HDU6023 Automatic Judge 2017-05-07 18:30 73人阅读 评论(0) 收藏
Automatic Judge
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 16 Accepted Submission(s): 11
A new automatic judge system is used for this competition. During the five-hour contest time, you can submit your code to the system, then the judge will reply you. Here is a list of the judge's replies and their meaning:
1. Accepted(AC):
Yes, your program is correct. You did a good job!
2. PresentationError(PE) :
Your program's output format is not exactly the same as required by the problem, although the output is correct. This usually means the existence of omitted or extra blank characters (white spaces, tab characters and/or new line characters) between any two
non-blank characters, and/or blank lines (a line consisting of only blank characters) between any two non-blank lines. Trailing blank characters at the end of each line and trailing blank lines at the of output are not considered format errors. Check the output
for spaces, blank lines, etc. against the problem's output specification.
3. WrongAnswer(WA) :
Correct solution not reached for the inputs. The inputs and outputs that we use to test the programs are not public (it is recomendable to get accustomed to a true contest dynamic :-)
4. RuntimeError(RE) :
Your program failed during the execution and you will receive the hints for the reasons.
5. TimeLimitExceeded(TLE) :
Your program tried to run during too much time.
6. MemoryLimitExceeded(MLE):
Your program tried to use more memory than the judge default settings.
7. OutputLimitExceeded(OLE):
Your program tried to write too much information. This usually occurs if it goes into a infinite loop.
8. CompilationError(CE):
The compiler fails to compile your program. Warning messages are not considered errors. Click on the judge's reply to see the warning and error messages produced by the compiler.
For each submission, if it is the first time that the judge returns ``AC'' on this problem, then it means you have passed this problem, and the current time will be added to the penalty of your team. In addition, every time you pass a problem, each unsuccessful
try for that problem before is counted as 20 minutes penalty, it should also be added to the penalty of your team.
Now given the number of problems in the contest and the submission records of a team. Please write a program to calculate the number of problems the team passed and their penalty.
denoting the number of test cases.
In each test case, there are two integers n(1≤n≤13) and m(1≤m≤100) in
the first line, denoting the number of problems and the number of submissions of a team. Problems are labeled by 1001, 1002, ..., 1000+n.
In the following m lines,
each line contains an integer x(1001≤x≤1000+n) and
two strings t(00:00≤t≤05:00) and s,
denoting the team submits problem x at
time t,
and the result is s. t is
in the format of HH:MM, while s is
in the set \{AC, PE, WA, RE, TLE, MLE, OLE\}. The team is so cautious that they never submit a CE code. It is guaranteed that all the t in
the input is in ascending order and every t is
unique.
denoting the number of problems the team passed and the penalty.
1
3 5
1002 00:02 AC
1003 00:05 WA
1003 00:06 WA
1003 00:07 AC
1002 04:59 AC
2 49
————————————————————————————————————
题意:算罚时
思路:先判断是否已AC,若AC不处理,最后把AC的题罚时加起来
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional> using namespace std; #define LL long long
const int INF=0x3f3f3f3f; int n,m;
struct node
{
int sum,flag;
} x[1200]; int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int id,h,mm;
char ch[20];
memset(x,0,sizeof x);
for(int i=1; i<=m; i++)
{
scanf("%d%d:%d%s",&id,&h,&mm,ch);
if(!strcmp(ch,"AC")&&!x[id].flag) x[id].flag=1,x[id].sum+=60*h+mm;
else if(!x[id].flag) x[id].sum+=20;
}
int cnt=0,ans=0;
for(int i=1001; i<=1000+n; i++)
if(x[i].flag) cnt++,ans+=x[i].sum;
printf("%d %d\n",cnt,ans);
}
return 0;
}
HDU6023 Automatic Judge 2017-05-07 18:30 73人阅读 评论(0) 收藏的更多相关文章
- winform 子窗体数据改变刷新父窗体 分类: WinForm 2014-05-06 18:30 246人阅读 评论(0) 收藏
两种方法实现: 第一种,传时间变量,主窗体要不停的刷新数据,占用资源比较大. 第二种,用this,感觉比较好用,建议用这种方法. 举例: 主窗体命名:FormA; 子窗体命名:FormB; 数据绑定方 ...
- c++map的用法 分类: POJ 2015-06-19 18:36 11人阅读 评论(0) 收藏
c++map的用法 分类: 资料 2012-11-14 21:26 10573人阅读 评论(0) 收藏 举报 最全的c++map的用法 此文是复制来的0.0 1. map最基本的构造函数: map&l ...
- [CS]C#操作word 2016-04-17 18:30 1506人阅读 评论(35) 收藏
最近在做的项目已经改了好几版,最近这一版用到了word,当然不是直接使用word,而是使用第三方的ActiveX控件:dsoframer.ocx,此控件的使用和其他控件的使用流程没有任何区别,接下来介 ...
- 第十二届浙江省大学生程序设计大赛-Lunch Time 分类: 比赛 2015-06-26 14:30 5人阅读 评论(0) 收藏
Lunch Time Time Limit: 2 Seconds Memory Limit: 65536 KB The 999th Zhejiang Provincial Collegiate Pro ...
- 移植QT到ZedBoard(制作运行库镜像) 交叉编译 分类: ubuntu shell ZedBoard OpenCV 2014-11-08 18:49 219人阅读 评论(0) 收藏
制作运行库 由于ubuntu的Qt运行库在/usr/local/Trolltech/Qt-4.7.3/下,由makefile可以看到引用运行库是 INCPATH = -I/usr//mkspecs/d ...
- Rebuild my Ubuntu 分类: ubuntu shell 2014-11-08 18:23 193人阅读 评论(0) 收藏
全盘格式化,重装了Ubuntu和Windows,记录一下重新配置Ubuntu过程. //build-essential sudo apt-get install build-essential sud ...
- highgui.h备查 分类: C/C++ OpenCV 2014-11-08 18:11 292人阅读 评论(0) 收藏
/*M/////////////////////////////////////////////////////////////////////////////////////// // // IMP ...
- HDU6026 Deleting Edges 2017-05-07 19:30 38人阅读 评论(0) 收藏
Deleting Edges Time ...
- Pots 分类: 搜索 POJ 2015-08-09 18:38 3人阅读 评论(0) 收藏
Pots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11885 Accepted: 5025 Special Judge D ...
随机推荐
- SVM支持向量机推导,工具介绍及python实现
支持向量机整理 参考: Alexandre KOWALCZYK大神的SVM Tutorial http://blog.csdn.net/alvine008/article/details/909711 ...
- Linux CentOS7中 设置IP地址、网关DNS
cd /etc/sysconfig/network-scripts/ #进入网络配置文件目录 vi ifcfg-eno16777736 #编辑配置文件,此处eno后边的编号因电脑而易 TYPE ...
- 把AVI存在资源中用TAnimate播放
Animate1.RESName := 'About'; Animate1.Active := True;
- localstorage是什么,它有哪些作用
localStorage作为HTML5本地存储web storage特性的API之一,主要作用是将数据保存在客户端中,而客户端一般是指上海网站设计用户的计算机.在移动设备上,由于大部分浏览器都支持 w ...
- conductor Kitchensink示例
一个示例的厨房工作流程,演示了所有模式构造的使用. 定义 { "name": "kitchensink", "description": & ...
- Group by 内部排序
1.right join # update_time gid=>sid, group_status => s_table select a.* from comment as a ri ...
- redis的五种常见数据类型的常用指令
一.String字符串,key-value 应用场景:string是redis的最基本数据类型,key-value格式,一个key对应一个值的情况下 1.设置key = value:set key ...
- JDK中rt.jar、tools.jar和dt.jar作用
dt.jar和tools.jar位于:{Java_Home}/lib/下,而rt.jar位于:{Java_Home}/jre/lib/下,其中: rt.jar是JAVA基础类库,也就是你在java d ...
- EasyUI多选的获取
function deletePRE() { var rows = $('#dg').datagrid('getSelections'); var ids = []; var other_ids = ...
- java实现24点游戏代码
import java.util.Arrays;import java.util.Scanner; public class Test07 { public static void main(S ...