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 (≤), the total number of users, K (≤), the total number of problems, and M (≤), 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 Mlines 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 − 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 -

题意:

对于每个测试用例,您应该按照以下格式输出ranklist:

user_id total_score s[1]…s [K]

其中根据total_score计算rank,具有相同total_score的所有用户获得相同的rank;s[i]是第i个问题得到的部分分数。如果用户从未提交过问题的解决方案,则必须在相应位置打印“-”。如果用户为解决一个问题提交了多个解决方案,那么将计算最高分。

ranklist必须按秩的非递减顺序打印。对于具有相同级别的用户,必须根据完美解决的问题的数量按非递增顺序排序。如果仍然有领带,则必须按其id的递增顺序打印。对于那些从来没有提交过任何可以通过编译器的解决方案的人,或者从来没有提交过任何解决方案的人,他们不能显示在ranklist上。可以保证至少有一个用户可以显示在ranklist上。

题解:

注意点: 
1、最重要最重要的一点,一定一定要仔细看题,每一个细节,每一句话,都要认真理解。 
2、按照总分排序高低排序。 
3、总分相同,那么按照答对题目个数(完全正确,拿满分)进行排序。 
4、满足2、3条件,那么就需要按照学号进行排序。 
5、最后一个输出,一道题都没有提交或者提交但没有一道题通过编译的同学不输出。(一定要注意,提交没有通过编译和提交得零分完全是两个概念,所以在这里输出的时候一定要注意,后者是需要输出的) 
6、最后一个注意点是我自己的错,没有注意到第一个数N是user个数,那么下面提交的学生id是1-N。注意到了这一点,不至于把题想得很复杂。 
7、输出的时候,提交没有通过编译的题目输出0,没有提交的题目输出“-”。 
8、做这个题还有一个感受,一门语言,需要好好理解,深入理解,会帮助我们解题。

思路:上面的注意点都想到了的话,那么这道题并不难。First,在输入各个提交情况的时候,简单处理一下,then,计算每个user提交正确的题目数,next,调用一个sort函数进行进行排序,last 
,按照格式输出。over…
————————————————
版权声明:本文为CSDN博主「鬼鬼珊」的原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接及本声明。
原文链接:https://blog.csdn.net/qq_41658889/article/details/82717767

AC代码:

#include<bits/stdc++.h>
using namespace std;
int n,m,k;
struct node{
int id;
int f=;
int fen[]={-,-,-,-,-,-,-,-,-,-};
int ok=;
int sum=;
int rank;
}a[];
int p[];
bool cmp(node x,node y){
if(x.sum==y.sum){
if(x.ok==y.ok){
return x.id<y.id;
}else return x.ok>y.ok;
}else return x.sum>y.sum; }
int main(){
cin>>n>>k>>m;
for(int i=;i<=k;i++) cin>>p[i];
for(int i=;i<=n;i++) a[i].id=i;
for(int i=;i<=m;i++){
int x,y,z;
cin>>x>>y>>z;
if(z>=) {//如果这个人已经得过分了,那么-1也要算作0
a[x].f=;
}else{
if(a[x].fen[y]==-) a[x].fen[y]=;//只要是提交了,原来是-1,要改为0
continue;
}
if(z==p[y] && a[x].fen[y]<z) a[x].ok++;//第一次完全答对本题
if(a[x].fen[y]==-){//计算总分
a[x].fen[y]=z;
a[x].sum+=z;
}else if(a[x].fen[y]<z){
a[x].sum=a[x].sum-a[x].fen[y]+z;
a[x].fen[y]=z;
}
}
sort(a+,a++n,cmp);
a[].rank=;
for(int i=;i<=n;i++){
if(a[i].f==) continue;
if(i!=){//计算排名
if(a[i].sum==a[i-].sum) a[i].rank=a[i-].rank;
else a[i].rank=i;
}
printf("%d %05d %d ",a[i].rank,a[i].id,a[i].sum);
for(int j=;j<=k;j++){
if(a[i].fen[j]!=-) cout<<a[i].fen[j];
else cout<<"-";
if(j!=k) cout<<" ";
else cout<<endl;
}
}
return ;
}

PAT 甲级 1075 PAT Judge (25分)(较简单,注意细节)的更多相关文章

  1. 【PAT甲级】1070 Mooncake (25 分)(贪心水中水)

    题意: 输入两个正整数N和M(存疑M是否为整数,N<=1000,M<=500)表示月饼的种数和市场对于月饼的最大需求,接着输入N个正整数表示某种月饼的库存,再输入N个正数表示某种月饼库存全 ...

  2. PAT 甲级 1020 Tree Traversals (25分)(后序中序链表建树,求层序)***重点复习

    1020 Tree Traversals (25分)   Suppose that all the keys in a binary tree are distinct positive intege ...

  3. PAT 甲级 1146 Topological Order (25 分)(拓扑较简单,保存入度数和出度的节点即可)

    1146 Topological Order (25 分)   This is a problem given in the Graduate Entrance Exam in 2018: Which ...

  4. PAT 甲级 1071 Speech Patterns (25 分)(map)

    1071 Speech Patterns (25 分)   People often have a preference among synonyms of the same word. For ex ...

  5. PAT 甲级 1063 Set Similarity (25 分) (新学,set的使用,printf 输出%,要%%)

    1063 Set Similarity (25 分)   Given two sets of integers, the similarity of the sets is defined to be ...

  6. PAT 甲级 1059 Prime Factors (25 分) ((新学)快速质因数分解,注意1=1)

    1059 Prime Factors (25 分)   Given any positive integer N, you are supposed to find all of its prime ...

  7. PAT 甲级 1051 Pop Sequence (25 分)(模拟栈,较简单)

    1051 Pop Sequence (25 分)   Given a stack which can keep M numbers at most. Push N numbers in the ord ...

  8. PAT 甲级 1048 Find Coins (25 分)(较简单,开个数组记录一下即可)

    1048 Find Coins (25 分)   Eva loves to collect coins from all over the universe, including some other ...

  9. PAT 甲级 1037 Magic Coupon (25 分) (较简单,贪心)

    1037 Magic Coupon (25 分)   The magic shop in Mars is offering some magic coupons. Each coupon has an ...

随机推荐

  1. 详解MongoDB中的多表关联查询($lookup) (转)

    一.  聚合框架 聚合框架是MongoDB的高级查询语言,它允许我们通过转换和合并多个文档中的数据来生成新的单个文档中不存在的信息. 聚合管道操作主要包含下面几个部分: 命令 功能描述 $projec ...

  2. 加速Github访问

    Github 仓库的数据传输很慢,甚至可能导致仓库拉取失败.例如: remote: Enumerating objects: , done. remote: Counting objects: % ( ...

  3. Java 出现cannot be resolved to a type

    package com.sysutil.util; /* thishi duo zhu */ // dan zhshi import com.sysutil.util.*; class Example ...

  4. el-table 多列显示隐藏造成内容错乱

  5. NOIP2019 PJ 对称二叉树

    题目描述 一棵有点权的有根树如果满足以下条件,则被轩轩称为对称二叉树: 二叉树: 将这棵树所有节点的左右子树交换,新树和原树对应位置的结构相同且点权相等. 下图中节点内的数字为权值,节点外的 id 表 ...

  6. 1220 Vue与Django前后端的结合

    目录 vue的安装 Vue前端的设置 页面的分布 后台数据的替换 css样式 Django的配置 国际化配置 axios插件安装 CORS跨域问题(同源策略) 处理跨域问题: cors插件 axios ...

  7. monkey+shell命令解析总结

    结束monkey方法 Linux下: adb shell top | grep monkey windows下: 1.adb shell top | find "monkey" 5 ...

  8. LeetCode 1105. Filling Bookcase Shelves

    原题链接在这里:https://leetcode.com/problems/filling-bookcase-shelves/ 题目: We have a sequence of books: the ...

  9. S1_搭建分布式OpenStack集群_07 nova服务配置 (计算节点)

    一.服务安装(计算节点)安装软件:# yum install openstack-nova-compute -y 编辑/etc/nova/nova.conf文件并设置如下内容:# vim /etc/n ...

  10. 如何在Unity中开发Leap Motion桌面版(Non-VR)APP

    最近因需要,翻出几年前的Leapmotion感测器,准备用Unity3D做个互动APP,于是连上官网下载SDK.等下载下来一安装调试,瞬间傻眼,居然要求VR设备.我们Lab倒是不缺VR,有几套VIVE ...