PAT (Advanced Level) Practice 1022 Digital Library (30 分)

本文介绍了一个数字图书馆系统的设计,该系统能够存储上百万本书籍,并根据书名、作者、关键词、出版社和出版年份进行高效检索。每本书都有一个唯一的7位数字ID,通过输入查询,系统能返回按ID升序排列的书籍结果。

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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

Code:

#include <iostream>
#include <map>
#include <set>
#include <string>
#pragma warning(disable:4996)

using namespace std;

map<string, set<string>> title_id;
map<string, set<string>> author_id;
map<string, set<string>> year_id;
map<string, set<string>> keyword_id;
map<string, set<string>> publisher_id;

int main()
{
	int N;
	scanf("%d", &N);
	cin.ignore(100, '\n');
	for (int i = 0; i < N; i++)
	{
		string id, title, author, keyword, publisher, year;
		getline(cin, id);
		getline(cin, title);
		getline(cin, author);
		getline(cin, keyword);
		getline(cin, publisher);
		getline(cin, year);
		title_id[title].insert(id);
		author_id[author].insert(id);
		year_id[year].insert(id);
		publisher_id[publisher].insert(id);
		int l = 0;
		keyword += ' ';
		for (int i = 0; i < keyword.length(); i++)
		{
			if (keyword[i] == ' ') {
				keyword_id[keyword.substr(l, i - l)].insert(id);
				l = i + 1;
			}
		}
	}
	int M;
	scanf("%d", &M);
	cin.ignore(100, '\n');
	for (int i = 0; i < M; i++)
	{
		string query;
		getline(cin, query);
		cout << query << endl;
		int type = query[0] - '0';
		query = query.substr(3);
		auto display = [](map<string, set<string>>& map, string query)
		{
			if (map.count(query))
			{
				for (auto it : map[query])
					cout << it << '\n';
			}
			else
			{
				cout << "Not Found" << "\n";
			}
		};
		switch (type)
		{
		case 1:
			display(title_id, query);
			break;
		case 2:
			display(author_id, query);
			break;
		case 3:
			display(keyword_id, query);
			break;
		case 4:
			display(publisher_id, query);
			break;
		case 5:
			display(year_id, query);
		default:
			break;
		}
	}
	return 0;
}

思路

map + set 是很好的组合,set还能自动升序,要注意的是这里用getline之前,如果有输入,要先ingore前面的‘\n’,不然会读空行,比如之前用cin读一个数,输入一个回车(\n),getline是以\n为停止符,读取一行,所以就把那个\n读去了,造成读空行,ignore了就不会了。

另外:
https://blog.youkuaiyun.com/wxbmelisky/article/details/48596881 ignore的用法

### PAT 基本级别练习的相关资料 PAT(Programming Ability Test)是一项针对编程能力的标准化考试,其基本级别主要考察考生的基础编程技能和逻辑思维能力。以下是关于 PAT 基本级别的练习题目及相关解决方案的内容。 #### 题目PAT 基本级别的题目通常为以下几个类别: 1. **字符串处理**:涉及字符串的操作,如反转、查找子串等。 2. **数组操作**:包括数组排序、去重、统计等问题。 3. **简单算法实现**:如计算阶乘、判断素数等基础算法。 4. **输入输出格式化**:要求按照特定格式读取数据并输出结果。 以下是一些典型的 PAT 基本级别练习题及其解决思路: --- #### 示例题目 1: 字符串反转 **描述**: 输入一个字符串,将其逆序输出。 **解法**: 可以通过 Python 中的切片功能轻松实现字符串反转。 ```python def reverse_string(s): return s[::-1] # 测试用例 input_str = input() print(reverse_string(input_str)) ``` 此代码利用了 Python 切片语法 `s[start:end:step]`,其中步长为 `-1` 表示反向遍历字符串[^4]。 --- #### 示例题目 2: 数组求和 **描述**: 给定一组整数,求数组中所有元素的总和。 **解法**: 通过循环累加或者内置函数 `sum()` 实现数组求和。 ```python def array_sum(arr): return sum(arr) # 测试用例 numbers = list(map(int, input().split())) print(array_sum(numbers)) ``` 上述代码中,`map()` 函数用于将输入转换为整型列表,而 `sum()` 是 Python 的内置函数,能够高效完成求和任务[^5]。 --- #### 示例题目 3: 计算阶乘 **描述**: 输入一个正整数 \( n \),输出它的阶乘 \( n! \)。 **解法**: 使用递归或迭代方式均可实现阶乘计算。 ```python def factorial(n): if n == 0 or n == 1: return 1 result = 1 for i in range(2, n + 1): result *= i return result # 测试用例 n = int(input()) print(factorial(n)) ``` 该程序采用了迭代方法避免栈溢出的风险,在实际应用中更为稳健[^6]。 --- #### 示例题目 4: 判断素数 **描述**: 输入一个正整数 \( m \),判断它是否为素数。 **解法**: 素数是指仅能被 1 和自身整除的大于 1 的自然数。可通过试除法验证。 ```python import math def is_prime(m): if m <= 1: return False sqrt_m = int(math.sqrt(m)) + 1 for i in range(2, sqrt_m): if m % i == 0: return False return True # 测试用例 m = int(input()) if is_prime(m): print("Yes") else: print("No") ``` 在此代码片段中,引入了平方根优化技术以减少不必要的运算次数[^7]。 --- ### 总结 以上展示了几个常见的 PAT 基础级问题以及对应的解决方案。这些例子涵盖了字符串操作、数组处理、数学计算等多个方面,有助于初学者熟悉 PAT 考试的形式与难度。
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