1022. Digital Library (30)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A Digital Library contains millions of books, stored according to their titles, authors, key words of their abstracts, publishers, and published years. Each book is assigned an unique 7-digit number as its ID. Given any query from a reader, you are supposed to output the resulting books, sorted in increasing order of their ID's.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the total number of books. Then N blocks follow, each contains the information of a book in 6 lines:

Line #1: the 7-digit ID number;

Line #2: the book title -- a string of no more than 80 characters;

Line #3: the author -- a string of no more than 80 characters;

Line #4: the key words -- each word is a string of no more than 10 characters without any white space, and the keywords are separated by exactly one space;

Line #5: the publisher -- a string of no more than 80 characters;

Line #6: the published year -- a 4-digit number which is in the range [1000, 3000].

It is assumed that each book belongs to one author only, and contains no more than 5 key words; there are no more than 1000 distinct key words in total; and there are no more than 1000 distinct publishers.

After the book information, there is a line containing a positive integer M (<=1000) which is the number of user's search queries. Then M lines follow, each in one of the formats shown below:

1: a book title

2: name of an author

3: a key word

4: name of a publisher

5: a 4-digit number representing the year

Output Specification:

For each query, first print the original query in a line, then output the resulting book ID's in increasing order, each occupying a line. If no book is found, print "Not Found" instead.

Sample Input:

31111111The Testing BookYue Chentest code debug sort keywordsZUCS Print20113333333Another Testing BookYue Chentest code sort keywordsZUCS Print220122222222The Testing BookCYLLkeywords debug bookZUCS Print2201161: The Testing Book2: Yue Chen3: keywords4: ZUCS Print5: 20113: blablabla

Sample Output:

1: The Testing Book111111122222222: Yue Chen111111133333333: keywords1111111222222233333334: ZUCS Print11111115: 2011111111122222223: blablablaNot Found

分析：

输入中包括书本的各种信息：标题、出版社、作者、keyword、出版年份等等。构造一个书本的结构体接收输入存储。然后再依据查询条件比較。当中。接收输入和keyword的存储是考点，前者考察get。getchar，scanf的各种使用方法。后者用结构体中包括vector来解决。难度适中。

AC代码：

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;struct Book{ char ID[8]; char title[81]; char author[81]; vector
         
          V;//结构体中有一个vector存放keyword的字符串 char publisher[81]; char pub_year[5];};vector
          
           buf;bool cmp(Book b1, Book b2){ return strcmp(b1.ID, b2.ID) < 0;}int main(void){ //freopen("F://Temp/input.txt", "r", stdin); int n, m; while (scanf("%d", &n) != EOF){ getchar();//取回车符 for (int i = 0; i < n; i++){ Book b1; gets(b1.ID); gets(b1.title); gets(b1.author); int j = 0; do{ char tmp[10]; scanf("%s", tmp);//读入keyword并放入vector中 b1.V.push_back(tmp); } while (getchar() != '\n');//当读入的接下来的字符都是空格而不是回车的时候 gets(b1.publisher); gets(b1.pub_year); buf.push_back(b1); } sort(buf.begin(), buf.end(),cmp);//按名称排序 scanf("%d", &m);//查询次数 for (int i = 0; i < m; i++){ int order; bool find = false; scanf("%d: ", &order); char q[82]; gets(q);//查询字符串 printf("%d: %s\n", order, q); for (int j = 0; j < n; j++){//依次对作者、出版社、出版年份、标题和keyword进行比較 if (strcmp(q, buf[j].author) == 0){ find = true; puts(buf[j].ID); } else if (strcmp(q, buf[j].publisher) == 0){ find = true; puts(buf[j].ID); } else if (strcmp(q, buf[j].pub_year) == 0){ find = true; puts(buf[j].ID); } else if (strcmp(q, buf[j].title) == 0){ find = true; puts(buf[j].ID); } else{ for (int k = 0; k

截图：

——Apie陈小旭

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