PAT 甲级 1141 PAT Ranking of Institutions
https://pintia.cn/problem-sets/994805342720868352/problems/994805344222429184
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
代码:
#include <bits/stdc++.h>
using namespace std; const int maxn = 1e5 + 10;
int N;
map<string, int> mp; string lowercase(string c) {
string ans = "";
int len = c.length();
for(int i = 0; i < len; i ++) {
if(c[i] >= 'A' && c[i] <= 'Z')
ans += (c[i] + 32);
else ans += c[i];
}
return ans;
} struct Node{
int st;
int num;
double score = 0;
string name;
}node[maxn]; bool cmp(const Node &a, const Node &b) {
if((int)a.score != (int)b.score)
return (int)a.score > (int)b.score;
else if((int)a.score == (int)b.score) {
if(a.num != b.num)
return a.num < b.num;
else return a.name < b.name;
}
} int main() {
scanf("%d", &N);
mp.clear();
int cnt = 0;
while(N --) {
char s[10]; int x; string namee;
scanf("%s%d", s, &x);
cin >> namee; namee = lowercase(namee);
if(mp[namee] == 0) {
cnt ++;
mp[namee] = cnt;
} node[mp[namee]].num ++;
node[mp[namee]].name = namee;
if(s[0] == 'B') node[mp[namee]].score += (2.0 * x / 3);
else if(s[0] == 'A') node[mp[namee]].score += 1.0 * x;
else if(s[0] == 'T') node[mp[namee]].score += (3.0 * x / 2);
} sort(node + 1, node + 1 + cnt, cmp); cout << cnt << endl; node[1].st = 1;
cout << node[1].st << " " << node[1].name << " " << (int)node[1].score << " " << node[1].num << endl;
for(int i = 2; i <= cnt; i ++) {
if((int)node[i].score == (int)node[i - 1].score)
node[i].st = node[i - 1].st;
else node[i].st = i; cout << node[i].st << " " << node[i].name << " " << (int)node[i].score << " " << node[i].num << endl;
}
return 0;
}
第一次交最后一个测试点没过 要用 double 算 score 然后强制类型转换 (int)
FHFHFH
PAT 甲级 1141 PAT Ranking of Institutions的更多相关文章
- PAT甲级1141 Ranking of Institutions
题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805344222429184 题意: 给定几个学生的PAT分数和学校 ...
- PAT 甲级 1025 PAT Ranking
1025. PAT Ranking (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Programmi ...
- PAT甲级——1025 PAT Ranking
1025 PAT Ranking Programming Ability Test (PAT) is organized by the College of Computer Science and ...
- PAT 甲级 1025.PAT Ranking C++/Java
Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Z ...
- PAT 甲级1025 PAT Ranking (25 分)(结构体排序,第一次超时了,一次sort即可小技巧优化)
题意: 给定一次PAT测试的成绩,要求输出考生的编号,总排名,考场编号以及考场排名. 分析: 题意很简单嘛,一开始上来就,一组组输入,一组组排序并记录组内排名,然后再来个总排序并算总排名,结果发现最后 ...
- PAT甲级——A1025 PAT Ranking
Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhe ...
- PAT甲级1075 PAT Judge
题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805393241260032 题意: 有m次OJ提交记录,总共有k道 ...
- PAT 甲级 1075 PAT Judge (25分)(较简单,注意细节)
1075 PAT Judge (25分) The ranklist of PAT is generated from the status list, which shows the scores ...
- PAT甲级——A1075 PAT Judge
The ranklist of PAT is generated from the status list, which shows the scores of the submissions. Th ...
随机推荐
- python模拟随机游走
在python中,可以利用数组操作来模拟随机游走. 下面是一个单一的200步随机游走的例子,从0开始,步长为1和-1,且以相等的概率出现.纯Python方式实现,使用了内建的 random 模块: # ...
- jzoj5341 捕老鼠
Description 为了加快社会主义现代化,建设新农村,农夫约(Farmer Jo)决定给农庄里的仓库灭灭鼠.于是,猫被农夫约派去捕老鼠. 猫虽然擅长捕老鼠,但是老鼠们太健美了,身手敏捷,于是猫想 ...
- python并发编程之多进程理论知识
一 什么是进程 进程:正在进行的一个过程或者说一个任务.而负责执行任务则是cpu. 举例(单核+多道,实现多个进程的并发执行): egon在一个时间段内有很多任务要做:python备课的任务,写书的任 ...
- iframe的简单使用方法
1.父页面调用子页面的元素(a代表iframe的id或者class,b代表子页面) $('a').contents().find("b") 2.子页面调用父页面的元素(c代表父页面 ...
- Python学习过程笔记整理(三)
函数 -函数使用 -函数需要先定义,定义不会执行函数 -使用函数,俗称调用 -定义函数 -格式:def 函数名称(参数=默认值):,函数名称不能用大驼峰,参数可以没有 -调用函数 -格式:函数名(参数 ...
- 15-RUN vs CMD vs ENTRYPOINT
RUN.CMD 和 ENTRYPOINT 这三个 Dockerfile 指令看上去很类似,很容易混淆.本节将通过实践详细讨论它们的区别. 简单的说: RUN 执行命令并创建新的镜像层,RUN 经常用于 ...
- PHP has encountered an Access Violation at 01F4A622解决方法
php搭建的网站出现以下问题的解决方法分享: Z-blog,DedeCMS,Dsicuz!,PhpWind,PhpCMS,帝国CMS等都有可能出现php访问冲突问题. 今天访问网站发现出现了一个错误& ...
- getField()与getDeclaredField()的区别
Java的反射机制中,用Class的getField(String name)或getDelaredField(String name)可以得到目标类的指定属性,返回类型是Field. 但这两个是有区 ...
- hwconfig命令详解
基础命令学习目录首页 转载自系统技术非业余研究 本文链接地址: hwconfig查看硬件信息 最近经常要测试新硬件,了解硬件的具体型号和参数就非常重要,过去经常透过lspci, dmidecode, ...
- Python图形界面开发—wxPython库的布局管理及页面切换
前言 wxPython是基于Python的跨平台GUI扩展库,对wxWidgets( C++ 编写)封装实现.GUI程序的开发中界面布局是很重要的一个部分,合理的页面布局能够给予用户良好使用体验.虽然 ...