1141 PAT Ranking of Institutions (25 分)

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 (≤10​5​​), 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, Athe 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

题目大意:给出学生的id、分数、学校了,然后计算学校的相关排名,通过计算本考场内考生的等级分数,等最后按照一个规则排名。

//这个题确实挺麻烦的。

#include <iostream>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
struct Sch{
int sb,sa,st;
int fin,stu;
string name;
Sch(string n){
name=n;
sb=;sa=;st=;stu=;}
void cal(){//但是如何取整数部分呢?如果直接赋值那应该是4舍5入的。
fin=sb/1.5+sa+st*1.5;
}
}; string lower(string s){
for(int i=;i<s.size();i++){
if(s[i]>='A'&&s[i]<='Z'){
s[i]=s[i]-'A'+'a';
}
}
return s;
} map<string,int> name2id;
vector<Sch> vsch; void add(string id,int index,int score){
if(id[]=='A')
vsch[index].sa+=score;
else if(id[]=='B')
vsch[index].sb+=score;
else vsch[index].st+=score;
} bool cmp(Sch& a,Sch& b){//排序函数。
if(a.fin>b.fin)return true;
else if(a.fin==b.fin&&a.stu<b.stu)return true;
else if(a.fin==b.fin&&a.stu==b.stu&&a.name<b.name)return true;
} int main() {
int n;
cin>>n;
string id,sh;
int sco,ct=;//ct表示有多少个学校。
for(int i=;i<n;i++){
cin>>id>>sco>>sh;
sh=lower(sh);//函数是如何转换大小写?
if(name2id.count(sh)==){//如果当前机构不存在的话
name2id[sh]=ct++;
vsch.push_back(Sch(sh));//把机构的名字传进去。
add(id,ct-,sco);
}else{//如果已经存在
int temp=name2id[sh];
add(id,temp,sco);
vsch[temp].stu++;//又多了一个学生。
}
}
int size=name2id.size();
cout<<size<<'\n';
//计算得分;
for(int i=;i<vsch.size();i++){
vsch[i].cal();
}
sort(vsch.begin(),vsch.end(),cmp);
int rank=;
for(int i=;i<size;i++){
if(i!=){
if(vsch[i].fin!=vsch[i-].fin)
rank++;//这时候排名才++;
//其他情况排名不++
}
cout<<rank<<" "<<vsch[i].name<<" "<<vsch[i].fin<<" "<<vsch[i].stu<<'\n'; }
return ;
}

wrong

//这个我写了将近一个小时,但是最终第一次提交结果如下:

真是心里凉凉,待会再看柳神的吧。

AC了:

#include <iostream>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
struct Sch{
int sb,sa,st;
int fin,stu;
string name;
Sch(string n){
name=n;
sb=;sa=;st=;stu=;}
void cal(){//但是如何取整数部分呢?如果直接赋值那应该是4舍5入的。
fin=sb/1.5+sa+st*1.5;
}
}; string lower(string s){
for(int i=;i<s.size();i++){
if(s[i]>='A'&&s[i]<='Z'){
s[i]=s[i]-'A'+'a';
}
}
return s;
} map<string,int> name2id;
vector<Sch> vsch; void add(string id,int index,int score){
if(id[]=='A')
vsch[index].sa+=score;//都是int型的操作。
else if(id[]=='B')
vsch[index].sb+=score;
else vsch[index].st+=score;
} bool cmp(Sch& a,Sch& b){//排序函数。
if(a.fin>b.fin)return true;
else if(a.fin==b.fin&&a.stu<b.stu)return true;
else if(a.fin==b.fin&&a.stu==b.stu&&a.name<b.name)return true;
return false;
} int main() {
int n;
cin>>n;
string id,sh;
int sco,ct=;//ct表示有多少个学校。
for(int i=;i<n;i++){
cin>>id>>sco>>sh;
sh=lower(sh);//函数是如何转换大小写?
if(name2id.count(sh)==){//如果当前机构不存在的话
name2id[sh]=ct++;
vsch.push_back(Sch(sh));//把机构的名字传进去。
add(id,ct-,sco);
}else{//如果已经存在
int temp=name2id[sh];
add(id,temp,sco);
vsch[temp].stu++;//又多了一个学生。
}
}
int sz=name2id.size();
cout<<sz<<'\n';
//计算得分;
for(int i=;i<vsch.size();i++){
vsch[i].cal();
}
sort(vsch.begin(),vsch.end(),cmp);
int rank=;
for(int i=;i<sz;i++){
if(i!=){
if(vsch[i].fin!=vsch[i-].fin)
rank=i+;//这时候排名才++;
//其他情况排名不++
}
cout<<rank<<" "<<vsch[i].name<<" "<<vsch[i].fin<<" "<<vsch[i].stu<<'\n'; }
return ;
}

1.在最终输出排名时,一开始写的rank++,明显不对,而应该是rank=i+1才对!

2.另外非常重要的是这个cmp函数里,最后一定要有一个return false,不然会出大问题!

3.tolower函数是对单个char来使用的。

1141 PAT Ranking of Institutions[难]的更多相关文章

  1. PAT 甲级 1141 PAT Ranking of Institutions

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

  2. [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 ...

  3. 1141 PAT Ranking of Institutions (25 分)

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

  4. PAT 1141 PAT Ranking of Institutions

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

  5. 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 ...

  6. PAT_A1141#PAT Ranking of Institutions

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

  7. A1141. PAT Ranking of Institutions

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

  8. 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 ...

  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. phpstrom xdebug wamp调试配置文档

    下载并安装phpstorm,下载地址如下 http://download-cf.jetbrains.com/webide/PhpStorm-9.0.2.exe 安装完成后,完成注册,注册方法如下   ...

  2. IOS 中微信 网页授权报 key[也就是code]失效 解决办法

    枪魂微信平台ios手机点击返回 网页授权失败,报key失效.已经解决,原因是授权key只能使用一次,再次使用就会失效. 解决办法:第一次从菜单中进行授权时,用session记录key和open_id. ...

  3. 002servlet生命周期以及有关servlet的各种知识

    4 Sevlet的生命周期(重点) 有关servlet的类有Servlet,HttpServlet以及GenericServlet. 其实我们要写一个Servlet只要写一个类去实现Servet就可以 ...

  4. iScroll框架的使用和修改

    iScroll 的诞生是因为手机 Webkit 浏览器(iPhone.iPod.Android 和 Pre)本身没有为固定宽度和高度的元素提供滚动内容的方法.这导致了很多网页使用 position:a ...

  5. 【BZOJ】1096: [ZJOI2007]仓库建设(dp+斜率优化)

    http://www.lydsy.com/JudgeOnline/problem.php?id=1096 首先得到dp方程(我竟然自己都每推出了QAQ)$$d[i]=min\{d[j]+cost(j+ ...

  6. Swift AVFoundation 二维码扫描和生成

    项目最终不须要支持iOS6了(泪崩),在二维码扫描这一块,可以全然的放弃ZXing库,改用系统的AVFoundation了,拿swift写了个Demo,效果例如以下: github地址:点这里 有关A ...

  7. JSON美化输出

    echo '{"a": 1, "b": 2}' | python -m json.tool 转自: http://blog.csdn.net/chosen0ne ...

  8. MP 及OMP算法解析

    转载自http://blog.csdn.net/pi9nc/article/details/18655239 1,MP算法[盗用2] MP算法是一种贪心算法(greedy),每次迭代选取与当前样本残差 ...

  9. WinForm------如何打开子窗体的同时关闭父窗体

    方法: 如何打开子窗体的同时关闭父窗体 this.Hide(); new Frm_Management().ShowDialog(); this.Close();

  10. IOS实现打电话后回调

    本文转载至 http://blog.csdn.net/cerastes/article/details/38340687   UIWebView *callWebview =[[UIWebView a ...