实验6

实验任务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) {
    printf("%-20s %-30s %-20s %-10s %-10d\n", "ISBN号", "书名", "作者", "售价", "销售册数");
    for (int i = 0; i < n; i++) {
        printf("%-20s%-30s%-20s%-10.lf%-10s\n", x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count);

    }

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

    }
    return total;
}

运行结果

 

实验任务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: 
// 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_s("%d-%d-%d", &pd->year, &pd->month, &pd->day);
}
int day_of_year(Date d) {
    int mouth_days[] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
    int total = 0;
    int i;
    if ((d.year % 4 == 0 && d.year % 100 != 0) || (d.year % 400 == 0)) {
        month_days[2] = 29;
    }
    for (i = 1; i < d.month; i++) {
        total += month_days[i];
    }
    total += d.day;
    return total;
}
int compare_dates(Dates 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;
        }
    }
}

运行结果

 

实验任务6

#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);
int main() {
    Account x[] = { {"A1001", "123456", student},
                   {"A1002", "123abcdef", student},
                   {"A1009", "xyz12121", student},
                   {"X1009", "9213071x", admin},
                   {"C11553", "129dfg32k", teacher},
                   {"X3005", "921fmg917", 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("%s\t", 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;
        default:
            printf("unknown\n");
            break;
        }
    }
}

运行结果

实验任务7

#include <stdio.h>
#include <string.h>
// 修正结构体名拼写（原Contact首字母大写，保持一致）
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;
// 函数声明修正（原output拼写错误）
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() {
    // 修正中文逗号为英文逗号
    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_s("%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("%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;
}

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[N];
    memcpy(temp, x, sizeof(temp));
    
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < n - 1 - i; j++) {
           
            if (temp[j].vip < temp[j + 1].vip) {
                Contact t = temp[j];
                temp[j] = temp[j + 1];
                temp[j + 1] = t;
            }
            
            else if (temp[j].vip == temp[j + 1].vip && strcmp(temp[j].name, temp[j + 1].name) > 0) {
                Contact t = temp[j];
                temp[j] = temp[j + 1];
                temp[j + 1] = t;
            }
        }
    }
   
    output(temp, n);
}

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

运行结果

如何解决第二张图片的问题