实验六

任务四

点击查看代码
#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) {
    int i;
    for (i = 0; i < n; i++) {
        printf("%s%s%s%.2f%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) {
    double total = 0;
    int i;
    for (i = 0; i < n; i++) {
        total += x[i].sales_price * x[i].sales_count;
    }
    return total;
}

联想截图_20251220212443

任务五

点击查看代码
#include <stdio.h>
typedef struct {
    int year;
    int month;
    int day;
} Date;
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\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 days_in_month[] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    int total = 0, i;
    if ((d.year % 4 == 0 && d.year % 100 != 0) || (d.year % 400 == 0)) {
        days_in_month[1] = 29;
    }
    for (i = 0; i < d.month - 1; i++) {
        total += days_in_month[i];
    }
    total += d.day;

    return total;
}
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;
}

联想截图_20251220213951

任务六

点击查看代码
#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", "921kfmg917", student} };
    int n;
    n = sizeof(x) / sizeof(Account);
    output(x, n);
    return 0;
}
void output(Account x[], int n) {
        int i;
        for (i = 0; i < n; i++) {
            int len = strlen(x[i].password);

            printf("用户名: %s, ", x[i].username);

            printf("密码: ");
            int j;
            for (j = 0; j < len; j++) {
                printf("*");
            }

            printf(", 账户类型: ");
            switch (x[i].type) {
            case admin:
                printf("管理员");
                break;
            case student:
                printf("学生");
                break;
            case teacher:
                printf("教师");
                break;
            default:
                printf("未知");
            }
            printf("\n");
        }
    }

联想截图_20251225222448

任务七

点击查看代码
#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: ");
    if (scanf_s("%d", &vip_cnt, (unsigned)sizeof(vip_cnt)) != 1) {
        printf("输入错误!\n");
        return 1;
    }


    printf("输入%d个紧急联系人姓名:\n", vip_cnt);
    for (i = 0; i < vip_cnt; ++i) {
        if (scanf_s("%s", name, (unsigned)sizeof(name)) != 1) {
            printf("输入姓名时发生错误!\n");
            return 1;
        }
        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[]) {
    int i;
    int found = 0;

    for (i = 0; i < n; i++) {
        if (strcmp(x[i].name, name) == 0) {
            x[i].vip = 1;
            found = 1;
            break;
        }
    }
}


void display(Contact x[], int n) {
    int i, j;
    Contact temp;

    for (i = 0; i < n - 1; i++) {
        for (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;
            }
        }
    }

    int vip_count = 0;
    for (i = 0; i < n; i++) {
        if (x[i].vip == 1) {
            vip_count++;
        }
    }

    for (i = 0; i < vip_count - 1; i++) {
        for (j = 0; j < vip_count - 1 - i; j++) {
            if (strcmp(x[j].name, x[j + 1].name) > 0) {
                temp = x[j];
                x[j] = x[j + 1];
                x[j + 1] = temp;
            }
        }
    }

    for (i = vip_count; i < n - 1; i++) {
        for (j = vip_count; j < n - 1 - (i - vip_count); j++) {
            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");
    }
}

联想截图_20251225222310

posted @ 2025-12-25 22:25  Asuka02Langely  阅读(0)  评论(0)    收藏  举报