Problem Description
Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file, then the system compare the two files. If the two files are absolutly same, then the Judge System return "Accepted", else if the only differences between the two files are spaces(' '), tabs('\t'), or enters('\n'), the Judge System should return "Presentation Error", else the system will return "Wrong Answer".
Given the data of correct output file and the data of user's result file, your task is to determine which result the Judge System will return.
 
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case has two parts, the data of correct output file and the data of the user's result file. Both of them are starts with a single line contains a string "START" and end with a single line contains a string "END", these two strings are not the data. In other words, the data is between the two strings. The data will at most 5000 characters.
 
Output
For each test cases, you should output the the result Judge System should return.
 
Sample Input
4
START
1 + 2 = 3
END
START
1+2=3
END
START
1 + 2 = 3
END
START
1 + 2 = 3
 
END
START
1 + 2 = 3
END
START
1 + 2 = 4
END
START
1 + 2 = 3
END
START
1 + 2 = 3
END
 
Sample Output
Presentation Error
Presentation Error
Wrong Answer
Presentation Error
 
 
//开始怀疑自己的智商了
//'\t'水平制表(跳到下一个tab位置)
 
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std; int main()
{
char data1[],ch[],data2[],p1[],p2[];
int t;
cin>>t;
getchar();
while(t--)
{
memset(data1,,sizeof(data1));
memset(data2,,sizeof(data2));
while(gets(ch))
{
if(strcmp(ch,"\0")==)
{ch[]=' ';
ch[]='\0';}
strcat(data1,ch);
if(strcmp(ch,"END")==)
break;
}
while(gets(ch))
{
if(strcmp(ch,"\0")==)
{ch[]=' ';
ch[]='\0';}
strcat(data2,ch);
if(strcmp(ch,"END")==)
break;
}
//cout<<data1<<endl;
//cout<<data2<<endl;
if(strcmp(data1,data2)==)
cout<<"Accepted"<<endl;
else
{
int j,k;
j=;k=;
for(int i=;i<strlen(data1);i++)
{
if(data1[i]!=' '&&data1[i]!='\t')
{
p1[j]=data1[i];
j++;
}
}
p1[j]='\0';
for(int i=;i<strlen(data2);i++)
{
if(data2[i]!=' '&&data2[i]!='\t')
{
p2[k]=data2[i];
k++;
}
}
p2[k]='\0';
//cout<<p1<<endl;
//cout<<p2<<endl;
if(strcmp(p1,p2)==)
cout<<"Presentation Error"<<endl;
else
cout<<"Wrong Answer"<<endl;
} }
return ;
}
 

HDU1073:Online Judge的更多相关文章

  1. Gym 101102C---Bored Judge(区间最大值)

    题目链接 http://codeforces.com/gym/101102/problem/C problem description Judge Bahosain was bored at ACM ...

  2. NOJ 1074 Hey Judge(DFS回溯)

    Problem 1074: Hey Judge Time Limits:  1000 MS   Memory Limits:  65536 KB 64-bit interger IO format: ...

  3. 【教程】如何正确的写一个Lemon/Cena的SPJ(special judge)

    转自:http://www.cnblogs.com/chouti/p/5752819.html Special Judge:当正确的输出结果不唯一的时候需要的自定义校验器 首先有个框架 #includ ...

  4. 九度 Online Judge 之《剑指 Offer》一书相关题目解答

    前段时间准备华为机试,正好之前看了一遍<剑指 Offer>,就在九度 Online Judge 上刷了书中的题目,使用的语言为 C++:只有3题没做,其他的都做了. 正如 Linus To ...

  5. UVa Online Judge 工具網站

    UVa Online Judge 工具網站   UVa中译题uHuntAlgorithmist Lucky貓的ACM園地,Lucky貓的 ACM 中譯題目 Mirror UVa Online Judg ...

  6. [swustoj 1021] Submissions of online judge

    Submissions of online judge(1021) 问题描述 An online judge is a system to test programs in programming c ...

  7. HDOJ/HDU 1073 Online Judge(字符串处理~)

    Problem Description Ignatius is building an Online Judge, now he has worked out all the problems exc ...

  8. write a macro to judge big endian or little endian

    Big endian means the most significant byte stores first in memory. int a=0x01020304, if the cpu is b ...

  9. UVA 489-- Hangman Judge(暴力串处理)

     Hangman Judge  In ``Hangman Judge,'' you are to write a program that judges a series of Hangman gam ...

随机推荐

  1. c++中小项堆声明和使用【转】

    c++默认是大顶堆,小顶堆有两种声明方法: 1.对于基本类型直接用 priority_queue<int, vector<int>, greater<int> >p ...

  2. 第二次讨论——响应式设计、布局技巧、css性能优化、css预处理

    第二次讨论 [响应式设计] 集中创建页面的图片排版大小,可以智能地根据用户行为以及使用的设备环境(系统平台.屏幕尺寸.屏幕定向等)进行相对应的布局. 响应式布局: meta标签的实用:设置布局宽度等于 ...

  3. Linux 相关的error处理

    1  dpkg: error: duplicate file trigger interest for filename Notice the first and last lines of /var ...

  4. wxpython 安装教程

    wxpython在windows 上的安装,需要在wxpython官网上下载对应的版本:Python分为32和64位系统不是系统的32位和64位 所以可以先在IDE 下输入Python看下当前是32还 ...

  5. On the first day here

    记录一下到这里的第一天 简单的自我介绍一下: 姓名: 郑超杰 昵称: 蝴蝶 English Nickname:   developerbfl     技能: OC  H5  Swift         ...

  6. 安卓Activity、service是否处于同一进程

    Activity与Service是否处于同一进程?   1)默认情况下(不写android:process的时候),此时同一个应用程序的所有组建位于同一进程里,Activity与service也处于同 ...

  7. 2016年团体程序设计天梯赛-决赛 L1-5. 是不是太胖了(5)

    据说一个人的标准体重应该是其身高(单位:厘米)减去100.再乘以0.9所得到的公斤数.已知市斤是公斤的两倍.现给定某人身高,请你计算其标准体重应该是多少?(顺便也悄悄给自己算一下吧……) 输入格式: ...

  8. setsockopt、getsockopt详细介绍(转)

    本文只讲解socket编程中的 setsockopt.getsockopt两个函数.关于其它函数的介绍可以参考下面链接:http://blog.sina.com.cn/s/blog_a459dcf50 ...

  9. Ambari删除服务

    ambari-cassandra-service上面推荐的方法: curl -u admin:$PASSWORD -i -H 'X-Requested-By: ambari' -X PUT -d '{ ...

  10. c++的复制构造函数

    在C++中,下面三种对象需要调用拷贝构造函数(有时也称“复制构造函数”): 1) 一个对象作为函数参数,以值传递的方式传入函数体: 2) 一个对象作为函数返回值,以值传递的方式从函数返回: 3) 一个 ...