1012. The Best Rank (25)

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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 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

提交代码

题意:给出N个学生的学号, 以及各项成绩, 然后给出M个学号, 查询他们各项成绩中, 最好的那一项是什么, 以及最好的那项排名多少, 如果有相同的, 优先次序为A>C>M>E, 找不到学号输出N/A.

我们可以用STL中的map<int, struct>来实现学号到各项成绩的映射, 如果 map.find(id) == map.end() 则说明找不到. 如果找到了, 我们可以把各项成绩预处理排好顺序, 直接输出即可.

#include <cstdio>
#include <cstring>
#include <map>

using namespace std;

struct node{
	int C, M, E, A;
};

map<int, node> mp;
const int N = 110;

void change(int* ary) {
	int total = 0;

	for (int i = 100; ~i; --i) {
		if (ary[i]) {
			int tmp = ary[i];
			ary[i] = total + 1;
			total += tmp;
		}
	}
}

int main() {
	int n, m, id;
	int aryA[N] = {0}, aryC[N] = {0}, aryM[N] = {0}, aryE[N] = {0};
	node Node;

	scanf("%d%d", &n, &m);

	for (int i = 0; i < n; ++i) {
		scanf("%d%d%d%d", &id, &Node.C, &Node.M, &Node.E);
		Node.A = (Node.C + Node.M + Node.E) / 3;
		mp[id] = Node;
		aryA[Node.A]++;
		aryC[Node.C]++;
		aryM[Node.M]++;
		aryE[Node.E]++;
	}

	change(aryA);
	change(aryC);
	change(aryM);
	change(aryE);

	for (int i = 0; i < m; ++i) {
		scanf("%d", &id);
		if (mp.find(id) == mp.end())
			puts("N/A");
		else {
			int ans = n + 1;
			char ch;
			if (ans > aryA[mp[id].A]) ans = aryA[mp[id].A], ch = 'A';
			if (ans > aryC[mp[id].C]) ans = aryC[mp[id].C], ch = 'C';
			if (ans > aryM[mp[id].M]) ans = aryM[mp[id].M], ch = 'M';
			if (ans > aryE[mp[id].E]) ans = aryE[mp[id].E], ch = 'E';
			printf("%d %c\n", ans, ch);
		}
	}

	return 0;
}