题目描述

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.

输入格式

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.

输出格式

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.

输入样例

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

输出样例

1 C
1 M
1 E
1 A
3 A
N/A

《算法笔记》中AC答案

#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std; struct Student {
int id; //存放6位整数的ID
int grade[4]; //存放4个分数
}stu[2010]; char course[4] = {'A', 'C', 'M', 'E'}; //按优先级顺序,方便输出
int Rank[10000000][4] = {0}; //Rank[id][0] ~ Rank[id][4]为4门课对应的排名
int now; //cmp函数中使用,表示当前按now号分数排序stu数组 bool cmp(Student a, Student b) { //stu数组按now分数递减排序
return a.grade[now] > b.grade[now];
} int main() {
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE}
int n, m;
scanf("%d%d", &n, &m);
//读入分数,其中grade[0] ~ grade[3]分别代表A, C, M, E
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++) { //枚举A, C, M, E 4个中的一个
sort(stu, stu + n, cmp); //对所有考生按该分数从大到小排序
Rank[stu[0].id][now] = 1; //排序完,将分数最高的设为rank1
for(int i = 1; i < n; i++) { //对于剩下的考生
//若与前一位考生分数相同
if(stu[i].grade[now] == stu[i - 1].grade[now]) {
Rank[stu[i].id][now] = Rank[stu[i - 1].id][now]; //则他们排名相同
} else {
Rank[stu[i].id][now] = i + 1; //否则,为其设置正确的排名
}
}
}
int query; //查询的考生ID
for(int i = 0; i < m; i++) {
scanf("%d", &query);
if(Rank[query][0] == 0) { //如果这个考生ID不存在,则输出“N/A”
printf("N/A\n");
} else {
int k = 0; //选出Rank[query][0~3]中最小的(rank值越小,排名越高)
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]);
}
}
return 0;
}

PAT A1012 Best Rank(25)的更多相关文章

  1. A1012 The Best Rank (25)(25 分)

    A1012 The Best Rank (25)(25 分) To evaluate the performance of our first year CS majored students, we ...

  2. 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 ...

  3. PAT 1085 PAT单位排行(25)(映射、集合训练)

    1085 PAT单位排行(25 分) 每次 PAT 考试结束后,考试中心都会发布一个考生单位排行榜.本题就请你实现这个功能. 输入格式: 输入第一行给出一个正整数 N(≤10​5​​),即考生人数.随 ...

  4. PAT The Best Rank[未作]

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

  5. PTA PAT排名汇总(25 分)

    PAT排名汇总(25 分) 计算机程序设计能力考试(Programming Ability Test,简称PAT)旨在通过统一组织的在线考试及自动评测方法客观地评判考生的算法设计与程序设计实现能力,科 ...

  6. pat1012. The Best Rank (25)

    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. 1012 The Best Rank (25分) vector与结构体排序

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

  8. PAT A1012 The Best Rank (25 分)——多次排序,排名

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

  9. 【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 ...

随机推荐

  1. 初学c++动态联编

    先看一下什么是C++联编? 我觉得通俗的讲,用对象来访问类的成员函数就是静态联编. 那什么是动态联编: 一般是通过虚函数实现动态联编. 看一个动态联编的例子: 我比较懒,所以直接粘贴了MOOC视频的图 ...

  2. tcp_wraper&xinetd 和telnet

    一.xinetd简介 1.什么是xinetd xinetd:eXtended InterNET Daemon  扩展的互联网守护程序   xinetd是新一代的网络守护进程服务程序,又叫超级守护进程, ...

  3. 2016 NEERC, Northern Subregional Contest G.Gangsters in Central City(LCA)

    G.Gangsters in Central City 题意:一棵树,节点1为根,是水源.水顺着边流至叶子.该树的每个叶子上有房子.有q个询问,一种为房子u被强盗入侵,另一种为强盗撤离房子u.对于每个 ...

  4. 安装mongodb-window10版

    第一.下载mongodb 官方地址:https://www.mongodb.com/ 第二步mongodb安装 运行mongodb-win32-x86_64-2008plus-ssl-v3.4-lat ...

  5. @Value和@PropertySource实现*.properties配置文件读取过程和实现原理

    @Value和@PropertySource实现*.properties 配置文件读取过程和实现原理 1       配置使用步骤 (1)右击resource目录添加*.prooerties配置文件

  6. JS 小脚本汇聚

    根据文件length展示文件大小 if (bytes === 0) return '0 B'; var k = 1024, sizes = ['B', 'KB', 'MB', 'GB', 'TB', ...

  7. legend3---8、烦请节约时间

    legend3---8.烦请节约时间 一.总结 一句话总结: 时间不要浪费在垃圾情绪和无效社交上面. 1.商标不能以个人的名义注册? 可以先注册个体工商户,然后再可以注册商标 2.注册一个商标大概花多 ...

  8. Flume-Replicating Channel Selector 单数据源多出口

    使用 Flume-1 监控文件变动,Flume-1 使用 Replicating Channel Selector 将变动内容传递给 Flume-2,Flume-2 负责存储到 HDFS.同时 Flu ...

  9. Twisted框架学习

    Twisted是用Python实现的基于事件驱动的网络引擎框架,是python中一个强大的异步IO库.理解twisted的一个前提是弄清楚twisted中几个核心的概念: reactor, Proto ...

  10. golang中mysql建立连接超时时间timeout 测试

    本文测试连接mysql的超时时间. 这里的"连接"是建立连接的意思. 连接mysql的超时时间是通过参数timeout设置的. 1.建立连接超时测试 下面例子中,设置连接超时时间为 ...