实验六

task 1

task 2

 

task 3

3.1需要区分链表是否为空，3.2的节点操作逻辑是统一的。

task 4

#include <stdio.h>
#include <string.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) {
    int i;
    for (i = 0; i < n; i++) {
        printf("第%d名：ISBN：%s，书名：%s，作者：%s，售价：%.2f，销量：%d\n",
               i + 1, 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 temp = x[j];
                x[j] = x[j + 1];
                x[j + 1] = temp;
            }
        }
    }
}

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;
}

task 5

#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

void test1() {
    Date d;
    int i; // C89标准：循环变量提前声明
    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; // C89标准：循环变量提前声明
    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();

    return 0;
}


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

int day_of_year(Date d) {

    int month_days[] = {31,28,31,30,31,30,31,31,30,31,30,31};
    int total = 0;
    int i;

    if((d.year % 4 == 0 && d.year % 100 != 0) || (d.year % 400 == 0)) {
        month_days[1] = 29;
    }
 
    for(i = 0; i < d.month - 1; i++) {
        total += month_days[i];
    } 
    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;  
    } else {
      
        if(d1.month < d2.month) {
            return -1;
        } else if(d1.month > d2.month) {
            return 1;
        } else {
          
            if(d1.day < d2.day) {
                return -1;
            } else if(d1.day > d2.day) {
                return 1;
            } else {
                return 0; 
            }
        }
    }
}

task 6

#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，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) {
    int i, j; 
    printf("用户名\t\t密码\t\t账户类型\n");
    printf("----------------------------------------\n");

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

task 7