PAT甲级1022

本文针对PAT甲级1022题“数字图书馆”进行详细解析,介绍了如何根据读者查询条件检索并排序图书ID的方法。该题涉及图书信息存储、字符串匹配及排序算法等内容。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

PAT甲级1022

原题:
1022 Digital Library (30分)
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 (≤10
​4
​​ ) 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: blablabla
Sample 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

题目大意:给出一些图书的信息,然后通过某个标签去查找图书的编号

注意点:
1、除了第4个关键词以外,其余信息全是完全匹配,关键词这个里面只要包括带查找的字符串就OK了,不需要完全一样
2、不要图方便省略if-else大括号(这个楼主刚试完毒)
3、查询的时候边输入边处理
4、凡是用到getline一定要记得清缓存(在scanf/cin之后用cin.ignore()或者fflush(这个以前用过现在用会报错,未解决)或者getchar())

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;

struct book{	//记录书籍的信息
	vector<string> info;	//0:ID,1:title,2:author,3:keywords,4:publisher,5:publishyear
};
struct query{	//记录查询信息
	int index;	//查询编号
	string content;	//查询内容
};
bool cmp(book a,book b){
	return a.info[0]<b.info[0];
}
int main(){
	int n,m;
	scanf("%d",&n);
	cin.ignore();
	book b[n];
	for(int i=0;i<n;i++)
		for(int j=0;j<6;j++){
			string tmp;
			getline(cin,tmp);
			b[i].info.push_back(tmp);
		}
	sort(b,b+n,cmp);
	scanf("%d",&m);
	cin.ignore();
	query q[m];
	for(int i=0;i<m;i++){
		scanf("%d:",&q[i].index);
		cin.ignore();
		getline(cin,q[i].content);
		printf("%d: ",q[i].index);
		cout<<q[i].content<<endl;
		bool flag=false;	//记录是否有找到
		for(int j=0;j<n;j++){
			if(q[i].index!=3){	//若查找的不是关键词
				if(b[j].info[q[i].index]==q[i].content) cout<<b[j].info[0]<<endl,flag=true;
			}
			else{
				if(b[j].info[3].find(q[i].content)!=string::npos) cout<<b[j].info[0]<<endl,flag=true;
			}
		}
		if(flag==false) printf("Not Found\n");
	}
	system("pause");
	return 0;
}
### 关于 PAT 甲级 1024 题目 PAT (Programming Ability Test) 是一项编程能力测试,其中甲级考试面向有一定编程基础的学生。对于 PAT 甲级 1024 题目,虽然具体题目描述未直接给出,但从相似类型的题目分析来看,这类题目通常涉及较为复杂的算法设计。 #### 数据结构的选择与实现 针对此类问题,常用的数据结构包括但不限于二叉树节点定义: ```cpp struct Node { int val; Node* lchild, *rchild; }; ``` 此数据结构用于表示二叉树中的节点[^1]。通过这种方式构建的二叉树能够支持多种遍历操作,如前序、中序和后序遍历等。 #### 算法思路 当处理涉及到图论的问题时,深度优先搜索(DFS)是一种常见的解题策略。特别是当需要寻找最优路径或访问尽可能多的节点时,结合贪心算法可以在某些情况下提供有效的解决方案[^2]。 #### 输入输出格式说明 根据以往的经验,在解决 PAT 类型的问题时,输入部分往往遵循特定模式。例如,给定 N 行输入来描述每个节点的信息,每行按照如下格式:“Address Data Next”,这有助于理解如何解析输入并建立相应的数据模型[^4]。 #### 数学运算示例 有时也会遇到基本算术表达式的求值问题,比如分数之间的加减乘除运算。下面是一些简单的例子展示不同情况下的计算结果: - \( \frac{2}{3} + (-2) = -\frac{7}{3}\) -2) = -\frac{4}{3}\) - \( \frac{2}{3} ÷ (-2) = -\frac{1}{3}\) 这些运算是基于样例提供的信息得出的结果[^3]。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值