记录书籍的结构

短命女 2023-05-21 13:58 12阅读 0赞

设计一个记录书籍的结构，其成员包括书号、书名、作者、出版社和出版日期。假设一批书籍的记录存放在一个单向链表中，编写程序输出与给定作者姓名相匹配的所有书籍的记录。给定5种书籍信息如下:  
Number: 1 2 3 4 5  
Name: b1 b2 b3 b4 b5  
Author: a1 a2 a2 a3 a3  
Press: gfkd gfkd tsinghua tsinghua gfkd  
Publication Date: 20150127 20150125 20140122 20130523 20150602  
输入格式  
第一步：输入以上书籍信息。  
第二步：输入0，以结束信息录入  
第三步：输入待查找作者姓名  
输出格式  
Number:1  
Name:b1  
Author:a1  
Press:gfkd  
Publication Date:20150127  
效果如下：  
输入：  
1 b1 a1 gfkd 20150127  
2 b2 a2 gfkd 20150125  
3 b3 a2 tsinghua 20140122  
4 b4 a3 tsinghua 20130523  
5 b5 a3 gfkd 20150602  
0  
a2  
输出:  
Number:2  
Name:b2  
Author:a2  
Press:gfkd  
Publication Date:20150125  
Number:3  
Name:b3  
Author:a2  
Press:tsinghua  
Publication Date:20140122

#include <iostream>
    #include<cstring>
    using namespace std; 
    struct book
    {  
     int Number; 
     char Name[10];
     char Author[10];
     char Press[10]; 
     int PublicationDate;
    }book[100];
    int main()
    { 
     int i;
       for( i=0;i<=99;i++)
       { 
         cin>>book[i].Number;
         if(book[i].Number==0)
         { 
          break; 
      }
         else
         { 
          cin>>book[i].Name>>book[i].Author>>book[i].Press>>book[i].PublicationDate;
      }
       }
       char au[10];
     for( i=0;i<=99;i++)
     { 
      cin>>au;
         if (strcmp(au,book[i].Author)==0)
         { 
            cout<<"Number:"<<book[i].Number<<endl;
            cout<<"Name:"<<book[i].Name<<endl;
            cout<<"Author:"<<book[i].Author<<endl;
             cout<<"Press:"<<book[i].Press<<endl;
             cout<<"Publication Date:"<<book[i].PublicationDate<<endl;
         }
     }
      return 0;
    }