After each PAT, the PAT Center will announce the ranking of institutions based on their students' performances. Now you are asked to generate the ranklist.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤), which is the number of testees. Then N lines follow, each gives the information of a testee in the following format:

ID Score School

where ID is a string of 6 characters with the first one representing the test level: B stands for the basic level, A the advanced level and T the top level; Score is an integer in [0, 100]; and School is the institution code which is a string of no more than 6 English letters (case insensitive). Note: it is guaranteed that ID is unique for each testee.

Output Specification:

For each case, first print in a line the total number of institutions. Then output the ranklist of institutions in nondecreasing order of their ranks in the following format:

Rank School TWS Ns

where Rank is the rank (start from 1) of the institution; School is the institution code (all in lower case); ; TWS is the total weighted score which is defined to be the integer part of ScoreB/1.5 + ScoreA + ScoreT*1.5, where ScoreX is the total score of the testees belong to this institution on level X; and Ns is the total number of testees who belong to this institution.

The institutions are ranked according to their TWS. If there is a tie, the institutions are supposed to have the same rank, and they shall be printed in ascending order of Ns. If there is still a tie, they shall be printed in alphabetical order of their codes.

Sample Input:

10
A57908 85 Au
B57908 54 LanX
A37487 60 au
T28374 67 CMU
T32486 24 hypu
A66734 92 cmu
B76378 71 AU
A47780 45 lanx
A72809 100 pku
A03274 45 hypu

Sample Output:

5
1 cmu 192 2
1 au 192 3
3 pku 100 1
4 hypu 81 2
4 lanx 81 2
#include<iostream>
#include<cstdio>
#include<string>
#include<map>
#include<algorithm>
using namespace std;
typedef struct NODE{
string scId;
int Asum, Bsum, Tsum, TWS;
int stuNum, rank;
NODE(){
Asum = ; Bsum = ; Tsum = ; stuNum = ;
TWS = ;
}
}info;
bool cmp(info a, info b){
if(a.TWS != b.TWS){
return a.TWS > b.TWS;
}else{
if(a.stuNum != b.stuNum)
return a.stuNum < b.stuNum;
else return a.scId < b.scId;
}
}
info school[];
map<string, int>mp;
int pt = ;
void str2Low(string &ss){
for(int i = ; i < ss.length(); i++){
if(ss[i] >= 'A' && ss[i] <= 'Z'){
ss[i] = ss[i] - 'A' + 'a';
}
}
}
int main(){
int N;
scanf("%d", &N);
for(int i = ; i < N; i++){
string id, sch;
int point;
cin >> id >> point >> sch;
str2Low(sch);
int index;
if(mp.count(sch) == ){
mp[sch] = pt++;
index = pt - ;
school[index].scId = sch;
}else{
index = mp[sch];
}
if(id[] == 'A'){
school[index].Asum += point;
}else if(id[] == 'B'){
school[index].Bsum += point;
}else{
school[index].Tsum += point;
}
school[index].stuNum++;
}
for(int i = ; i < pt; i++){
school[i].TWS = (double)school[i].Bsum / 1.5 + (double)school[i].Asum + (double)school[i].Tsum * 1.5;
}
sort(school, school + pt, cmp);
school[].rank = ;
cout << pt << endl;
for(int i = ; i < pt; i++){
if(school[i].TWS == school[i - ].TWS){
school[i].rank = school[i - ].rank;
}else{
school[i].rank = i + ;
}
}
for(int i = ; i < pt; i++){
cout << school[i].rank << " " << school[i].scId << " " << school[i].TWS << " "<<school[i].stuNum << endl;
}
cin >> N;
return ;
}

A1141. PAT Ranking of Institutions的更多相关文章

  1. PAT A1141 PAT Ranking of Institutions (25 分)——排序,结构体初始化

    After each PAT, the PAT Center will announce the ranking of institutions based on their students' pe ...

  2. PAT_A1141#PAT Ranking of Institutions

    Source: PAT A1141 PAT Ranking of Institutions (25 分) Description: After each PAT, the PAT Center wil ...

  3. 1141 PAT Ranking of Institutions[难]

    1141 PAT Ranking of Institutions (25 分) After each PAT, the PAT Center will announce the ranking of ...

  4. PAT 甲级 1141 PAT Ranking of Institutions

    https://pintia.cn/problem-sets/994805342720868352/problems/994805344222429184 After each PAT, the PA ...

  5. [PAT] 1141 PAT Ranking of Institutions(25 分)

    After each PAT, the PAT Center will announce the ranking of institutions based on their students' pe ...

  6. 1141 PAT Ranking of Institutions (25 分)

    After each PAT, the PAT Center will announce the ranking of institutions based on their students' pe ...

  7. PAT 1141 PAT Ranking of Institutions

    After each PAT, the PAT Center will announce the ranking of institutions based on their students' pe ...

  8. 1141 PAT Ranking of Institutions

    题意:给出考生id(分为乙级.甲级和顶级),取得的分数和所属学校.计算各个学校的所有考生的带权总成绩,以及各个学校的考生人数.最后对学校进行排名. 思路:本题的研究对象是学校,而不是考生!因此,建立学 ...

  9. PAT Ranking (排名)

    PAT Ranking (排名) Programming Ability Test (PAT) is organized by the College of Computer Science and ...

随机推荐

  1. Flutter之 LimitedBox、Offstage、OverflowBox、SizedBox详解

    1. LimitedBox A box that limits its size only when it's unconstrained. 1.1 简介 LimitedBox,通过字面意思,也可以猜 ...

  2. RDD特性

  3. socket基础编程-2

    client端: import socket while True: client=socket.socket(socket.ANET,socket.SOCK_STREAM) client.conne ...

  4. windows php7 安装redis扩展

    1,首先查看phpinfo 这会决定扩展文件版本(特别注意以php版本的architecture是x86还是64为准,不能以操作系统为准): 2.根据PHP版本号,编译器版本号和CPU架构  一定要根 ...

  5. vscode運行vue和html

    html 选中html文件,右键选择view in broswer.

  6. java 中 System

    package cn.zhou.com; /* * System 类 * * 不能实列化 * * long t=System.currentTimeMillis();//用于计算程序的执行时间! * ...

  7. KKT条件

    kkt条件背下来容易.理解上还有问题 主要是lambda≥0和lambda*f(x)=0这两个条件懵逼. 下面说明一下为什么 参考:https://blog.csdn.net/newthinker_w ...

  8. Codeforces#543 div2 A. Technogoblet of Fire(阅读理解)

    题目链接:http://codeforces.com/problemset/problem/1121/A 真·阅读理解 题意就是 有n个人 pi表示他们的强度 si表示他们来自哪个学校 现在Arkad ...

  9. 进程间通信IPC与Binder机制原理

    1, Intent隐式意图携带数据 2, AIDL(Binder) 3, 广播BroadCast 4, 内容提供者ContentProvider 5,Messager(内部通过binder实现) 6, ...

  10. Calendar代替Date常用方法小记

    业务逻辑中遇到日期的相关操作,整理一下小做总结 日期操作离不开 java.util.Calendar 和 java.util.Date ,实体类中日期字段设计也可能用到java.sql.Date.但目 ...