实验6

实验任务五

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));
    
    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 %-10g %-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)
{
    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;
    for(i = 0;i < n;i++)
    total = total + x[i].sales_price*x[i].sales_count;
    
    return total;
}

代码运行结果：

 实验任务六：

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 i;
    int total = 0;
    int months[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

    if((d.year%4==0&&d.year&100!=0)||(d.year%400==0))
    months[2] = 29;
    
    for(i = 1;i < d.month;i++)
    total = total + months[i];
    total = total + d.day;
    
    return total; 
}

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;
        }
    }
}

代码运行结果：

 实验任务六

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

代码运行结果：

 实验任务七

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：(按名字字典序升序排列，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(x[i].name == name)
        {
        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)
{
    int i;
    int j;
    for(i = 0;i < n-1;i++)
    for(j = 0;j <n-i-1;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))
        {
            Contact t;
            t = x[j];
            x[j] = x[j+1];
            x[j+1] = t;
        }
    }
    output(x,n);
}

代码运行结果：