实验6

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) {
    printf("%-22s%-28s%-15s%-20s%-12s\n","isbn","书名","作者","售价","销量");
    
    for(int i=0;i<n;i++)
    printf("%-22s%-28s%-15s%-20.2f%-12d\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_count*x[i].sales_price;
    
    return sum;
}

运行结果截图

任务5

源代码

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

int day_of_year(Date d) {
    int month_days[12]={31,28,31,30,31,30,31,31,30,31,30,31};
    int sum=d.day;
    
    for(int i=0;i<d.month-1;i++)
        sum+=month_days[i];
        
    int is_leap=0;
    if((d.year%4==0&&d.year%100!=0)||(d.year%400==0))
        is_leap=1;
    if(is_leap&&d.month>2)
        sum+=1;
        
    return sum;            
}

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;                
}

运行结果截图

任务6

源代码

#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, j;
    int pwd_len;
    char role_str[10];

    for(i = 0; i < n; i++) {
        printf("%-10s", x[i].username);
        pwd_len = strlen(x[i].password);
        for(j = 0; j < pwd_len; j++) {
            printf("*");
        }
            printf("     ");
        switch(x[i].type) {
            case admin:
                strcpy(role_str, "admin");
                break;
            case student:
                strcpy(role_str, "student");
                break;
            case teacher:
                strcpy(role_str, "teacher");
                break;
            default:
                strcpy(role_str, "unknown");
        }
        printf("%s\n", role_str);
    }
}

运行结果截图

任务7

源代码

#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：");
    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 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");
    }
}

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

运行结果截图