实验六

task4

#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 sort(Book x[],int n){
int i,j;
Book t;
    for(i=0;i<n-1;i++)
    for(j=0;j<n-i-1;j++){
        if(x[j].sales_count<x[j+1].sales_count){
        
         t=x[j];
        x[j]=x[j+1];
        x[j+1]=t;
    }
        
    }
    
    }
    
void output(Book x[],int n){
    int i;
    
    printf("%-20s%-30s%-20s%-10s%s\n","ISBN号","书名","作者","售价","销售册数");
    for(i=0;i<n-1;i++)
    printf("%-20s%-30s%-20s%-10.2f%d\n",x[i].isbn,x[i].name,x[i].author,x[i].sales_price,x[i].sales_count);
}


double sales_amount(Book x[],int n){
    int i,j;
    
    double t=0;
    for(i=0;i<n;i++)
{
t+=x[i].sales_price*x[i].sales_count;
    
    
}
    
    return t;
    
}

 task5

#include <stdio.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: 
                                        // 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2024-12-16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2024-12-16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 n,i,t=0;
    if((d.year%4==0&&d.year%100!=0)||d.year%400==0)
   n=29;
   else
  n=28;
   int x[12]={31,n,31,30,31,30,31,31,30,31,30,31};
   for(i=0;i<d.month-1;i++){
       t+=x[i];
       
   }
   t=t+d.day;
   
  return t;
}


int compare_dates(Date d1, Date d2) {

    
    if(d1.year<d2.year)
    return -1;
   else if(d1.year>d2.year)
    return 1;
    else {
        if(day_of_year(d1)>day_of_year(d2))
        return 1;
        if(day_of_year(d1)<day_of_year(d2))
          return -1;
    else
        {
        return 0;}
        }
        
    }
    

 task6

#include <stdio.h>
#include <string.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);

    return 0;
}


void output(Account x[], int n) {
    int i,j,t;
    char a[10]="admin",s[10]="student",tea[10]="teacher";
    for(i=0;i<n;i++){
    
        printf("%-20s",x[i].username);
        t=strlen(x[i].password);
        for(j=0;j<t;j++){
            printf("*");
        }
        
        t=0;
        j=0;
        printf("\t");
        switch(x[i].type){
            case admin:printf("%20s",a);break;
            case student:printf("%20s",s);break;
            case teacher:printf("%20s",tea);break;
        
        }
        
    printf("\n");
    }

}

 task7

#include <stdio.h>
#include <string.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);

    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,k;
    Contact t;
    for(i=0,j=0;i<n;i++){
        if(x[i].vip==1){
            t=x[j];
            x[j]=x[i];
            x[i]=t;
            j++;
        }    
    }
    for(k=0;k<j-1;k++){
        for(i=0;i<j-k-1;i++)
            if(strcmp(x[i].name,x[i+1].name)>0){
                t=x[i];
                x[i]=x[i+1];
                x[i+1]=t;
            }        
    }
    for(k=0;k<n-j;k++){
        for(i=j;i<n-k-1;i++)
            if(strcmp(x[i].name,x[i+1].name)>0){
                t=x[i];
                x[i]=x[i+1];
                x[i+1]=t;
            }
    }
    for(i=0;i<n;i++){
        if(x[i].vip==1)
            printf("%-10s%-20s*\n",x[i].name,x[i].phone);
        else
            printf("%-10s%-20s\n",x[i].name,x[i].phone);
    }
   
}

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");
    }
}

 

posted @ 2024-12-22 17:11  夏庆兴  阅读(7)  评论(0)    收藏  举报