题目链接：http://www.patest.cn/contests/pat-a-practise/1022
题目：
1022. Digital Library (30)
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:3 1111111 The Testing Book Yue Chen test code debug sort keywords ZUCS Print 2011 3333333 Another Testing Book Yue Chen test code sort keywords ZUCS Print2 2012 2222222 The Testing Book CYLL keywords debug book ZUCS Print2 2011 6 1: The Testing Book 2: Yue Chen 3: keywords 4: ZUCS Print 5: 2011 3: blablablaSample Output:
1: The Testing Book 1111111 2222222 2: Yue Chen 1111111 3333333 3: keywords 1111111 2222222 3333333 4: ZUCS Print 1111111 5: 2011 1111111 2222222 3: blablabla Not 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<stdio.h>
#include<vector>
#include<string.h>
#include<algorithm>
using namespace std;
struct Book{
char ID[8];
char title[81];
char author[81];
vector<string>V;//结构体中有一个vector存放keyword的字符串
char publisher[81];
char pub_year[5];
};
vector<Book>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() != '
');//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
", 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<buf[j].V.size(); k++){
if (strcmp(q, buf[j].V[k].c_str()) == 0){
find = true;
puts(buf[j].ID);
break;
}
}
}
}
if (!find){
puts("Not Found");
}
}
}
return 0;
}截图：
——Apie陈小旭