Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate the final rank.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive number N (≤), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (≤), the number of testees, and then K lines containing the registration number (a 13-digit number) and the total score of each testee. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in one line the total number of testees. Then print the final ranklist in the following format:

registration_number final_rank location_number local_rank

The locations are numbered from 1 to N. The output must be sorted in nondecreasing order of the final ranks. The testees with the same score must have the same rank, and the output must be sorted in nondecreasing order of their registration numbers.

Sample Input:

2
5
1234567890001 95
1234567890005 100
1234567890003 95
1234567890002 77
1234567890004 85
4
1234567890013 65
1234567890011 25
1234567890014 100
1234567890012 85

Sample Output:

9
1234567890005 1 1 1
1234567890014 1 2 1
1234567890001 3 1 2
1234567890003 3 1 2
1234567890004 5 1 4
1234567890012 5 2 2
1234567890002 7 1 5
1234567890013 8 2 3
1234567890011 9 2 4
简单的排序题,使用结构体
 #include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
struct Node
{
string No;
int score, final_rank, location_num, local_rank;
}; bool cmp(Node* a, Node* b)
{
return (a->score == b->score) ? (a->No < b->No) : (a->score > b->score);
}
int N, K;
vector<Node*>allPerson;
int main()
{
cin >> N;
for (int i = ; i <= N; ++i)
{
cin >> K;
vector<Node*>temp;
for (int j = ; j < K; ++j)
{
Node* node = new Node;
cin >> node->No >> node->score;
node->location_num = i;
temp.push_back(node);
}
sort(temp.begin(), temp.end(), cmp);
for (int j = ; j < temp.size(); ++j)
{
if (j > && temp[j]->score == temp[j - ]->score)
temp[j]->local_rank = temp[j - ]->local_rank;
else
temp[j]->local_rank = j + ;
}
allPerson.insert(allPerson.end(), temp.begin(), temp.end());
}
sort(allPerson.begin(), allPerson.end(), cmp);
cout << allPerson.size() << endl;
for (int j = ; j < allPerson.size(); ++j)
{
if (j> && allPerson[j]->score == allPerson[j - ]->score)
allPerson[j]->final_rank = allPerson[j - ]->final_rank;
else
allPerson[j]->final_rank = j + ;
cout << allPerson[j]->No << " " << allPerson[j]->final_rank << " " <<
allPerson[j]->location_num << " " << allPerson[j]->local_rank << endl;
}
return ;
}

PAT甲级——A1025 PAT Ranking的更多相关文章

  1. PAT 甲级 1141 PAT Ranking of Institutions

    https://pintia.cn/problem-sets/994805342720868352/problems/994805344222429184 After each PAT, the PA ...

  2. PAT 甲级 1025 PAT Ranking

    1025. PAT Ranking (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Programmi ...

  3. PAT甲级——1025 PAT Ranking

    1025 PAT Ranking Programming Ability Test (PAT) is organized by the College of Computer Science and ...

  4. PAT 甲级 1025.PAT Ranking C++/Java

      Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Z ...

  5. PAT 甲级1025 PAT Ranking (25 分)(结构体排序,第一次超时了,一次sort即可小技巧优化)

    题意: 给定一次PAT测试的成绩,要求输出考生的编号,总排名,考场编号以及考场排名. 分析: 题意很简单嘛,一开始上来就,一组组输入,一组组排序并记录组内排名,然后再来个总排序并算总排名,结果发现最后 ...

  6. PAT甲级1075 PAT Judge

    题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805393241260032 题意: 有m次OJ提交记录,总共有k道 ...

  7. PAT 甲级 1075 PAT Judge (25分)(较简单,注意细节)

    1075 PAT Judge (25分)   The ranklist of PAT is generated from the status list, which shows the scores ...

  8. PAT甲级——A1075 PAT Judge

    The ranklist of PAT is generated from the status list, which shows the scores of the submissions. Th ...

  9. A1025 PAT Ranking (25)(25 分)

    A1025 PAT Ranking (25)(25 分) Programming Ability Test (PAT) is organized by the College of Computer ...

随机推荐

  1. 类的反射实例(servlet的抽取)

    类的反射实例 具体以后我们写的时候不用写BaseServlet,因为各种框架都已经给我们写好了 所以,user对应的servlet的界面长这样:

  2. System.Convert.cs

    ylbtech-System.Convert.cs 1. 程序集 mscorlib, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c56 ...

  3. Mybatis笔记 - 原始Dao开发方法

    使用Mybatis开发Dao,通常有两个方法,即原始Dao开发方法和Mapper接口开发方法.原始Dao的开发方式是基于入门程序的基础上,对 控制程序 进行分层开发,程序员需要 编写 Dao接口 和 ...

  4. spring 中常用的设计模式

    一. Spring 中常见的设计模式 工厂模式 : BeanFactory 装饰器模式: BeanWrapper 代理模式: AopProxy 单例模式: ApplicationContext 委派模 ...

  5. PAT甲级——A1128 N Queens Puzzle【20】

    The "eight queens puzzle" is the problem of placing eight chess queens on an 8 chessboard ...

  6. 深夜Python - 第2夜 - 爬行

    深夜Python - 第2夜 - 爬行 我曾经幻想自己是一只蜗牛,有自己的一只小壳,不怕风,不怕雨,浪荡江湖,游历四方……夜猫兄一如既往地打断了我不切实际的幻想:“浪荡?游历?等你退休了都爬不出家门口 ...

  7. 《代码整洁之道》ch1~ch4读书笔记 PB16110698 (~3.8 第一周)

    <代码整洁之道>ch1~ch4读书笔记  <clean code>正如其书名所言,是一本关于整洁代码规范的“教科书”.作者在书中通过实例阐述了整洁代码带来的种种利处以及混乱代码 ...

  8. more 和less 命令简单介绍以及使用

    一.more命令 more功能类似 cat ,cat命令是整个文件的内容从上到下显示在屏幕上. more会以一页一页的显示方便使用者逐页阅读,而最基本的指令就是按空白键(space)就往下一页显示,按 ...

  9. python学院体系

  10. MySQL之从忘记密码到重置密码

    在对MySQL的应用中,难免会有忘记登陆密码的情况:接下来,将简单介绍下MySQL忘记密码如何登陆和重置密码的操作过程. 首先来说下新版MySQL(5.7+)的重置密码过程: 由于忘记登陆密码,所以正 ...