PAT甲级 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

分析：

题目要求：给定N个学生，及其C语言（C）、数学（M）、英语（E）三门课的成绩，对包括平均分（A）在内的四个成绩分别进行排名。

给定M个学生的id，求他们的最好排名及对应的课程（或者平均分）

当有多个科目取得相同的最好排名时，按照A > C > M > E的顺序，输出其中优先级最大的课程。

若学生id不存在，则输出N/A。

需要注意的地方：1.平均分要四舍五入；2.若分数相同，则排名并列。

#include <cstdio>
#include <algorithm>
using namespace std;

struct Student {
	int id;
	// 0-A,1-C,2-M,3-E
	int score[4];
	int rank[4];
	int best;
};

int flag = 0;
// 快速判断id是否存在
int exists[1000000];
char course[5] = { 'A', 'C', 'M', 'E' };

bool cmp(struct Student s1, struct Student s2) {
	return s1.score[flag] > s2.score[flag];
}

int main() {
	int n, m, id;
	scanf("%d %d", &n, &m);
	struct Student stu[n];
	// 输入
	for (int i = 0; i < n; i++) {
		scanf("%d", &stu[i].id);
		int sum = 0;
		for (int j = 1; j <= 3; j++) {
			scanf("%d", &stu[i].score[j]);
			sum += stu[i].score[j];
		}
		// 平均成绩四舍五入
		stu[i].score[0] = sum / 3.0 + 0.5;
	}

	// 按照不同的科目分别排序
	for (flag = 0; flag < 4; flag++) {
		sort(stu, stu + n, cmp);
		stu[0].rank[flag] = 1;
		for (int j = 1; j < n; j++) {
			stu[j].rank[flag] = j + 1;
			// 若分数相同，则排名并列
			if (stu[j].score[flag] == stu[j - 1].score[flag]) {
				stu[j].rank[flag] = stu[j - 1].rank[flag];
			}
		}
	}
	// 计算最佳排名
	for (int i = 0; i < n; i++) {
		exists[stu[i].id] = i + 1;
		int rank = n + 1;
		for (int j = 0; j < 4; j++) {
			if (stu[i].rank[j] < rank) {
				rank = stu[i].rank[j];
				stu[i].best = j;
			}
		}
	}

	for (int i = 0; i < m; i++) {
		scanf("%d", &id);
		int index = exists[id];
		if (index) {
			index -= 1;
			int best = stu[index].best;
			printf("%d %c\n", stu[index].rank[best], course[best]);
		} else {
			printf("N/A\n");
		}
	}
	return 0;
}