浙大pat 1025题解
1025. PAT Ranking (25)
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
#include"iostream"
#include "algorithm"
#include "string"
#include "vector"
using namespace std;
struct Student
{
string id;
int grade;
int final_rank;
int local;
int local_rank;
};
bool gradecompare(Student a,Student b)
{
if(a.grade !=b.grade)
return a.grade > b.grade;
else return a.id<b.id;
}
vector<Student> stu;
int main()
{
int n,m,count=0;
int start=0;
cin >> n;
Student t;
while(count<n)
{
cin >> m;
for(int i=0;i<m;i++)
{
cin >> t.id >> t.grade;
stu.push_back(t);
}
sort(stu.begin()+start,stu.end(),gradecompare);
for(int i=0+start;i<stu.size();i++)
{
stu[i].local = count+1;
stu[i].local_rank = i+1-start;
if(i>0)
{
if(stu[i].grade==stu[i-1].grade)
stu[i].local_rank = stu[i-1].local_rank;
}
}
start +=m;
count++;
}
sort(stu.begin(),stu.end(),gradecompare);
for(int i=0;i<stu.size();i++)
{
stu[i].final_rank = i+1;
if(i>0)
{
if(stu[i].grade==stu[i-1].grade)
stu[i].final_rank = stu[i-1].final_rank;
}
}
cout<<start<<endl;
vector<Student>::iterator it=stu.begin();
while(it!=stu.end())
{
cout<<(*it).id<<" "<<(*it).final_rank<<" "<<(*it).local<<" "<<(*it).local_rank<<endl;
it++;
}
return 0;
}
浙大pat 1025题解的更多相关文章
- 浙大pat 1035题解
1035. Password (20) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue To prepare f ...
- 浙大pat 1011题解
With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excite ...
- 浙大PAT 7-06 题解
#include <stdio.h> #include <iostream> #include <algorithm> #include <math.h> ...
- 浙大pat 1012题解
1012. The Best Rank (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue To eval ...
- 浙大 pat 1003 题解
1003. Emergency (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue As an emerg ...
- 浙大 pat 1038 题解
1038. Recover the Smallest Number (30) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHE ...
- 浙大 pat 1047题解
1047. Student List for Course (25) 时间限制 400 ms 内存限制 64000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Y ...
- 浙大pat 1054 题解
1054. The Dominant Color (20) 时间限制 100 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard Behind the scen ...
- 浙大pat 1059 题解
1059. Prime Factors (25) 时间限制 50 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 HE, Qinming Given ...
随机推荐
- Makefile.am编写规则
概念 Makefile.am是比Makefile更高层次的规则只需要指定要生成什么目标,依赖于什么文件,和要安装到什么目录.automake会根据Makefile.am来自动生成Makefile.in ...
- tcp异常终止连接
服务端: #include <sys/socket.h> #include <unistd.h> #include <sys/types.h> #include & ...
- crontab 添加sh文件定时
(1)编写sh文件,比如/orcl/test/export.sh 编写crond文件 chmod 755 /orcl/test/* //复制所有权限 [root@postest test]# ...
- 改造vim
1.安装Vim和Vim基本插件首先安装好Vim和Vim的基本插件.这些使用apt-get安装即可: lingd@ubuntu:~/arm$sudo apt-get install vim vim-sc ...
- Qt实现悬浮窗效果
当鼠标移动到头像控件时,显示悬浮窗,当鼠标离开时,悬浮窗隐藏. 1.控件选择 悬浮窗可以从QDialog派生,并将窗口的属性设置为无边框 this->setWindowFlags(this- ...
- 学习自动化工具gulp
<什么是gulp>官网地址:http://gulpjs.com/ gulp是可以自动化执行任务的工具,在开发流程里,一定有一些动作需要手工的重复的去执行,例如: ·把一个文件拷贝到另外一个 ...
- schema change + ogg 变更手册
Check OGG until no data queuing in replication process:testRO:a)login test5 –l oggmgrb)oggc)#ggsci ...
- 4. printf 命令
1. printf 命令的语法 printf format-string [arguments...] 参数说明: format-string: 为格式控制字符串 arguments: 为参数列表. ...
- 使用MyEclipse构建MAVEN项目
这里用的是MyEclpise的自带的MAVEN插件.Maven最好配置成你自己安装的那个,MyEclipse自带会有些许Bug.用nexus代理Maven的中央仓库,setting.xml的配置文件修 ...
- js的日期、定时器
第一.js的日期 js的日期是用内置对象Date来操作,首先先创建一个日期 var date=new Date();,然后就可以调用它的API来获取年月日.时分秒等等,这不是本文重点,本文重点讲如何把 ...