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 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 Algebra), 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

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
/*建立索引，通过数组建立索引，一个结构体的id设置为数组的index,通过exist[index]来确定新的index 遍历一遍后就可以快速查询，快速找到位置*/
struct stu {
	int id,score[4], rank[4];
	int best;
}s[2005];
int exist[10000000];
int flag = -1;
bool cmp(stu a, stu b) {
	return a.score[flag] > b.score[flag];
}
int main() {
#ifdef JUDGE
	freopen("D:\\a.txt", "r", stdin);
	//freopen("D:\b.txt", "r", stdout);
#endif
	fill(exist, exist + 10000000, -1);
	int n, m;
	scanf("%d%d", &n, &m);
	stu t;
	for (int i = 0; i < n; i++) {
		int sum=0;
		scanf("%d", &s[i].id);
		for (int j = 1; j <= 3; j++) {
			scanf("%d", &s[i].score[j]);
			sum += s[i].score[j];
		}
		s[i].score[0] = sum / 3.0 + 0.5;
	}
	for (int i = 0; i < 4; i++) {
		flag = i;
		sort(s, s + n, cmp);
		s[0].rank[flag] = 1;
		for (int j = 1; j < n; j++) {
			s[j].rank[flag] = j + 1;
			if (s[j].score[flag] == s[j - 1].score[flag]) {
				s[j].rank[flag] = s[j - 1].rank[flag];
			}
		}

	}
	for (int i = 0; i < n; i++) {
		exist[s[i].id] = i;
		s[i].best = 0;
		int minn = s[i].rank[0];
		for (int j = 0; j < 4; j++) {
			if (minn > s[i].rank[j]) {
				minn = s[i].rank[j];
				s[i].best = j;
				//cout << j << endl;
			}
		}
	}
	char str[5] = "ACME";
	for (int i = 0; i < m; i++) {
		int tmp;
		scanf("%d", &tmp);
		//cout << tmp;
		if (exist[tmp]>=0) {
			printf("%d %c", s[exist[tmp]].rank[s[exist[tmp]].best],str[s[exist[tmp]].best]);
		}
		else {
			printf("N/A");
		}
		if (i != m - 1)
			printf("\n");
	}
}