A1075 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 (≤104), the total number of users, K (≤5), the total number of problems, and M (≤105), 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 -
出错点:sort范围弄错,必须maxn,不能用m,注意细节!!!
#include<bits/stdc++.h>
using namespace std; const int maxn=; int n,k,m; int problem[]; struct Record{
int stuId;
int pId;
int score;
}rec[maxn]; struct Student{
int stuId;
int total;
int grade[];
bool flag[];
int needPrint;
int solve;
int rank; }stu[maxn]; bool cmp(Student a,Student b){
if(a.needPrint!=b.needPrint)
return a.needPrint>b.needPrint;
else if(a.total!=b.total)
return a.total>b.total;
else if(a.solve!=b.solve)
return a.solve>b.solve;
else if(a.stuId!=b.stuId)
return a.stuId<b.stuId;
} int main(){ //cin>>n>>k>>m; scanf("%d%d%d",&n,&k,&m); for(int i=;i<=k;i++)
//cin>>problem[i];
scanf("%d",&problem[i]); Record temp; for(int i=;i<m;i++){
//cin>>temp.stuId>>temp.pId>>temp.score; int stuId,pId,score; scanf("%d%d%d",&stuId,&pId,&score); stu[stuId].flag[pId]=true; if(score!=-){
stu[stuId].stuId=stuId;
stu[stuId].needPrint=; if(score>stu[stuId].grade[pId]){
stu[stuId].total+=(score-stu[stuId].grade[pId]);
stu[stuId].grade[pId]=score; if(score==problem[pId])
stu[stuId].solve++; } }else{
continue;
} } sort(stu,stu+maxn,cmp); //必须用maxn,不能用m stu[].rank=; for(int i=;i<m;i++){
if(stu[i].total==stu[i-].total)
stu[i].rank=stu[i-].rank;
else
stu[i].rank=i+;
} for(int i=;i<m;i++){
if(stu[i].needPrint){
printf("%d %05d %d",stu[i].rank,stu[i].stuId,stu[i].total); // printf("%d %d %05d %d",stu[i].needPrint,stu[i].rank,stu[i].stuId,stu[i].total); for(int j=;j<=k;j++){
if(stu[i].flag[j]==false){
printf(" -");
}else{
printf(" %d",stu[i].grade[j]);
}
} printf("\n"); }else{
break;
}
} return ;
}
A1075 PAT Judge (25 分)的更多相关文章
- A1075 PAT Judge (25)(25 分)
A1075 PAT Judge (25)(25 分) The ranklist of PAT is generated from the status list, which shows the sc ...
- PTA 10-排序5 PAT Judge (25分)
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/677 5-15 PAT Judge (25分) The ranklist of PA ...
- 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 ...
- PATA1075 PAT Judge (25 分)
The ranklist of PAT is generated from the status list, which shows the scores of the submissions. Th ...
- 10-排序5 PAT Judge (25 分)
The ranklist of PAT is generated from the status list, which shows the scores of the submissions. Th ...
- 【PAT甲级】1075 PAT Judge (25 分)
题意: 输入三个正整数N,K,M(N<=10000,K<=5,M<=100000),接着输入一行K个正整数表示该题满分,接着输入M行数据,每行包括学生的ID(五位整数1~N),题号和 ...
- PTA 5-15 PAT Judge (25分)
/* * 1.主要就用了个sort对结构体的三级排序 */ #include "iostream" #include "algorithm" using nam ...
- 1040 有几个PAT (25 分)
题目链接:1040 有几个PAT (25 分) 做这道题目,遇到了新的困难.解决之后有了新的收获,甚是欣喜! 刚开始我用三个vector数组存储P A T三个字符出现的位置,然后三层for循环,根据字 ...
- 1025 PAT Ranking (25分)
1025 PAT Ranking (25分) 1. 题目 2. 思路 设置结构体, 先对每一个local排序,再整合后排序 3. 注意点 整体排序时注意如果分数相同的情况下还要按照编号排序 4. 代码 ...
随机推荐
- 树形dp+贪心+增量法+排序——cf1241E(好题)
/* 给定一棵树,每个结点最多选和其相连的k条边,问使边权和最大的策略 dp[u][0|1]用来表示u没连父边|连了父边 时u子树下的最优解 如果u不和任意一个儿子连边,那么u下的收益是tot=sum ...
- contest-20191021
文化课读的真不开心 回来竞赛 假人 sol 根据不等式有 abs(a-b)+abs(b-c)>=abs(a-c) 那么每一个都会选. 可以发现每一段只会选在端点上(否则移到端点更优). 那么dp ...
- pycharm中evaluate expression的用法
pycharm中evaluate expression的用法 首先要用debug调试模式运行程序,在代码编辑处右键debug,或着选择右上角的小虫子图标点击. 然后保证整个程序运行的时候可以中断,然后 ...
- Codeforces 340B - Maximal Area Quadrilateral (计算几何)
Codeforces Round #198 (Div. 2) 题目链接:Maximal Area Quadrilateral Iahub has drawn a set of \(n\) points ...
- HDU 6651 Final Exam (思维)
2019 杭电多校 7 1006 题目链接:HDU 6651 比赛链接:2019 Multi-University Training Contest 7 Problem Description Fin ...
- 第八篇 编写spider爬取jobbole的所有文章
通过scrapy的Request和parse,我们能很容易的爬取所有列表页的文章信息. PS:parse.urljoin(response.url,post_url)的方法有个好处,如果post_ur ...
- 矢量切片应用中geoserver与geowebcache分布式部署方案
在进行GIS项目开发中,常使用Geoserver作为开源的地图服务器,Geoserver是一个JavaEE项目,常通过Tomcat进行部署.而GeoWebCache是一个采用Java实现用于缓存WMS ...
- Vue之自建管理后台(三)登录页面
在做登录页面之前,我们必须得完成路由的设定... 按照之前的设计我们路由的文件夹是src/router 官方默认的index.js,如下: import Vue from 'vue' import R ...
- codeforces750E New Year and Old Subsequence 矩阵dp + 线段树
题目传送门 思路: 先看一个大牛的题解 题解里面对矩阵的构造已经写的很清楚了,其实就是因为在每个字符串都有固定的很多中状态,刚好可以用矩阵来表达,所以$(i,j)$这种状态可以通过两个相邻的矩阵的$m ...
- Hadoop–Task 相关
在MapReduce计算框架中,一个应用程序被划分为Map和Reduce两个计算阶段.他们分别由一个或多个Map Task 和Reduce Task组成. Map Task: 处理输入数据集合中的一片 ...