实验6

实验任务4

源代码

#include <stdio.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){
     int i;
     printf("ISBN号              书名                          作者                售价          销售册数\n");
     for(i=0;i<n;i++){
         printf("%-20s%-30s%-20s%-6g\t%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){
    Book t={0};
    int i,j;
    for(i=0;i<n;i++){
        for(j=0;j<n-i;j++){
            if(x[j].sales_count<x[j+1].sales_count){
            t=x[j];
            x[j]=x[j+1];
            x[j+1]=t;
            }
        }
    }
}
double sales_amount(Book x[], int n){
    int i;
    double sum=0,s[N]={0};
    for(i=0;i<n;i++){
        s[i]=x[i].sales_count*x[i].sales_price;
        sum+=s[i];
    }
    return sum;
}

运行结果截图

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");
}
void input(Date *pd) {
    scanf("%d-%d-%d",&pd->year,&pd->month,&pd->day);
}
int day_of_year(Date d) {
    int a=0;
    switch(d.month){
    case 12:a+=30;
    case 11:a+=31;
    case 10:a+=30;
    case 9:a+=31;
    case 8:a+=31;
    case 7:a+=30;
    case 6:a+=31;
    case 5:a+=30;
    case 4:a+=31;
    case 3:if((d.year%4==0&&d.year%100!=0)||(d.year%400==0)){
              a+=29;
           }else
               a+=28;
    case 2:a+=31;
    case 1:a+=0;break;}
    a+=d.day;
    return a;
}
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;
}

运行结果截图

实验任务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);
    system("pause");
    return 0;
}
void output(Account x[], int n) {
    int len,i,j;
    for(i=0;i<n;i++){
        len=strlen(x[i].password);
        for(j=0;j<len;j++){
            x[i].password[j]='*';
        }
    }
    for(i=0;i<n;i++){
        printf("%s\t\t%-20s",x[i].username,x[i].password);
        switch(x[i].type){
        case(student):printf("student\n");break;
        case(admin):printf("admin\n");break;
        case(teacher):printf("teacher\n");break;
        
        }
    }
}

运行结果截图

实验任务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[]) {
    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) {
     int i,j;
     Contact t,temp[N+1]={0};
     for(i=0;i<n-1;i++){
         for(j=0;j<n-i-1;j++){
             if(strcmp(x[j].name,x[j+1].name)>0){
             t=x[j+1];
             x[j+1]=x[j];
             x[j]=t;
             }
         }
     }
     j=0;
     for(i=0;i<n;i++){
         if(x[i].vip==1){
           temp[j++]=x[i];
         }
     }
     for(i=0;i<n;i++){
         if(x[i].vip==0){
           temp[j++]=x[i];
         }
     }
     for(i=0;i<n;i++){
           x[i]=temp[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 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");
    }
}

运行结果截图