实验六

task4 code

#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] = { {"978-7-229-14156-1", "源泉", "安.兰德", 84, 59},
    {"978-7-5133-5261-1", "李白来到旧金山", "谭夏阳", 48, 16},
    {"978-7-5617-4347-8", "陌生人日记", "周怡芳", 72.6, 27},
    {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49},
    {"978-7-5046-9568-0", "数据化决策", "道格拉斯·W·哈伯德", 49,42},
    {"978-7-5133-4388-6", "美好时代的背后", "凯瑟琳.布", 34.5, 39},
    {"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},
    {"978-7-5321-5691-7", "何为良好生活", "陈嘉映", 29.5 , 31},
    {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118,42},
    {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5,44} };

    printf("图书销量排名: \n");
    sort(x, N);
    output(x, N);
    printf("\n图书销售总额: %.2f\n", sales_amount(x, N));

    return 0;
}

void sort(Book x[], int n) {
    int i, j;
    Book temp;
    Book* ptr = x;

    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;

            }
        }
}

void output(Book x[], int n) {
    Book* ptr;
    int i;

    printf("%-25s%-25s%-25s%-25s%-25s\n", "ISBN号", "书名", "作者", "售价", "销售册数");
    for (ptr = x; ptr < x + n; ptr++)
        printf("%-25s%-25s%-25s%-25.1lf%-25d\n", ptr->isbn, ptr->name, ptr->author, ptr->sales_price,
            ptr->sales_count);
}

double sales_amount(Book x[], int n) {
    double total = 0;
    int i;

    for (i = 0; i < n; i++)
        total += (x[i].sales_price * x[i].sales_count);

}

task4 result

task5 code

#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2023-12-11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2023-12-11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) {
    int t1, t2;
    scanf("%d", &pd->year);
    scanf("%d", &t1);
    pd->month = -t1;
    scanf("%d", &t2);
    pd->day = -t2;
}
// 补足函数day_of_year实现 31 
// 功能：返回日期d是这一年的第多少天
int day_of_year(Date d) {
    int leap = 0;
    int sum = 0;

    if (d.year % 4 == 0 && d.year % 100 != 0 || d.year % 400 == 0)
        leap = 1;

    if (leap) {
        switch (d.month) {
        case 1:sum = 0; break;
        case 2:sum = 31; break;
        case 3:sum = 60; break;
        case 4:sum = 91; break;
        case 5:sum = 121; break;
        case 6:sum = 152; break;
        case 7:sum = 182; break;
        case 8:sum = 213; break;
        case 9:sum = 244; break;
        case 10:sum = 274; break;
        case 11:sum = 305; break;
        case 12:sum = 334; break;
        }
        sum += d.day;
    }
    else {
        switch (d.month) {
        case 1:sum = 0; break;
        case 2:sum = 31; break;
        case 3:sum = 59; break;
        case 4:sum = 90; break;
        case 5:sum = 120; break;
        case 6:sum = 151; break;
        case 7:sum = 181; break;
        case 8:sum = 212; break;
        case 9:sum = 243; break;
        case 10:sum = 273; break;
        case 11:sum = 304; break;
        case 12:sum = 334; break;
        }
        sum += d.day;
    }

    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;
    }

}

task5 result

task6 code

 

#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); // 输出账户数组x中n个账户信息，其中，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;
}
// 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) {
    int i, j, k;
    char t[20] = {'\0'};

    for (i = 0; i < n; i++) {
        k = strlen(x[i].password);
        for (j = 0; j < k; j++)
            t[j] = '*';
        t[k] = '\0';

        printf("%-9s %-9s ", x[i].username, t);
        switch (x[i].type) {
        case 0: printf("    admin"); break;
        case 1: printf("    student"); break;
        case 2: printf("    teacher"); break;
        }
        printf("\n");

        for (j = 0; j < 20; j++)
            t[j] = '\0';
    }
}

task6 result