1. 本题用java极容易超时,提交了好几次才成功
  2. 另外90 88 77 77 50,名次应该是1 2 3 3 5 ,不是1 2 3 3 4
import java.io.*;

public class Main {
@SuppressWarnings("unchecked")
public static void main(String[] args) throws IOException {
StreamTokenizer st = new StreamTokenizer(new InputStreamReader(System.in));
int n, m;
st.nextToken();
n = (int) st.nval;
st.nextToken();
m = (int) st.nval;
int[] id = new int[n];
int[][] score = new int[n][4];
int[][] pl = new int[4][101];
for (int i = 0; i < n; i++) {
st.nextToken();
id[i] = (int) st.nval;
int avg = 0;
for (int j = 0; j < 3; j++) {
st.nextToken();
score[i][j] = (int) st.nval;
avg += score[i][j];
pl[j][score[i][j]]++;
}
score[i][3] = (int) (Math.round((double) avg / 3));
pl[3][score[i][3]]++;
}
int[][] rank = new int[n][4];
for (int i = 0; i < 4; i++) {
int[] sum = new int[101];
sum[100] = 0;
for (int j = 99; j >= 0; j--) {
sum[j] = sum[j + 1] + pl[i][j + 1];
} for (int j = n - 1; j >= 0; j--) {
rank[j][i] = sum[score[j][i]] + 1;
} } StringBuilder[] rankmap = new StringBuilder[1_000_000];
for (int i = 0; i < n; i++) {
int min = n + 1;
for (int j = 0; j < 4; j++) {
if (min > rank[i][j]) {
min = rank[i][j];
}
} StringBuilder sb = new StringBuilder();
sb.append(min).append(" ");
if (rank[i][3] == min) {
sb.append("A");
rankmap[id[i]] = sb;
} else if (rank[i][0] == min) {
sb.append("C");
rankmap[id[i]] = sb;
} else if (rank[i][1] == min) {
sb.append("M");
rankmap[id[i]] = sb;
} else if (rank[i][2] == min) {
sb.append("E");
rankmap[id[i]] = sb;
}
} PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
StringBuilder sb = new StringBuilder();
for (int i = 0; i < m; i++) {
st.nextToken();
int searchedid = (int) st.nval;
if (rankmap[searchedid] == null) {
sb.append("N/A");
} else {
sb.append(rankmap[searchedid]);
}
sb.append("\n");
}
pw.print(sb);
pw.flush();
}
}

简单优化下:

import java.io.*;

public class Main {
static char[] arr = new char[]{'C', 'M', 'E', 'A'};
@SuppressWarnings("unchecked")
public static void main(String[] args) throws IOException {
StreamTokenizer st = new StreamTokenizer(new InputStreamReader(System.in));
int n, m;
st.nextToken();
n = (int) st.nval;
st.nextToken();
m = (int) st.nval;
int[] id = new int[n];
int[][] score = new int[n][4];
int[][] pl = new int[4][101];
for (int i = 0; i < n; i++) {
st.nextToken();
id[i] = (int) st.nval;
int avg = 0;
for (int j = 0; j < 3; j++) {
st.nextToken();
score[i][j] = (int) st.nval;
avg += score[i][j];
pl[j][score[i][j]]++;
}
score[i][3] = (int) (Math.round((double) avg / 3));
pl[3][score[i][3]]++;
}
int[][] rank = new int[n][4];
for (int i = 0; i < 4; i++) {
int[] sum = new int[101];
sum[100] = 0;
for (int j = 99; j >= 0; j--) {
sum[j] = sum[j + 1] + pl[i][j + 1];
} for (int j = n - 1; j >= 0; j--) {
rank[j][i] = sum[score[j][i]] + 1;
} } StringBuilder[] rankmap = new StringBuilder[1_000_000]; for (int i = 0; i < n; i++) {
int min = n + 1;
int index = 0;
for (int j = 0; j < 4; j++) {
if (min > rank[i][j] || (j == 3 && min == rank[i][j])) {
min = rank[i][j];
index = j;
}
} StringBuilder sb = new StringBuilder();
sb.append(min).append(" ").append(arr[index]);
rankmap[id[i]] = sb;
} PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
StringBuilder sb = new StringBuilder();
for (int i = 0; i < m; i++) {
st.nextToken();
int searchedid = (int) st.nval;
if (rankmap[searchedid] == null) {
sb.append("N/A");
} else {
sb.append(rankmap[searchedid]);
}
sb.append("\n");
}
pw.print(sb);
pw.flush();
}
}

  

PAT 甲级【1012 The Best Rank】的更多相关文章

  1. PAT甲级1012. The Best Rank

    PAT甲级1012. The Best Rank 题意: 为了评估我们第一年的CS专业学生的表现,我们只考虑他们的三个课程的成绩:C - C编程语言,M - 数学(微积分或线性代数)和E - 英语.同 ...

  2. PAT 甲级 1012 The Best Rank

    https://pintia.cn/problem-sets/994805342720868352/problems/994805502658068480 To evaluate the perfor ...

  3. PAT 甲级 1012 The Best Rank (25 分)(结构体排序)

    题意: 为了评估我们第一年的CS专业学生的表现,我们只考虑他们的三个课程的成绩:C - C编程语言,M - 数学(微积分或线性代数)和E - 英语.同时,我们鼓励学生强调自己的最优秀队伍 - 也就是说 ...

  4. PAT甲级——1012 The Best Rank

    PATA1012 The Best Rank To evaluate the performance of our first year CS majored students, we conside ...

  5. PAT——甲级1012:The Best Rank(有坑)

    1012 The Best Rank (25 point(s)) To evaluate the performance of our first year CS majored students, ...

  6. PAT甲 1012. The Best Rank (25) 2016-09-09 23:09 28人阅读 评论(0) 收藏

    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 eval ...

  7. pat甲级1012

    1012 The Best Rank (25)(25 分) To evaluate the performance of our first year CS majored students, we ...

  8. 【PAT】1012. The Best Rank (25)

    题目链接: http://pat.zju.edu.cn/contests/pat-a-practise/1012 题目描述: To evaluate the performance of our fi ...

  9. PAT甲级——A1012 The Best Rank

    To evaluate the performance of our first year CS majored students, we consider their grades of three ...

  10. PAT甲级1012题解——选择一种合适数据存储方式能使题目变得更简单

    题目分析: 本题的算法并不复杂,主要是要搞清楚数据的存储方式(选择一种合适的方式存储每个学生的四个成绩很重要)这里由于N的范围为10^6,故选择结构体来存放对应下标为学生的id(N只有2000的范围, ...

随机推荐

  1. 6.用户输入和 while 循环--《Python编程:从入门到实践》

    6.1 input 函数 函数input()接受一个参数:即要向用户显示的提示或说明.input 将用户输入解释为字符串. name = input("Please enter your n ...

  2. Burnside引理和Pólya定理

    不想写很多冗杂的群论定义,所以本博客不是用来入门的. 如果你想要入门,请点这里. 概要 对于一个作用在集合 \(X\) 上的有限群 \(G\) ,对于每个 \(g\in G\) 令 \(X^g\) 表 ...

  3. Ubuntu22.04 将EFI启动分区迁移到另一块硬盘

    机器上有两块硬盘, 一块已经安装了Win10, 另一块新装Ubuntu22.04, 在新硬盘上划分分区的时候, 有分出256M给 BOOT EFI, 但是安装的时候没注意, 启动分区不知道怎的跑到 W ...

  4. PLSQL编译存储过程无响应

    解决方法如下: 1:查V$DB_OBJECT_CACHE SELECT * FROM V$DB_OBJECT_CACHE WHERE name='CRM_LASTCHGINFO_DAY' AND LO ...

  5. js与java使用AES加密算法实现前后端加密解密

    AES加密算法入门:https://blog.csdn.net/IndexMan/article/details/87284833 第三方crypto.js下载地址:https://download. ...

  6. 石子合并(区间dp+记忆化搜索)

    经典例题:石子合并 题目链接 N 堆石子排成一行,现要将石子有次序地合并成一堆,规定每次只能选相邻的2堆合并成新的一堆,并将新的一堆的石子数,记为该次合并的得分.计算合并最小得分. 方法一.区间dp ...

  7. git回退至指定版本,并更新远程仓库

    1. git log   查到commit记录 2.复制 commit 后面的id 3. git reset --hard  commit 后面的id   // 回退 4. 强制更新远程仓库  git ...

  8. CentOS 8安装RabbitMQ

    第一步:安装yum仓库 导入签名KEY: ## primary RabbitMQ signing key ## 这一步如果因为网络问题下载不成功,可以先将签名文件下载下来,本地导入 rpm --imp ...

  9. 阿里云 SMS 短信 Java SDK 封装

    Github & Issues: https://github.com/cn-src/aliyun-sms 官方文档:https://help.aliyun.com/document_deta ...

  10. Flex 弹性盒子布局

    可以少去理解一些不必要的概念,而多去思考为什么会有这样的东西,它解决了什么问题,或者它的运行机制是什么? 1. 弹性盒子布局概念 Flex 是 Flexible Box 的缩写,意为"弹性布 ...