PAT Ranking (排名)
PAT Ranking (排名)
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 (<=100), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (<=300), 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
先分组排名,再全部排名
#include <iostream> #include <string> #include <algorithm> #include <vector> using namespace std; struct stu { int loc,frank,lrank,gra; string name; }; bool cmp(stu s1,stu s2) { if(s1.frank==s2.frank) return s1.name<s2.name; return s1.frank<s2.frank; } bool cmpg(stu s1,stu s2) { return s1.gra>s2.gra; } int main() { int n;int k;int rank,count; int i,j; while(cin>>n) { vector<stu> total; for(i=;i<=n;i++) { cin>>k; vector<stu> s(k); for(j=;j<k;j++) { cin>>s[j].name>>s[j].gra; s[j].loc=i; } sort(s.begin(),s.end(),cmpg); count=; s[].lrank=; total.push_back(s[]); for(j=;j<k-;j++) { if(s[j].gra==s[j+].gra) { s[j+].lrank=s[j].lrank; count++; } else { s[j+].lrank=s[j].lrank+count; count=; } total.push_back(s[j+]); } } sort(total.begin(),total.end(),cmpg); count=; total[].frank=; for(j=;j<total.size()-;j++) { if(total[j].gra==total[j+].gra) { total[j+].frank=total[j].frank; count++; } else { total[j+].frank=total[j].frank+count; count=; } } sort(total.begin(),total.end(),cmp); cout<<total.size()<<endl; for(i=;i<total.size();i++) cout<<total[i].name<<" "<<total[i].frank<<" "<<total[i].loc<<" "<<total[i].lrank<<endl; } return ; }
PAT Ranking (排名)的更多相关文章
- 1025 PAT Ranking[排序][一般]
1025 PAT Ranking (25)(25 分) Programming Ability Test (PAT) is organized by the College of Computer S ...
- 1141 PAT Ranking of Institutions[难]
1141 PAT Ranking of Institutions (25 分) After each PAT, the PAT Center will announce the ranking of ...
- pat1025. PAT Ranking (25)
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 ...
- PAT_A1141#PAT Ranking of Institutions
Source: PAT A1141 PAT Ranking of Institutions (25 分) Description: After each PAT, the PAT Center wil ...
- PAT_A1025#PAT Ranking
Source: PAT A1025 PAT Ranking Description: Programming Ability Test (PAT) is organized by the Colleg ...
- PAT甲级:1025 PAT Ranking (25分)
PAT甲级:1025 PAT Ranking (25分) 题干 Programming Ability Test (PAT) is organized by the College of Comput ...
- 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 ...
- A1025 PAT Ranking (25)(25 分)
A1025 PAT Ranking (25)(25 分) Programming Ability Test (PAT) is organized by the College of Computer ...
- 1025 PAT Ranking (25分)
1025 PAT Ranking (25分) 1. 题目 2. 思路 设置结构体, 先对每一个local排序,再整合后排序 3. 注意点 整体排序时注意如果分数相同的情况下还要按照编号排序 4. 代码 ...
随机推荐
- struts2 后台action向前端JSP传递参数的问题
后台有个方法,里面部分代码如下: System.out.print("alter:"+user.getId()); User temp=userservice.getUserByU ...
- spring+hibernate+struts整合(1)
spring+hibernate:整合 步骤1:引入类包 如下图:这里是所有的类包,为后面的struts整合考虑
- [ImportNew]Java线程面试题
Java语言一个重要的特点就是内置了对并发的支持,让Java大受企业和程序员的欢迎.大多数待遇丰厚的Java开发职位都要求开发者精通多线程技术并且有丰富的Java程序开发.调试.优化经验,所以线程相关 ...
- Web前端学习笔记2
一.开发工具sublime的常用快捷键. 1. 快捷键 功能 ctrl+shift+D 快速复制 ctrl+L 快 ...
- sql常识-top
TOP 子句 TOP 子句用于规定要返回的记录的数目. 对于拥有数千条记录的大型表来说,TOP 子句是非常有用的. 注释:并非所有的数据库系统都支持 TOP 子句. SQL Server 的语法: S ...
- Cocos开发中Visual Studio下HttpClient开发环境设置
Cocos2d-x 3.x将与网络通信相关的类集成到libNetwork类库工程中,这其中包括了HttpClient类.我们需要在Visual Studio解决方案中添加libNetwork类库工程. ...
- 关于iOS9中配置App Transport Security(应用程序传输安全协议)
在 在info.plist中,进行上面的配置就行了,注意的是,那个网址,你需要访问什么网址,就填写什么网址就行了. NSTemporaryExceptionAllowsInsecureHTTPLoad ...
- ios 微信细节
1.登录后,下次登录保存其用户名. * 官方的登录实现 * 1.把用户名和密码放在沙盒 NSString *user = self.userField.text; NSStri ...
- MyBatis参数传入集合之foreach动态sql
foreach的主要用在构建in条件中,它可以在SQL语句中进行迭代一个集合.foreach元素的属性主要有item,index,collection,open,separator,close.ite ...
- UML静态类图
0,主要分为类.接口.协作.关系,这四种元素.作用:a,显示类.接口以及他们之间的静态结构和关系:b,用于描述系统的结构化设计. 1,类 CStudent +m_strName : string +S ...