实验6

任务4

#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) {
    int i;
    for (i = 0; i < n; ++i) {
        printf("%-20s %-30s %-20s %-10f %-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 s = 0.0;
    int i;
    for (i = 0; i < n; i++) {
        s += x[i].sales_price * x[i].sales_count;
    }
    return s;
}

 

任务5

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>

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

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

// 测试1：计算日期是一年中的第几天
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));
}
}

// 测试2：比较两个人的出生日期大小
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();

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[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int total_days = 0;
if ((d.year % 4 == 0 && d.year % 100 != 0) || (d.year % 400 == 0)) {
month_days[2] = 29;
}
for (int i = 1; i < d.month; i++) {
total_days += month_days[i];
}
total_days += d.day;

return total_days;
}
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;
}
}
}
}

 

任务6

#define  _CRT_SECURE_NO_WARNINGS
#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;

void output(Account x[], int n);  // 输出账户信息，密码用*替代

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("%-8s", 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\n");
            break;
        case student:
            printf("student\n");
            break;
        case teacher:
            printf("teacher\n");
            break;
        }
    }
}

 

任务7

 

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

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

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;
}

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) {
    Contact temp;
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < n - 1 - i; 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");
    }
}