【PAT】A1012 The Best Rank【排序】

最新推荐文章于 2024-03-15 08:30:00 发布

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最新推荐文章于 2024-03-15 08:30:00 发布 · 1.1k 阅读

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#PAT #PAT甲级 #排序

该博客介绍了PAT甲级考试中的一道题目，要求找出学生在C、M、E三门课程及平均成绩(A)中的最佳排名及对应的科目。题目给出了输入输出规范及示例，并提供了两种解题思路：一是通过四次排序确定最佳排名；二是利用空间换时间的方法优化处理。

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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

题目大意

求每个学生在所有科目（三门学科加一个平均分）中的最佳排名及对应科目。如果一个学生的最佳排名存在多个，则根据A > C > M > E的优先级定最佳科目。

思路1

对于平均值，我们采取将三门学科加起来的方式，这样可以避免除法的舍入问题。进行四次排序，记录每个学生各科的排名，要注意相同分数并列排名。使用一个一维数组id来存储各个学号的学生在students数组中的索引值，方便查询，对于不存在的学生，默认值为-1。

代码1

#include <iostream>
#include <string>
#include <algorithm>
#include <cstdio>
using namespace std;
struct Student{
   
   
    int grade[4];
    int rank[4];
    int id;