To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algrbra), and E - English.  At the mean time, we encourage students by emphasizing on their best ranks -- that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.

For example, The grades of C, M, E and A - Average of 4 students are given as the following:

StudentID  C  M  E  A
310101 98 85 88 90
310102 70 95 88 84
310103 82 87 94 88
310104 91 91 91 91

Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.

Input

Each input file contains one test case.  Each case starts with a line containing 2 numbers N and M (<=2000), which are the total number of students, and the number of students who would check their ranks, respectively.  Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E.  Then there are M lines, each containing a student ID.

Output

For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.

The priorities of the ranking methods are ordered as A > C > M > E.  Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.

If a student is not on the grading list, simply output "N/A".

Sample Input

5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999

Sample Output

1 C
1 M
1 E
1 A
3 A
N/A
// 1012pat1.cpp : 定义控制台应用程序的入口点。
//
#include <fstream>
#include <iostream>
#include <vector>
#include <map>
#include <string>
#include <algorithm>
using namespace std; const int INF=0x7fffffff;
struct Edge
{
string id;
int C,M,E,A;
}; struct R
{
int r;
char c;
bool flag;
R(){r=INF;c='A';flag=false;}
}; class TA:public binary_function<Edge,Edge,bool>
{
public:
bool operator()(const Edge& lhs,const Edge& rhs) const
{
return lhs.A>rhs.A;
}
};
class TC:public binary_function<Edge,Edge,bool>
{
public:
bool operator()(const Edge& lhs,const Edge& rhs) const
{
return lhs.C>rhs.C;
}
};
class TM:public binary_function<Edge,Edge,bool>
{
public:
bool operator()(const Edge& lhs,const Edge& rhs) const
{
return lhs.M>rhs.M;
}
};
class TE:public binary_function<Edge,Edge,bool>
{
public:
bool operator()(const Edge& lhs,const Edge& rhs) const
{
return lhs.E>rhs.E;
}
};
int main()
{
int n,m;
while(cinf>>n>>m)
{
Edge temp;
vector<Edge> vec;
map<string,R> rp;
for(int i=;i<=n;++i)
{
cinf>>temp.id>>temp.C>>temp.M>>temp.E;
rp[temp.id].flag=true;
temp.A=(temp.C+temp.M+temp.E)/;
vec.push_back(temp);
}
sort(vec.begin(),vec.end(),TA());
int rank=;
int cnt=;
for(vector<Edge>::iterator iter=vec.begin();iter!=vec.end();++iter)
{
if((iter+)!=vec.end())
{
if(iter->A==(iter+)->A)
{
++cnt;
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='A';
}
if(rp[(iter+)->id].r>rank)
{
rp[(iter+)->id].r=rank;
rp[(iter+)->id].c='A';
}
}
else
{
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='A';
}
rank=rank+cnt;
cnt=;
}
}
else
{
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='A';
}
}
}
sort(vec.begin(),vec.end(),TC());
rank=;
cnt=;
for(vector<Edge>::iterator iter=vec.begin();iter!=vec.end();++iter)
{
if((iter+)!=vec.end())
{
if(iter->C==(iter+)->C)
{
++cnt;
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='C';
}
if(rp[(iter+)->id].r>rank)
{
rp[(iter+)->id].r=rank;
rp[(iter+)->id].c='C';
}
}
else
{
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='C';
}
rank=rank+cnt;
cnt=;
}
}
else
{
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='C';
}
}
}
sort(vec.begin(),vec.end(),TM());
rank=;
cnt=;
for(vector<Edge>::iterator iter=vec.begin();iter!=vec.end();++iter)
{
if((iter+)!=vec.end())
{
if(iter->M==(iter+)->M)
{
++cnt;
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='M';
}
if(rp[(iter+)->id].r>rank)
{
rp[(iter+)->id].r=rank;
rp[(iter+)->id].c='M';
}
}
else
{
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='M';
}
rank=rank+cnt;
cnt=;
}
}
else
{
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='M';
}
}
}
sort(vec.begin(),vec.end(),TE());
rank=;
cnt=;
for(vector<Edge>::iterator iter=vec.begin();iter!=vec.end();++iter)
{
if((iter+)!=vec.end())
{
if(iter->E==(iter+)->E)
{
++cnt;
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='E';
}
if(rp[(iter+)->id].r>rank)
{
rp[(iter+)->id].r=rank;
rp[(iter+)->id].c='E';
}
}
else
{
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='E';
}
rank=rank+cnt;
cnt=;
}
}
else
{
if(rp[iter->id].r>rank)
{
rp[iter->id].r=rank;
rp[iter->id].c='E';
}
}
}
string id;
for(int i=;i<=m;++i)
{
cinf>>id;
if(rp[id].flag==true)
cout<<rp[id].r<<" "<<rp[id].c<<endl;
else
cout<<"N/A"<<endl;
}
}
return ;
}

PAT 1012. The Best Rank的更多相关文章

  1. PAT 1012 The Best Rank 排序

    To evaluate the performance of our first year CS majored students, we consider their grades of three ...

  2. PAT甲级1012. The Best Rank

    PAT甲级1012. The Best Rank 题意: 为了评估我们第一年的CS专业学生的表现,我们只考虑他们的三个课程的成绩:C - C编程语言,M - 数学(微积分或线性代数)和E - 英语.同 ...

  3. PAT甲 1012. The Best Rank (25) 2016-09-09 23:09 28人阅读 评论(0) 收藏

    1012. The Best Rank (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue To eval ...

  4. PAT 1012

    1012. The Best Rank (25) To evaluate the performance of our first year CS majored students, we consi ...

  5. 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 ...

  6. 1012 The Best Rank (25 分)

    1012 The Best Rank (25 分) To evaluate the performance of our first year CS majored students, we cons ...

  7. 1012 The Best Rank

    1012 The Best Rank 1. 注意点 一名同学同样排名下的科目优先级问题 不同同学分数相同时排名相同,注意 排名不是 1 1 2 3 4 这种, 而是 1 1 3 4 5 注意到有些同学 ...

  8. 1012 The Best Rank (25分) vector与结构体排序

    1012 The Best Rank (25分)   To evaluate the performance of our first year CS majored students, we con ...

  9. 【PAT】1012. The Best Rank (25)

    题目链接: http://pat.zju.edu.cn/contests/pat-a-practise/1012 题目描述: To evaluate the performance of our fi ...

随机推荐

  1. 关于APlayer播放器在打包安装后提示“没有注册类”的解决办法

    1.首先需要确定必要的DLL文件都已经在正确的安装目录下了: 2.项目中引用的DLL必须是Debug目录下的: 3.若后续修改或者重新注册了APlayer组件,那么所有的DLL都需要替换成最新的. 关 ...

  2. xml转array

    1.字串 $xml = simplexml_load_string($data);$array = json_decode(json_encode($xml),TRUE); 2.文件$xml = si ...

  3. insert遭遇阻塞

    insert的阻塞确实不常见,今天碰到了一个,看书又了解一个,整理下.1.多个会话同时向unique字段插入相同的值session1:首先建测试表test,并在字段id上创建一个主键索引(唯一键也可以 ...

  4. 在ubuntu中获得root权限

    在终端中输入:(1)sudo passwd rootEnter new UNIX password: (在这输入你的密码)Retype new UNIX password: (确定你输入的密码)pas ...

  5. NET笔记——Delegate

    对于初学者,委托是很容易让人晕的,一是晕它如何起作用,二是晕它有什么用. 最近回过头来又看了下委托,又有些不同的感觉,写之自用. 声明方面,委托可以被声明在类内,也可以与类同级,并且声明时没有方法体: ...

  6. PHP中的urlencode和urldecode的理解

    平时在工作中经常要写 $xxx = urldecode($_GET['xxx']);的类似代码,大部分的情况都是没有问题的.也能很好的工作. 所以也没有怎么在意.但是突然有一天我想到 $xxx =$_ ...

  7. ECMall注册机制简要分析

    ecmall的注册流程index.php?app=member&act=register. 首先app是member,act是register方法. index.php中.通过ecmall的s ...

  8. 【HDU 2853】Assignment (KM)

    Assignment Problem Description Last year a terrible earthquake attacked Sichuan province. About 300, ...

  9. [置顶] 【Git入门之一】Git是神马?

    1.Git是神马? 一个开源的分布式版本控制系统,可以有效的高速的控制管理各种从小到大的项目版本.他的作者就是大名鼎鼎的Linux系统创始人Linus. 2.分布式又是神马? 先看看集中式.简单说来, ...

  10. 3.android下Makefile编写规范

    随着移动互联网的发展,移动开发也越来越吃香了,目前最火的莫过于android,android是什么就不用说了,android自从开源以来,就受到很多人的追捧.当然,一部人追捧它是因为它是Google开 ...