实验6

任务四

#include <stdio.h>
#define N 10

typedef struct {
    char isbn[20];
    char name[100];
    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", "门罗之光", "罗伯特·韦", 42, 51},
                 {"978‑7‑308‑17947‑5", "自由与爱之地：入以色列记", "云也退", 49, 301},
                 {"978‑7‑5404‑9346‑8", "伦敦人", "克莱格泰勒", 68, 27},
                 {"978‑7‑5642‑5246‑8", "软件体的生命周期", "特德姜", 35, 90},
                 {"978‑7‑5722‑5476‑7", "花方简史", "汪玟", 74.9, 49},
                 {"978‑7‑5133‑5750‑0", "注视战争", "苏莱曼·阿勒蒙", 128, 42},
                 {"978‑7‑1081‑4661‑1", "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", 116, 42},
                 {"978‑7‑5133‑0509‑5", "无穷的开始:世界进步的本源", "戴维·多伊奇", 37.5, 55},
                 {"978‑7‑229‑04156‑1", "源泉", "安·兰德", 84, 59}};

    printf("图书销量排名(按销量降序)：\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 %.1f %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) {
    int i;
    double sum = 0;
    for (i = 0; i < n; i++) {
        sum += x[i].sales_price * x[i].sales_count;
    }
    return sum;
}

 任务五

#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 s;
    printf("输入日期(以形如2025‑12‑19这样的形式输入)\n");
    input(&d);
    s = day_of_year(d);
    printf("该日期是这一年中第%d天\n", s);
}

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];
    }
    return sum;
}

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

 任务六

#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;
    for(i = 0; i < n; i++)
    {
        printf("%s ", x[i].username)
        for(j = 0; j < strlen(x[i].password); j++)
        {
            printf("*");
        }
        printf(" ");
        if(x[i].type == admin)
            printf("admin\n");
        else if(x[i].type == student)
            printf("student\n");
        else
            printf("teacher\n");
    }
}

 任务七

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

void display(Contact x[], int n) {
    Contact temp[N];
    int k = 0;
    for(int i = 0; i < n; i++)
    {
        if(x[i].vip == 1)
            temp[k++] = x[i];
    }
    for(int i = 0; i < n; i++)
    {
        if(x[i].vip == 0)
            temp[k++] = x[i];
    }
    for(int i = 0; i < n‑1; i++)
    {
        for(int j = 0; j < n‑1‑i; j++)
        {
            if(strcmp(temp[j].name, temp[j+1].name) > 0)
            {
                Contact t = temp[j];
                temp[j] = temp[j+1];
                temp[j+1] = t;
            }
        }
    }
    output(temp, 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");
    }
}