实验六

实验任务4：

#include <stdio.h>
#include <stdlib.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));
    system("pause");
    return 0;
}

void output(Book x[], int n)
{
    int i;
    for(i = 0; i < n; i++)
    {
        printf("%s %s %s %.2lf %d\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;
    Book temp;
    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)
            {
                temp = x[j];
                x[j] = x[j+1];
                x[j+1] = temp;
            }
        }
    }
}

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

 

实验任务五

#include <stdio.h>

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

// 函数声明
void input(Date *pd);
int day_of_year(Date d);
int compare_dates(Date d1, Date d2);

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", 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[12] = {31,28,31,30,31,30,31,31,30,31,30,31};
    int sum = d.day;
    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++)
        sum += month_days[i];
    return sum;
}

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;
        }
    }
}

 

 

实验任务六：

#include <stdio.h>
#include <string.h>
#include <stdlib.h> 
enum Role {admin, student, teacher};

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

void output(Account x[], int n);

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);
    system("pause"); 
    return 0;
}

void output(Account x[], int n)
{
    int i, j;
    int len;
    char role[16];
    for(i = 0; i < n; i++)
    {
        if(x[i].type == admin)
            strcpy(role, "管理员");
        else if(x[i].type == student)
            strcpy(role, "学生");
        else
            strcpy(role, "教师");

        len = strlen(x[i].password);
        printf("用户名：%s\t密码：", x[i].username);
        for(j = 0; j < len; j++)
        {
            putchar('*');
        }
        printf("\t账户类型：%s\n", role);
    }
}

 

实验任务七：

 

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

typedef struct {
    char name[20];
    char phone[12];
    int vip;
} Contact;

// 函数声明
void set_vip_contact(Contact x[], int n, char name[]);
void output(Contact x[], int n);
void display(Contact x[], int n);

#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:(按姓名字典序排列,紧急联系人最先显示)\n");
    display(list, N);

    return 0;
}

// 补足函数set_vip_contact实现
// 功能: 将联系人数组x中，联系人姓名与name一样的人，设置为紧急联系人（即成员vip值设为1）
void set_vip_contact(Contact x[], int n, char name[]) {
    int i;
    for(i = 0; i < n; i++)
    {
        if(strcmp(x[i].name, name) == 0)
        {
            x[i].vip = 1;
        }
    }
}

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");
    }
}

// 补足函数display实现
// 功能: 显示联系人数组x中的联系人信息
//      按姓名字典升序显示，紧急联系人显示在最前面
void display(Contact x[], int n) {
    Contact temp[N];
    Contact t;
    int i, j, k = 0;
    // 先复制数组，不修改原数据
    memcpy(temp, x, sizeof(temp));
    // 冒泡排序：优先vip=1在前，同vip按姓名升序
    for(i = 0; i < n - 1; i++)
    {
        for(j = 0; j < n - 1 - i; j++)
        {
            // 前面非vip、后面vip → 交换
            if(temp[j].vip == 0 && temp[j+1].vip == 1)
            {
                t = temp[j];
                temp[j] = temp[j+1];
                temp[j+1] = t;
            }
            // 同为vip或同为普通，按姓名字典序升序
            else if(temp[j].vip == temp[j+1].vip)
            {
                if(strcmp(temp[j].name, temp[j+1].name) > 0)
                {
                    t = temp[j];
                    temp[j] = temp[j+1];
                    temp[j+1] = t;
                }
            }
        }
    }
    // 遍历输出排序后结果
    for(i = 0; i < n; i++)
    {
        printf("%-10s%-15s", temp[i].name, temp[i].phone);
        if(temp[i].vip)
            printf("%5s", "*");
        printf("\n");
    }
}