实验6

任务四

源代码

#include <stdio.h>
#define N 10
#include <stdlib.h>
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){
    int i=0;
    printf("%-18s %-25s %-15s %-6s %s\n","ISBN号","书号","作者","售价","销售册数");
    for(i;i<n;i++){
        printf("%-18s %-25s %-15s %-6.1f %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){
    double total=0.0;
    int i=0;
    for(i;i<n;i++){
        total+=x[i].sales_price*x[i].sales_count;
    }
    return total;
}

结果

 任务五

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("%d-%d-%d",&pd->year,&pd->month,&pd->day);
}
int day_of_year(Date d){
    int days_in_month[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
    int i,total=0;
    if((d.year%4==0 && d.year%100!=0)||(d.year%400==0)){
        days_in_month[2]=29;
    }
    for(i=1;i<d.month;i++){
        total+=days_in_month[i];
    }
    total+=d.day;
    return total;
}
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;
}

结果

 任务六

源代码

#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;
    n = sizeof(x) / sizeof(Account);
    output(x, n);
    system("pause");
    return 0;
}


void output(Account x[], int n) {
    int i,j;
    for ( i = 0; i < n; i++) {
        int pwd_len = strlen(x[i].password);
        printf("%-10s ", x[i].username);

       
        
        for ( j = 0; j < pwd_len; j++) {
            printf("*");
        }
        
        printf("%*s", 12 - pwd_len, "");

       
        switch (x[i].type) {
            case admin:
                printf("admin\n");
                break;
            case student:
                printf("student\n");
                break;
            case teacher:
                printf("teacher\n");
                break;
            default:
                printf("unknown\n");
        }
    }
}

结果

 

任务七

源代码

#include <stdio.h>
#include <string.h>
#include <stdlib.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
void swap(Contact *a, Contact *b) {
    Contact temp = *a;
    *a = *b;
    *b = temp;
}
void set_vip_contact(Contact x[], int n, char name[]) {
    int i ;
    for ( i = 0; i < n; i++) {
        if (strcmp(x[i].name, name) == 0) {
            x[i].vip = 1;
        }
    }
}
void display(Contact x[], int n) {
    Contact temp[N];
    int i ,j;
    for ( i = 0; i < n; i++) {
        temp[i] = x[i];
    }
    
    for (i = 0; i < n - 1; i++) {
        for ( j = 0; j < n - 1 - i; j++) {
            if (temp[j].vip < temp[j+1].vip ||
                (temp[j].vip == temp[j+1].vip && strcmp(temp[j].name, temp[j+1].name) > 0)) {
                swap(&temp[j], &temp[j+1]);
            }
        }
    }
    
    for (i = 0; i < n; i++) {
        printf("%-10s%-15s", temp[i].name, temp[i].phone);
        if (temp[i].vip) {
            printf("%5s", "*");
        }
        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");
    }
}

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;
}

 

结果