实验六

任务4：

源代码：

#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 s[], int n);
double sales_amount(Book s[], 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;
}

// 待补足：函数output()实现
// ×××
void output(Book x[],int n)
{
    printf("%-30s","ISBN"),printf("%-30s","书名"),printf("%-20s","作者"),printf("%-20s","售价"),printf("%-20s","销售册数");
    printf("\n");
    int i;
    for(i=0;i<n;i++)
     {
       printf("%-30s%-30s%-20s%-20d%-20d",x[i].isbn,x[i].name,x[i].author,x[i].sales_price,x[i].sales_count);
       printf("\n");    
}
}

// 待补足：函数sort()实现
// ×××
void sort(Book s[],int n)
{
    Book temp;
    int i,j;
    for(i=0;i<n;i++)
      for(j=0;j<n-i-1;j++)
          if(s[i].sales_count>s[i+1].sales_count)
          {
        temp=s[i];
          s[i]=s[i+1];
          s[i+1]=temp;
               }
}

// 待补足：函数sales_count()实现
// ×××
double sales_amount(Book s[],int n)
{
    int i;
    double much=0;
    for(i=0;i<n;i++)
      much+=s[i].sales_price*s[i].sales_count;
    return much;
}

 

 

运行结果：

 

 

任务5：

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 compre_dates(Date d1,Date d2);
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是这一年的第%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的出生日期:(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("测试一:输入日期,打印输出这是一年中的第多少天\n");
    test1();
    printf("\n测试二:两个人年龄大小\n");
    test2();
}
void input(Date*pd)
{
    scanf("%d-%d-%d",&(pd->year),&(pd->month),&(pd->day));
}
int day_of_year(Date d)
{
    int s;
    if(d.month==1)
      return d.day;
    else if(d.month==2)
      return 31+d.day;
    else if(d.month==3)
      s=31+28+d.day;
    else if(d.month==4)
      s=31*2+28+d.day;
    else if(d.month==5)
      s=31*2+30+28+d.day;
    else if(d.month==6)
      s=31*3+30+28+d.day;
    else if(d.month==7)  
      s=31*3+30*2+28+d.day;
    else if(d.month==8)
      s=31*4+30*2+28+d.day;
    else if(d.month==9)    
      s=31*5+30*2+28+d.day;
    else if(d.month==10)
      s=31*5+30*3+28+d.day; 
    else if(d.month==11)  
      s=31*6+30*3+28+d.day;
    else
     s=31*6+30*4+28+d.day;
    if(d.year==d.year/100*100&&d.year%400==0)
     return s+1;
    else if(d.year%4==0&&d.year%100!=0)
     return s+1;
    else
     return s;       
}
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;
        else if(day_of_year(d1)>day_of_year(d2))
          return 1;
        else 
          return 0;  
    }
}

 

 

运行结果：

 

 

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 i,j,len,m;
    char s[3][10]={"admin","student","teacher"};
    for(i=0;i<n;i++)
     {
         len=strlen(x[i].password);
         printf("%-20s",x[i].username);
         for(j=0;j<len;j++)
         printf("*");
         for(j=0;j<20-len;j++)
         printf(" ");
         printf("%s",s[x[i].type]);
         printf("\n");
     }
}

 

 

运行结果：

 

 

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:(按姓名字典升序排列，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)
{
    Contact temp;
    int m;
    int i,j=0;
    for(i=0;i<n;i++)
     {
     if(x[i].vip==1)
     {
     temp=x[i];
     x[i]=x[j];
     x[j]=temp;
     j++;
}
  m=j;
     }
    for(i=0;i<n-1-m;i++)
      for(j=m;j<n-1-i;j++)
    {
        if(strcmp(x[j].name,x[j+1].name)>0)
    {
        temp=x[j];
        x[j]=x[j+1];
        x[j+1]=temp;
}
    }
    output(x,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：