1075 PAT Judge (25 分)

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 positive integers, N (≤10​4​​), the total number of users, K (≤5), the total number of problems, and M (≤10​5​​), the total number of submissions. It is then assumed that the user id's are 5-digit numbers from 00001 to N, and the problem id's are from 1 to K. The next line contains K positive integers p[i] (i=1, ..., K), where p[i] corresponds to the full mark of the i-th problem. Then M lines follow, each gives the information of a submission in the following format:

user_id problem_id partial_score_obtained

where partial_score_obtained is either −1 if the submission cannot even pass the compiler, or is an integer in the range [0, p[problem_id]]. All the numbers in a line are separated by a space.

Output Specification:

For each test case, you are supposed to output the ranklist in the following format:

rank user_id total_score s[1] ... s[K]

where rank is calculated according to the total_score, and all the users with the same total_score obtain the same rank; and s[i] is the partial score obtained for the i-th problem. If a user has never submitted a solution for a problem, then "-" must be printed at the corresponding position. If a user has submitted several solutions to solve one problem, then the highest score will be counted.

The ranklist must be printed in non-decreasing order of the ranks. For those who have the same rank, users must be sorted in nonincreasing order according to the number of perfectly solved problems. And if there is still a tie, then they must be printed in increasing order of their id's. For those who has never submitted any solution that can pass the compiler, or has never submitted any solution, they must NOT be shown on the ranklist. It is guaranteed that at least one user can be shown on the ranklist.

Sample Input:

7 4 20
20 25 25 30
00002 2 12
00007 4 17
00005 1 19
00007 2 25
00005 1 20
00002 2 2
00005 1 15
00001 1 18
00004 3 25
00002 2 25
00005 3 22
00006 4 -1
00001 2 18
00002 1 20
00004 1 15
00002 4 18
00001 3 4
00001 4 2
00005 2 -1
00004 2 0

Sample Output:

1 00002 63 20 25 - 18
2 00005 42 20 0 22 -
2 00007 42 - 25 - 17
2 00001 42 18 18 4 2
5 00004 40 15 0 25 -

题目大意:输入包括n个学生,编号从00001到N,包括K个题目,m次提交,对它们进行排序:

1.一个学生多次对一个题目提交,要取分数最高的一次;

2.如果一个学生没有提交过,或者是提交了均未通过编译,也就是输入未-1,那么不计入统计中,注意:这里提交了得分为0,和-1是不一样的,前者是通过了编译但是得分为0,后者是未通过编译。

我的AC:

#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
using namespace std; struct Stu{
int id,total,perfect,pass;//pass记录通过编译的题目数,如果为0,那么不进入list。
int task[];//先存上5道题的数组。
Stu(){
fill(task,task+,-);
total=;perfect=;pass=;
}
}stu[];
int score[];
vector<Stu> vt;
bool cmp(Stu&a,Stu&b){
if(a.total>b.total)return true;
else if(a.total==b.total&&a.perfect>b.perfect)return true;
else if(a.total==b.total&&a.perfect==b.perfect) return a.id<b.id;
return false;
}
int main() {
int n,k,m;
cin>>n>>k>>m;
for(int i=;i<=k;i++){
cin>>score[i];
}
int id,tid,sco;
for(int i=;i<m;i++){
cin>>id>>tid>>sco;
if(sco!=-)stu[id].pass+=;//此处只要!=0就可以计入排序。
if(sco==-)sco=;
stu[id].task[tid]=max(sco,stu[id].task[tid]);//分数每次都取最高的那个。
stu[id].id=id;
//提交多次的要取得分最高的那次。
// if(sco!=-1)
// stu[id].total+=sco;
// if(sco==score[tid])//AC数目也不能在这里判断,有可能会多次AC
// stu[id].perfect+=1;//记录AC的题目数。
}
for(int i=;i<=n;i++){
for(int j=;j<=k;j++){
if(stu[i].task[j]!=-)
stu[i].total+=stu[i].task[j];
if(stu[i].task[j]==score[j])
stu[i].perfect+=;
}
if(stu[i].pass!=){
vt.push_back(stu[i]);
}
}
sort(vt.begin(),vt.end(),cmp);
int rank=;
printf("1 %05d %d",vt[].id,vt[].total);
for(int i=;i<=k;i++){
if(vt[].task[i]==-)
printf(" -");
else
printf(" %d",vt[].task[i]);
}
printf("\n");
for(int i=;i<vt.size();i++){
if(vt[i].total!=vt[i-].total){
rank=i+;
}
printf("%d %05d %d",rank,vt[i].id,vt[i].total);
for(int j=;j<=k;j++){
if(vt[i].task[j]==-)
printf(" -");
else
printf(" %d",vt[i].task[j]);
}
printf("\n");
}
return ;
}

//总体来说就是根据题目的要求来写,整个过程思路清晰的话,还是比较简单的。

PAT 1075 PAT Judge[比较]的更多相关文章

  1. PAT 1075. PAT Judge (25)

    题目地址:http://pat.zju.edu.cn/contests/pat-a-practise/1075 此题主要考察细节的处理,和对于题目要求的正确理解,另外就是相同的总分相同的排名的处理一定 ...

  2. PAT 1075. PAT Judge

    The ranklist of PAT is generated from the status list, which shows the scores of the submittions. Th ...

  3. 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 ...

  4. 【转载】【PAT】PAT甲级题型分类整理

    最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel P ...

  5. 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 ...

  6. [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 ...

  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. 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 ...

  9. PTA(Advanced Level)1075.PAT Judge

    The ranklist of PAT is generated from the status list, which shows the scores of the submissions. Th ...

随机推荐

  1. select2 选择框插件

    <select id="selBusi_type"><select> //初始化业务类型下拉 var initBusiTypeSel = function( ...

  2. cesium入门1

    本教程将获得所有技能水平的开发人员和他们的第一个铯应用程序运行. 验证Cesium在您的Web浏览器中工作的最简单的方法是单击此处运行Hello World示例 (打开一个新窗口).如果你看到像下面的 ...

  3. 复习前面一个月的学习C#感觉道路好艰难啊

    今天是复习前面学习的内容,感觉这一个月来真的学习了很多,但是掌握的不好,好多都是在老师讲完课后做起来练习感觉这知识用起来蛮轻松地,但是经过昨天和今天的复习发现好多还是给忘记啦,甚是失落啊,刚开始就知道 ...

  4. Nginx遇上Access Denied提示怎么解决

    这几天在摆弄linux下面的各种服务器,对nginx非常有兴趣. 于是把phpmyadmin传上去了,先是phpmyadmin配了半天,结果配好之后发现phpmyadmin一些logo.css.js文 ...

  5. 使用Node.js完成路由

    首先先看一下文件的结构: 我想通过改变不同的路由进不同的页面, 先看这几个HTML页面: 404: <!DOCTYPE html> <html lang="en" ...

  6. Size Balanced Tree

    Size Balanced Tree(SBT)是目前速度最快的平衡二叉搜索树,且能够进行多种搜索操作,区间操作:和AVL.红黑树.伸展树.Treap类似,SBT也是通过对节点的旋转来维持树的平衡,而相 ...

  7. JS-完美运动框架(封装)

    function getStyle(obj, name) { if(obj.currentStyle) { return obj.currentStyle[name]; } else { return ...

  8. Jboss AS 7 部署web应用程序时无法初始化spring的bean的解决办法

    Jboss AS 7 在部署web应用程序的时候无法初始化spring的bean(在tomcat下边不会出现这个问题) 原因是web应用程序没有导入jboss对spring的支持的jar包 解决方法: ...

  9. 【BZOJ4149】[AMPPZ2014]Global Warming 单调栈+RMQ+二分

    [BZOJ4149][AMPPZ2014]Global Warming Description 给定一个序列a[1],a[2],...,a[n].请从中选出一段连续子序列,使得该区间最小值唯一.最大值 ...

  10. innodb的innodb_buffer_pool_size和MyISAM的key_buffer_size(转自:http://www.java123.net/898181.html)

    一. key_buffer_size 对MyISAM表来说非常重要. 如果只是使用MyISAM表,可以把它设置为可用内存的 -%.合理的值取决于索引大小.数据量以及负载 -- 记住,MyISAM表会使 ...