A1012 PAT甲级JAVA题解 The Best Bank

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:

复制代码
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:

复制代码
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:

复制代码
1 C
1 M
1 E
1 A
3 A
N/A

核心题意:

给出学生数量和查询数量

输入学生学号和三门课的分数

对学生的单科排名以及平均分进行排名

查询时输出学生的最好一门科目以及排名

如学生不存在则输出'N/A'

JAVA存在运行超时情况

java 复制代码
import java.util.*;

public class Main {
    static class Student {
        int id;
        int[] grade = new int[4]; // grade[0]为平均分,grade[1-3]为CME成绩

        public Student(int id, int c, int m, int e) {
            this.id = id;
            grade[1] = c;
            grade[2] = m;
            grade[3] = e;
            // 修正:先转换为double再计算平均分,避免整数除法截断
            grade[0] = (int) Math.round((c + m + e) / 3.0);
        }
    }

    // 科目顺序:A(平均分)、C、M、E
    static char[] course = {'A', 'C', 'M', 'E'};
    static Map<Integer, int[]> rankMap = new HashMap<>(); // 存储每个学生的各科排名

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();

        Student[] students = new Student[n];
        for (int i = 0; i < n; i++) {
            int id = sc.nextInt();
            int c = sc.nextInt();
            int mScore = sc.nextInt();
            int e = sc.nextInt();
            students[i] = new Student(id, c, mScore, e);
        }

        // 计算各科排名
        for (int now = 0; now < 4; now++) {
            final int currentSubject = now; // 为Lambda表达式创建final变量
            // 按当前科目降序排序
            Arrays.sort(students, (a, b) -> b.grade[currentSubject] - a.grade[currentSubject]);
            
            // 计算排名(处理同分情况)
            int[] ranks = new int[n];
            ranks[0] = 1;
            for (int i = 1; i < n; i++) {
                if (students[i].grade[currentSubject] == students[i-1].grade[currentSubject]) {
                    ranks[i] = ranks[i-1];
                } else {
                    ranks[i] = i + 1;
                }
            }
            
            // 存储排名到Map
            for (int i = 0; i < n; i++) {
                int[] studentRank = rankMap.computeIfAbsent(students[i].id, k -> new int[4]);
                studentRank[currentSubject] = ranks[i];
            }
        }

        // 处理查询
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < m; i++) {
            int queryId = sc.nextInt();
            int[] ranks = rankMap.get(queryId);
            if (ranks == null) {
                sb.append("N/A\n");
            } else {
                int bestRank = Integer.MAX_VALUE;
                int bestCourseIndex = 0;
                // 找出最优排名(多个科目并列时按A>C>M>E优先级)
                for (int j = 0; j < 4; j++) {
                    if (ranks[j] < bestRank) {
                        bestRank = ranks[j];
                        bestCourseIndex = j;
                    }
                }
                sb.append(bestRank).append(" ").append(course[bestCourseIndex]).append("\n");
            }
        }
        System.out.print(sb);
        sc.close();
    }
}

C++版本

cpp 复制代码
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

struct Student{
	int id;
	int grade[4];
}stu[2010]; 

char course[4] = {'A','C','M','E'};
int Rank[10000000][4]={0};
int now; //cmp函数中使用

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

int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	//读入分数
	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;
	}
	for(now=0;now<4;now++){
		sort(stu,stu+n,cmp);
		Rank[stu[0].id][now]=1;//分数最大的 
		for(int i=1;i<n;i++){//对于剩下的学生 
			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
	for(int i=0;i<m;i++){
		scanf("%d",&query);
		if(Rank[query][0]==0){
			printf("N/A\n");
		}else{
			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]);
		}
	} 
}
相关推荐
BK贝壳3 小时前
亿级用户签到位图选型复盘:用 bool-hybrid-array 扛冷启动打卡,也把它和 Roaring/Redis 的边界测清楚了
数据结构·python·性能优化·numpy
2601_966949654 小时前
如何根据研究的起止时间截取分钟级行情?Python 实战与 QuantDash 数据处理指南
开发语言·python·pandas·量化交易·1024程序员节·股票数据·quantdash
晨曦中的暮雨6 小时前
限流算法——固定窗口、滑动窗口、漏桶、令牌桶
java·开发语言·算法
喵了几个咪7 小时前
GoWind CMS 拆解:在 Go 里给 TipTap 搭一套 Yjs 协同后端
开发语言·后端·golang
程序员清风7 小时前
数据处理脚本工程化:日志、配置、命令行与测试
服务器·开发语言·数据库
在世修行8 小时前
智学文枢:页面切换与跨页数据预填
python·桌面应用·signal·1024程序员节·pyside·跨页跳转·状态预填
2601_962885729 小时前
如何用 Python 把 K 线重采样成任意周期?
开发语言·python
打工仔折腾 AI9 小时前
在 Ubuntu ARM64 部署 Q1 V3.0 MuJoCo 模型:加载、单步推理与执行器响应解析
linux·运维·人工智能·python·深度学习·ubuntu·性能优化
Wang's Blog9 小时前
Java 项目实战: 交友项目 - 圈子功能实现:总结
java·开发语言·交友
信誓旦旦的程序猿10 小时前
【Python量化实战宝典 #21】北向资金跟随手册:聪明钱择时信号(Python 可跑)
python·1024程序员节·股票数据api·股票数据api接口·股票api数据接口·股票量化数据api·股票量化数据接口