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 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
就是输入输出,简单的判断PE,WA,AC这几种情况。
START是开始输入了,
END是结束输入了。数据在这个之间!
每一组有2个 START和END。
就是判断这2个之间的数据是PE,WA还是AC。
分析:
用2字符串分别存储这2个需要比较的数据,遇到换行需要在字符串中加入’\n’。
先看这2个字符串是不是相等,如果相等,就是AC。
不相等,再来判断是PE还是WA。
import java.util.Scanner;
/**
* @author 陈浩翔
*
* 2016-5-26
*/
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
sc.nextLine();
while (t-- > 0) {
String a = "";
String b = "";
while (true) {
String str = sc.nextLine();
if ("END".equals(str)) {
break;
}
if ("START".equals(str)) {
continue;
}
a=a+str;
a=a+"\n";
}
while (true) {
String str = sc.nextLine();
if ("END".equals(str)) {
break;
}
if ("START".equals(str)) {
continue;
}
b=b+str;
b=b+"\n";
}
int con = 0;// -1--PE,0--WA,1--AC
// System.out.println(a);
// System.out.println(a.length());
// System.out.println(b);
// System.out.println(b.length());
if(a.equals(b)){
con=1;
}else{
con=-1;
String stra="";
for(int i=0;i<a.length();i++){
if(a.charAt(i)==' '||a.charAt(i)=='\t'||a.charAt(i)=='\n'){
continue;
}
stra=stra+a.charAt(i);
}
String strb="";
for(int i=0;i<b.length();i++){
if(b.charAt(i)==' '||b.charAt(i)=='\t'||b.charAt(i)=='\n'){
continue;
}
strb=strb+b.charAt(i);
}
// System.out.println(stra);
// System.out.println("-------------");
// System.out.println(strb);
if(stra.equals(strb)){
con=-1;
}else{
con=0;
}
}
if(con==0){
System.out.println("Wrong Answer");
}else if(con==-1){
System.out.println("Presentation Error");
}else{
System.out.println("Accepted");
}
}
}
}
HDOJ/HDU 1073 Online Judge(字符串处理~)的更多相关文章
- HDU 1073 Online Judge (字符串处理)
题目链接 Problem Description Ignatius is building an Online Judge, now he has worked out all the problem ...
- HDU 1073 Online Judge(字符串)
Online Judge Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Tot ...
- 解题报告:hdu 1073 Online Judge
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1073 Problem Description Ignatius is building an Onli ...
- HDOJ/HDU 1062 Text Reverse(字符串翻转~)
Problem Description Ignatius likes to write words in reverse way. Given a single line of text which ...
- 【HDOJ】1073 Online Judge
这道题TLE了N多次,完全不明白为什么,稍微改了一下,居然过了.使用gets过的,看讨论帖有人还推荐用hash. #include <stdio.h> #include <strin ...
- HDU 1073 - Online Judge
模拟评测机判断答案 先判断有没有不一样的 有的话再提取出 有效子列 看看有没有错的 #include <iostream> #include <cstdio> #include ...
- HDOJ(HDU).1059 Dividing(DP 多重背包+二进制优化)
HDOJ(HDU).1059 Dividing(DP 多重背包+二进制优化) 题意分析 给出一系列的石头的数量,然后问石头能否被平分成为价值相等的2份.首先可以确定的是如果石头的价值总和为奇数的话,那 ...
- HDOJ(HDU).1258 Sum It Up (DFS)
HDOJ(HDU).1258 Sum It Up (DFS) [从零开始DFS(6)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双 ...
- HDOJ(HDU).1035 Robot Motion (DFS)
HDOJ(HDU).1035 Robot Motion [从零开始DFS(4)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DF ...
随机推荐
- 初学HTML5系列一:简单介绍
最近很闲,就想着学点东西,然后就瞄中了html5,以前只看过很简单的一些,这次是系统的学下,顺便也记录下.废话不多说,开始正题. 稍微介绍下html5,html5是W3C和WHATWG 合作的结果. ...
- js EasyUI前台 全选的实现
aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAAWcAAAEQCAIAAADj/SKjAAAgAElEQVR4nO1dz48ry1W+/5N3swaFEP ...
- PL/SQL 中文转换为\uxxxx格式(unicode)
用plsql生成json的时候,经常要处理中文的情况. SELECT replace(asciistr('你好,,hello'),'\','\u') FROM dual 输出:\u4F60\u597D ...
- CI 笔记,使用 json的参考文档(废弃)
Json的处理转换, Json转换步骤, a) 先设置json为空字符串, b) While循环,遍历, While(!!$row = mysql_fetch_arra ...
- (五)Struts2 标签
所有的学习我们必须先搭建好Struts2的环境(1.导入对应的jar包,2.web.xml,3.struts.xml) 第一节:Struts2 标签简介 Struts2 自己封装了一套标签,比JSTL ...
- 完全步卸载oracle11g步骤
完全 步 卸载oracle11g骤: . 开始->设置->控制面板->管理工具->服务 停止所有Oracle服务. . 开始->程序->Oracle - OraH ...
- C# 之【获取网页响应码200】
做了个监控网站的C#项目,判断网站是否运行正常,采用的就是获取HTTP头返回的状态码. 比如状态码为200的就是网站正常,403是Forbidden,404是网页未找到这样. 代码如下: 首先是使用库 ...
- Linux 下的 blog客户端
apt-cahce search metaweblog found gnome-blog blogilo try blogilo
- 将[{},{}]转为dict
经常遇到一种需求,需要把从数据库取出的数据,转为dict对象([{}, {},...]-->dict). rs = [{, , "name":"edf"} ...
- JavaScript学习总结【2】、JS基础
1.JS 命名规范 命名规范是很有必要的,可增强代码的可读性,一眼就能看懂要表达的意思,规范就是符合规则,使代码有利于后期维护,也能很大程度的提高开发效率.一个正常的网站有很多 JS 代码,如果在编写 ...