实验6

实验任务4

源代码：

#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) {
    for (int i = 0; i < n; i++) {
        printf("ISBN: %-18s 书名: %-30s 作者: %-20s 售价: %-6.1f 销量: %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) {
    for (int i = 0; i < n - 1; i++) {
        int max_index = i;
        for (int j = i + 1; j < n; j++) {
            if (x[j].sales_count > x[max_index].sales_count) {
                max_index = j;
            }
        }
        if (max_index != i) {
            Book temp = x[i];
            x[i] = x[max_index];
            x[max_index] = temp;
        }
    }
}


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

源代码：

#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2025-06-01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2025-06-01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("%d-%d-%d", &pd->year, &pd->month, &pd->day);
}


int day_of_year(Date d) {
 int days_in_month[] = {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)) {
        days_in_month[2] = 29;
    }
    for (int i = 1; i < d.month; i++) {
        total_days += days_in_month[i];
    }
    total_days += d.day;
    return total_days;
}

int compare_dates(Date d1, Date d2) {
     if (d1.year < d2.year) return -1;
    if (d1.year > d2.year) return 1;
    if (d1.month < d2.month) return -1;
    if (d1.month > d2.month) return 1;
    if (d1.day < d2.day) return -1;
    if (d1.day > d2.day) return 1;
    return 0;
}

运行结果：

 

实验任务6

源代码：

#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，其中，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) {
    for (int i = 0; i < n; i++) {
        printf("%-8s  ", x[i].username);
        int len = strlen(x[i].password);
        for (int j = 0; j < len; j++) {
            printf("*"); 
        }
        int total_chars = 8 + 2 + len; 
        if (total_chars < 20) { 
            for (int j = total_chars; j < 20; j++) {
                printf(" ");
            }
        }
        const char *typeStr;
        switch (x[i].type) {
            case admin: typeStr = "admin"; break;
            case student: typeStr = "student"; break;
            case teacher: typeStr = "teacher"; break;
            default: typeStr = "unknown"; break;
        }
        printf("%s\n", typeStr);
    }
}

运行结果：

 

 

实验任务7

源代码：

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

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
void set_vip_contact(Contact x[], int n, char name[]);  // 设置紧急联系人
void output(Contact x[], int n);    // 输出x中联系人信息
void display(Contact x[], int n);   // 按联系人姓名字典序升序显示信息，紧急联系人最先显示


#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 j=0; j < n; ++j) {
        if (strcmp(name, x[j].name) == 0) {
            x[j].vip = 1;
        }
    }
}
void display(Contact x[], int n) {
   Contact temp;
    for (int i = 0; i < n-1; ++i) {
        for (int j = 0; j < n - i - 1; ++j) {
            if (strcmp(x[j].name,x[j+1].name)>0) {
                temp = x[j];
                x[j] = x[j + 1];
                x[j + 1] = temp;
            }
        }
    }
    // ×××
    for (int k = 0; k < n; ++k) {
        if (x[k].vip == 1) {
            printf("%-10s%-15s    *\n", x[k].name, x[k].phone);
        }
    }
    for (int m = 0; m < n; ++m) {
        if (x[m].vip == 0) {
            printf("%-10s%-15s\n", x[m].name, x[m].phone);
        }
    }
}

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");
    }
}

运行结果：