-20s %-10s %-10s\n", 
           "ISBN号", "书名", "作者", "售价", "销售册数");
    
     
    for (int i = 0; i < n; i++) {
        printf("%-20s %-25s %-20s %-10.1f %-10d\n", 
               x[i].isbn,        
               x[i].name,        
               x[i].author,      
               x[i].sales_price, 
               x[i].sales_count  
        );
    }
}

void sort(Book x[], int n){
    int i, j;
    for(i = 0; i < n - 1; i++)
        for(j = 0; j < n - 1 - i; j++){
            if(x[j].sales_count < x[j+1].sales_count){
                Book t = x[j];
                x[j] = x[j+1];
                x[j+1] = t;
            }
        }
}

double sales_amount(Book x[], int n){
    double total = 0.0;
    int i;
    for(i = 0; i < n; i++){
        total += x[i].sales_price * x[i].sales_count;
    }
    return total;
}

 

 task5

#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();               
}

void input(Date *pd){
    scanf("%d-%d-%d", &(pd->year), &(pd->month), &(pd->day));
}

int day_of_year(Date d){
    int mon[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int total = 0;
    int Run = (d.year % 4 == 0 && d.year % 100 != 0) || (d.year % 400 == 0);
    
    for(int i = 0; i < d.month - 1; i++){
        total += mon[i];
    }
    if(Run && d.month > 2){
        total += 1;
    }
    total += d.day;
    
    return total;
}

int compare_dates(Date d1, Date d2){
    if (d1.year < d2.year) {
        return -1; 
    } 
    else if (d1.year > d2.year) {
        return 1;  
    }
    
    if (d1.month < d2.month) {
        return -1; 
    } 
    else if (d1.month > d2.month) {
        return 1;  
    }
    
    if (d1.day < d2.day) {
        return -1; 
    } 
    else if (d1.day > d2.day) {
        return 1; 
    }
    
    return 0;
}

 task6

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

enum Role {admin, student, teacher};

typedef struct{
    char username[20];
    char password[20];
    enum Role type;
} Account;

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;
}

void output(Account x[], int n){
    for(int i = 0; i < n; i++){
        printf("%-10s", x[i].username);
        int pwd_len = strlen(x[i].password);
        for(int j = 0; j < pwd_len; j++){
            printf("*");
        }
        printf("\t");
        
        switch(x[i].type){
            case admin:
                printf("admin");
                break;
            case student:
                printf("student");
                break;
            case teacher:
                printf("teacher");
                break;
            default:
                printf("unknown");
        }
        printf("\n");
    }
}

 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);                      // 按姓名字典序+紧急优先显示

#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;
}

// 补全set_vip_contact函数：将姓名匹配的联系人设为紧急联系人（vip=1）
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;
        }
    }
}

// 补全display函数：按“紧急优先+姓名字典序”排序后输出
void display(Contact x[], int n) {
    
    Contact temp;
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < n - i - 1; j++) {
         
            if (x[j].vip < x[j+1].vip) {
                temp = x[j];
                x[j] = x[j+1];
                x[j+1] = temp;
            }
          
            else if (x[j].vip == x[j+1].vip) {
                if (strcmp(x[j].name, x[j+1].name) > 0) {
                    temp = x[j];
                    x[j] = x[j+1];
                    x[j+1] = temp;
                }
            }
        }
    }
    
    output(x, n);
}


void output(Contact x[], int n) {
    int i;
    
    for(i = 0; i < n; ++i) {
        printf("%-10s%-15s", x[i].name, x[i].phone);
        if(x[i].vip)
            printf("%5s", "*");
        printf("\n");
    }
}