实验6 C语言结构体和枚举应用编程

实验任务4：

源代码：

#include <stdio.h>
#include <stdlib.h>
#define N 10

typedef struct {
    char isbn[20];    
    char name[80];    
    char author[80];  
    double sales_price;
    int sales_count;  
} 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));

    system("pause");
    return 0;
}
void output(Book x[], int n) {
    for (int i = 0; i < n; i++) {
        printf("%-20s %-30s %-20s %-8.2f %-6d\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 sum = 0.0;
    for (int i = 0; i < n; i++) {
        sum += x[i].sales_price * x[i].sales_count;
    }
    return sum;
}

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实验任务5：

源代码：

#include <stdio.h>
#include <stdlib.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();
    system("pause");
    return 0;
}
void input(Date* pd) {
    scanf_s("%d-%d-%d", &pd->year, &pd->month, &pd->day);
}
int isLeap(int y) {
    if ((y % 4 == 0 && y % 100 != 0) || (y % 400 == 0))
        return 1;
    return 0;
}
int day_of_year(Date d) {
    int days[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
    int sum = 0;
    if (isLeap(d.year))
        days[2] = 29;

    for (int i = 1; i < d.month; i++) {
        sum += days[i];
    }
    sum += d.day;
    return sum;
}
int compare_dates(Date d1, Date d2) {
    if (d1.year != d2.year) {
        return d1.year < d2.year ? -1 : 1;
    }
    if (d1.month != d2.month) {
        return d1.month < d2.month ? -1 : 1;
    }
    if (d1.day != d2.day) {
        return d1.day < d2.day ? -1 : 1;
    }
    return 0;
}

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实验任务6：

源代码：

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <string.h>
#include <stdlib.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 = sizeof(x) / sizeof(Account);
    output(x, n);
    system("pause");
    return 0;
}
void output(Account x[], int n) {
    for (int i = 0; i < n; i++) {
        printf("%-10s ", x[i].username);
        int len = strlen(x[i].password);
        for (int j = 0; j < len; j++) {
            printf("*");
        }
        for (int j = len; j < 12; j++) {
            printf(" ");
        }
        printf(" "); 
        switch (x[i].type) {
        case admin:
            printf("%-8s\n", "admin");
            break;
        case student:
            printf("%-8s\n", "student");
            break;
        case teacher:
            printf("%-8s\n", "teacher");
            break;
        }
    }
}

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实验任务7：

源代码：

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef struct {
    char name[20];   
    char phone[12];  
    int vip;         
} Contact;
#define N 10
void set_vip_contact(Contact x[], int n, char name[]);
void output(Contact x[], int n);
void display(Contact x[], int n);
// 辅助排序：按姓名字典序
void sortByName(Contact arr[], int len);
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);
    system("pause");
    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;
        }
    }
}
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("   *");
        printf("\n");
    }
}
void sortByName(Contact arr[], int len) {
    int i, j;
    Contact temp;
    for (i = 0; i < len - 1; i++) {
        for (j = 0; j < len - 1 - i; j++) {
            if (strcmp(arr[j].name, arr[j + 1].name) > 0) {
                temp = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = temp;
            }
        }
    }
}
void display(Contact x[], int n) {
    Contact temp[N];
    int idx = 0;
    for (int i = 0; i < n; i++) {
        if (x[i].vip == 1) {
            temp[idx++] = x[i];
        }
    }
    for (int i = 0; i < n; i++) {
        if (x[i].vip == 0) {
            temp[idx++] = x[i];
        }
    }
    sortByName(temp, n);
    for (int i = 0; i < n; i++) {
        printf("%-10s%-15s", temp[i].name, temp[i].phone);
        if (temp[i].vip)
            printf("   *");
        printf("\n");
    }
}

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