实验六 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];          // 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));
    system("pause");
    return 0;
}


 
 void output(Book x[], int n){
     printf("%-30s%-30s%-20s%-10s%-10s\n","ISBN号","书名","作者","售价","销售册数");
     Book *ptr;
     for(ptr=x;ptr<x+n;++ptr){
             printf("%-30s%-30s%-20s%-10.2lf%-10d\n",ptr->isbn,ptr->name,ptr->author,ptr->sales_price,ptr->sales_count);
     }
 }
 
 double sales_amount(Book x[], int n){
     double s=0;
     Book *ptr;
     for(ptr=x;ptr<x+n;++ptr){
         s+=(ptr->sales_price)*(ptr->sales_count);
     }
    return s;
 } 

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

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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;

// 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: 
int runyear(int year);
                                        // 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();
    system("pause");
    return 0;
}

void input(Date *pd) {
    scanf("%d-%d-%d",&pd->year,&pd->month,&pd->day);
}

int day_of_year(Date d) {
    int month_day[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    if (runyear(d.year)) 
        month_day[1] = 29;
    
    int sum = 0;
    for (int i = 0; i < d.month - 1; i++) {
        sum += month_day[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;
}

int runyear(int year){
    return((year%4==0&&year%100!=0)||(year%400==0));
}

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实验任务6

源代码

#include <stdio.h>
#include <string.h>
#include<stdlib.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);
    system("pause");
    return 0;
}

void output(Account x[], int n) {
     for (int i = 0; i < n; i++) {
        printf("%-7s\t", x[i].username);
        int len = strlen(x[i].password);
        for (int j = 0; j < len; j++) 
            printf("*");
        printf("\t");
        switch (x[i].type) 
        {
            case admin: printf("admin"); break;
            case student: printf("student"); break;
            case teacher: printf("teacher"); break;
        }
        printf("\n");
    }
}

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实验任务7

源代码

#include <stdio.h>
#include <string.h>
#include<stdlib.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);
    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;
            break; 
        }
    }
}

void display(Contact x[], int n) {
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < n - 1 - i; j++) {
            if (x[j].vip < x[j + 1].vip) {
                Contact t = x[j];
                x[j] = x[j + 1];
                x[j + 1] = t;
            }
            else if (x[j].vip == x[j + 1].vip) {
                if (strcmp(x[j].name, x[j + 1].name) > 0) {
                    Contact t = x[j];
                    x[j] = x[j + 1];
                    x[j + 1] = t;
                }
            }
        }
    }

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

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