pat甲级 1022 Digital Library

这篇具有很好参考价值的文章主要介绍了pat甲级 1022 Digital Library。希望对大家有所帮助。如果存在错误或未考虑完全的地方,请大家不吝赐教,您也可以点击"举报违法"按钮提交疑问。

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 (≤104) 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

代码长度限制

16 KB

时间限制

1200 ms

内存限制

64 MB文章来源地址https://www.toymoban.com/news/detail-851962.html

#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
using namespace std;
const int N = 10010;
int n;
int m;
struct st {
    string id;
    string title;
    string author;
    string words;
    string publisher;
    string year;
}q[N];

bool cmp(st a, st b) {
    return a.id < b.id;
}
int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> q[i].id;
        getchar();
        getline(cin, q[i].title);
        getline(cin, q[i].author);
        getline(cin, q[i].words);
        getline(cin, q[i].publisher);
        getline(cin, q[i].year);
    }
    sort(q + 1, q + 1 + n, cmp);
    cin >> m;
    for (int i = 1; i <= m; i++) {
        int w;
        char op;
        string s;
        cin >> w >> op;
        getchar();
        getline(cin, s);
        if (w == 1) {
            bool flag = false;
            cout << w << op << " " << s << endl;
            for (int i = 1; i <= n; i++) {
                if (q[i].title == s) {
                    cout << q[i].id << endl;
                    flag = true;
                }
            }
            if (!flag) cout << "Not Found" << endl;
        }
        if (w == 2) {
            cout << w << op << " " << s << endl;
            bool flag = false;
            for (int i = 1; i <= n; i++) {
                if (q[i].author == s) {
                    flag = true;
                    cout << q[i].id << endl;
                }
            }
            if (!flag) cout << "Not Found" << endl;
        }
        if (w == 3) {
            cout << w << op << " " << s << endl;
            bool flag = false;;
            for (int i = 1; i <= n; i++) {
                if (q[i].words.find(s) !=-1) {
                    flag = true;
                    cout << q[i].id << endl;
                }
            }
            if (!flag) cout << "Not Found" << endl;
        }
        if (w == 4) {
            cout << w << op << " " << s << endl;
            bool flag = false;
            for (int i = 1; i <= n; i++) {
                if (q[i].publisher == s) {
                    flag = true;
                    cout << q[i].id << endl;
                }
            }
            if (!flag) cout << "Not Found" << endl;
        }
        if (w== 5) {
            cout << w << op << " " << s << endl;
            bool flag = false;
            for (int i = 1; i <= n; i++) {
                if (q[i].year == s) {
                    flag = true;
                    cout << q[i].id << endl;
                }
            }
            if (!flag) cout << "Not Found" << endl;
        }
    }
    return 0;
}

到了这里,关于pat甲级 1022 Digital Library的文章就介绍完了。如果您还想了解更多内容,请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章,希望大家以后多多支持TOY模板网!

本文来自互联网用户投稿,该文观点仅代表作者本人,不代表本站立场。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如若转载,请注明出处: 如若内容造成侵权/违法违规/事实不符,请点击违法举报进行投诉反馈,一经查实,立即删除!

领支付宝红包 赞助服务器费用

相关文章

  • 菜鸟记录PAT甲级1003--Emergency

    久违的PAT,由于考研408数据结构中有一定需要,同时也是对先前所遗留的竞赛遗憾进行一定弥补 ,再次继续PAT甲级1003.。 As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the l

    2023年04月13日
    浏览(45)
  • 1114 Family Property (PAT甲级)

    This time, you are supposed to help us collect the data for family-owned property. Given each person\\\'s family members, and the estate(房产)info under his/her own name, we need to know the size of each family, and the average area and number of sets of their real estate. Input Specification: Each input file contains one test case. For each case, the fir

    2024年02月06日
    浏览(51)
  • 1028 List Sorting (PAT甲级)

    Excel can sort records according to any column. Now you are supposed to imitate this function. Input Specification: Each input file contains one test case. For each case, the first line contains two integers N (≤105) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then N lines follow, eac

    2024年02月15日
    浏览(41)
  • 1072 Gas Station (PAT甲级)

    A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range. Now given the map of the city and several candidate locations for the gas station, you are supposed to give the best recommendatio

    2024年02月09日
    浏览(42)
  • 1111 Online Map (PAT甲级)

    这道题我读题不仔细导致踩了个大坑,一个测试点过不了卡了好几个小时:第二个dijkstra算法中,题目要求是“In case the fastest path is not unique, output the one that passes through the fewest intersections”,我却想当然地认为在fastest path is not unique的时候,判断标准是最短距离…… Input our

    2024年02月07日
    浏览(42)
  • 1083 List Grades (PAT甲级)

    Given a list of N student records with name, ID and grade. You are supposed to sort the records with respect to the grade in non-increasing order, and output those student records of which the grades are in a given interval. Input Specification: Each input file contains one test case. Each case is given in the following format: where  name[i]  and  ID[i

    2024年02月08日
    浏览(48)
  • PAT甲级图论相关题目

    PAT甲级图论相关题目: 分数 25 As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some o

    2024年01月21日
    浏览(54)
  • PAT 甲级【1007 Maximum Subsequence Sum】

    本题是考察动态规划与java的快速输入: max[i]表示第i个结尾的最大的连续子串和。b begin[i]表示第[begin[i],i]为最大和的开始位置 超时代码: 未超时: 能用动态规划解决的问题,需要满足三个条件:最优子结构,无后效性和子问题重叠。 具有最优子结构也可能是适合用贪心的

    2024年02月08日
    浏览(41)
  • 1074 Reversing Linked List (PAT甲级)

    Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6. Input Specification: Each input file contains one test case. For each case, the first line

    2024年02月09日
    浏览(41)
  • PAT 甲级1005【1005 Spell It Right】

    用JAVA可以用BigInteger解决。       太长不看版:结尾自取模板…… 高精度计算(Arbitrary-Precision Arithmetic),也被称作大整数(bignum)计算,运用了一些算法结构来支持更大整数间的运算(数字大小超过语言内建整型)。 高精度问题包含很多小的细节,实现上也有很多讲究。

    2024年02月08日
    浏览(88)

觉得文章有用就打赏一下文章作者

支付宝扫一扫打赏

博客赞助

微信扫一扫打赏

请作者喝杯咖啡吧~博客赞助

支付宝扫一扫领取红包,优惠每天领

二维码1

领取红包

二维码2

领红包