1012 The Best Rank （25 分）

1012 The Best Rank （25 分）

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 Specification:

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 Specification:

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

/*
解题思路：
N个考生，M次查询
有3门课C，M，E，平均分A
这4个成绩，每项成绩对N个考生从高到低排序，这样对每个考生来说就会有4个排名。
接下来是M次查询，每个查询输入一个考生ID，输出该考生的四个排名中最高的那个排名，及对应的科目是C,M,E,A中的哪一个
如果对不同课程有相同排名的情况，则按优先级A>C>M>E输出，如果查询的学生的ID不存在，则输出N/A

1.考虑到优先级为A>C>M>E，不妨在设置数组时就按这个顺序分配序号为0~3的元素，即0对应A,1对应C,2对应M,3对应E
以结构体类型Student存放6位整数的ID和4个分数(grade[0]~grade[3]分别代表A,C,M,E)
由于ID是6位的整数，不妨设置Rank[1000000][4]数组，其中Rank[id][0]~Rank[id][3]表示编号为ID的考生的4个分数各自在所有考生中的排名

*/
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

struct student{
    int id; //存放6位整数的id
    int grade[4]; //存放4个分数

}stu[2010];

char course[4]={'A','C','M','E'}; //按优先级顺序，方便输出
int Rank[1000000][4]={0}; //Rank[id][0]~Rank[id][3]为4门课对应的排名
int now; //cmp函数中使用，表示当前按now号分数排序stu数组

bool cmp(student a,student b)
{
    return a.grade[now]>b.grade[now];
}

int main()
{
    int N,M; //N为考生总数，L为最低录取分数线，K为最优录取分数线
    scanf("%d%d",&N,&M);
    //读入分数，其中grade[0]~grade[3]分别代表A，C，M,E
    for(int i=0;i<N;i++)
    {
        scanf("%d%d%d%d",&stu[i].id,&stu[i].grade[1],&stu[i].grade[2],&stu[i].grade[3]);
        stu[i].grade[0]=(round(stu[i].grade[1]+stu[i].grade[2]+stu[i].grade[3])/3.0)+0.5; //round函数用于将double型变量四舍五入，返回类型也是double型，需进行取整。

    }
    for(now=0;now<4;now++) //枚举A,C,M,E 4个中的一个
    {
        sort(stu,stu+N,cmp); //对所有考生按该分数从大到小排序
        Rank[stu[0].id][now]=1; //排完序，将分数最高的设为1
        for(int i=1;i<N;i++)
        {
            if(stu[i].grade[now]==stu[i-1].grade[now])
            {
                Rank[stu[i].id][now]= Rank[stu[i-1].id][now];
            }
            else
            {
                 Rank[stu[i].id][now]=i+1;

            }

        }
    }
    int query; //查询的ID考生
    for(int i=0;i<M;i++)
    {
        scanf("%d",&query);
        if(Rank[query][0]==0) //如果这个ID的考生不存在
        {
            printf("N/A\n");
        }
        else  //选出Rank[query][0~3]中值最小的，rank越小，排名越高
        {
            int k=0;
            for(int j=0;j<4;j++)
            {
                if(Rank[query][j]<Rank[query][k])
                k=j;
            }
            printf("%d %c\n",Rank[query][k],course[k]);
        }
    }

    return 0;
}