Poj1298_The Hardest Problem Ever(水题)
一、Description
sound, that no one could figure it out without knowing how it worked.
You are a sub captain of Caesar's army. It is your job to decipher the messages sent by Caesar and provide to your general. The code is simple. For each letter in a plaintext message, you shift it five places to the right to create the secure message (i.e.,
if the letter is 'A', the cipher text would be 'F'). Since you are creating plain text out of Caesar's messages, you will do the opposite:
Cipher text
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Plain text
V W X Y Z A B C D E F G H I J K L M N O P Q R S T U
Only letters are shifted in this cipher. Any non-alphabetical character should remain the same, and all alphabetical characters will be upper case.
Input
be uppercase.
A single data set has 3 components
- Start line - A single line, "START"
- Cipher message - A single line containing from one to two hundred characters, inclusive, comprising a single message from Caesar
- End line - A single line, "END"
Following the final data set will be a single line, "ENDOFINPUT".
Output
二、问题分析
最近两天,哥们被三伏天折磨的想死的心都有了。于是乎,我决定找几道水题来降降温。这道号称是“史上最难的问题”的题目就是一道标准的水题。
题目很简单,密码转译问题。开始的时候以为是加密,这也算是我所唯一检查了的地方。用字符串存放输入,然后把字符串转换为字符数组。把字母分为两部分,A~F字符转换为Ascii码后加21,G~Z转换为AscII后减5。遍历每个字符,按上面的规则改变字母。
刚刚看到网友爆料,此题居然和3749一样。我还以为是和它一样水呢,没想到是一样的题,不过3749是中文的。哎呀,还省了翻译的时间,要知道水题最难之处就是翻译啊!!!
三、Java代码
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader; public class Main { public static void main(String[] args) throws IOException {
BufferedReader read=new BufferedReader(new InputStreamReader(System.in));
while(!read.readLine().equals("ENDOFINPUT")){
String s=read.readLine();
char[] c=s.toCharArray();
for(int i=0;i<c.length;i++){
int a=(int)c[i];
if(a>= 70 && a<=90){
c[i]=(char) (c[i]-5);
}else if(a>=65 && a<=69){
c[i]=(char) (c[i]+21);
}
}
System.out.println(c);
read.readLine();
}
}
}
版权声明:本文为博主原创文章,未经博主允许不得转载。
Poj1298_The Hardest Problem Ever(水题)的更多相关文章
- fzuoj Problem 2182 水题
http://acm.fzu.edu.cn/problem.php?pid=2182 Problem 2182 水题 Accept: 188 Submit: 277Time Limit: 100 ...
- 烟大 Contest1024 - 《挑战编程》第一章:入门 Problem A: The 3n + 1 problem(水题)
Problem A: The 3n + 1 problem Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 14 Solved: 6[Submit][St ...
- [POJ 1000] A+B Problem 经典水题 C++解题报告 JAVA解题报告
A+B Problem Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 311263 Accepted: 1713 ...
- HDU 4716 A Computer Graphics Problem (水题)
A Computer Graphics Problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (J ...
- poj1298 The Hardest Problem Ever 简单题
链接:http://poj.org/problem?id=1298&lang=default&change=true 简单的入门题目也有这么强悍的技巧啊!! 书上面的代码: 很厉害有没 ...
- POJ1298_The Hardest Problem Ever_最难的问题_Caesar 密码_C++
题目:http://poj.org/problem?id=1298 好吧,给了题目也看不懂……给出翻译(题目名翻译是:最难的问题,233333) 这一看就是老师给出题解: 然而没有什么用哈 最快的办法 ...
- HDOJ(HDU) 1898 Sempr == The Best Problem Solver?(水题、、、)
Problem Description As is known to all, Sempr(Liangjing Wang) had solved more than 1400 problems on ...
- zzulioj--1634--Happy Thanksgiving Day - A + B Problem(模拟水题)
1634: Happy Thanksgiving Day - A + B Problem Time Limit: 1 Sec Memory Limit: 128 MB Submit: 136 ...
- HDU 5443 The Water Problem (水题,暴力)
题意:给定 n 个数,然后有 q 个询问,问你每个区间的最大值. 析:数据很小,直接暴力即可,不会超时,也可以用RMQ算法. 代码如下: #include <cstdio> #includ ...
随机推荐
- zookeeper curator CRUD
目录 Curator客户端的基本操作 写在前面 1.1.1. Curator客户端的依赖包 1.1.2. Curator 创建会话 1.1.3. CRUD 之 Create 创建节点 1.1.4. C ...
- mysql mariadb 乱码
mysql 创建临时表 CREATE TEMPORARY TABLE tmp_table SELECT COUNT(*) AS num FROM student_info GROUP BY LEFT( ...
- java面向对象入门之方法参数的传递
/* Name : Power by :Stuart Date:2015.4.25 */ class PassOn{ //创建show方法,把i传入,输出i+1的结果 public void show ...
- hive查询注意及优化tips
Hive是将符合SQL语法的字符串解析生成可以在Hadoop上执行的MapReduce的工具.使用Hive尽量按照分布式计算的一些特点来设计sql,和传统关系型数据库有区别, 所以需要去掉原有关系型数 ...
- nginx 搭建虚拟主机
一.排错三部曲 第一步在客户端上ping服务端ip ping 10.0.0.8 第二部在客户端上telnet服务器端IP.端口 telnet 10.0.0.8 第三部在客户端使用wget命令检测 ...
- 20145229吴姗珊 《Java程序设计》小总结
20145229吴姗珊 <Java程序设计>小总结 教材学习内容总结 由于今天考试考到了操作题,根本无从下手,然后才意识到原来之前的学习都是蜻蜓点水,一味的把学习建立在给老师学,为家长学的 ...
- CustomizaitonSpec Clone_VM
克隆虚拟机可以加上CustomizationSpec来自动配置好:IP地址.DNS.Domain等信息 1.可以利用PyVmimo中的vim模块在python中完全自定义CustomizationSp ...
- Ubuntu14 下安装jdk1.8
1.oracle官网下载jdk-8u45-linux-i586.gz 此为32位 2.放在任意目录下,如/home/hongxf 3.直接解压该文件 sudo tar -xvf jdk-8u45-li ...
- HTML5学习记录——3
HTML媒体 1.HTML多媒体 视频格式 .avi 微软开发 .wmv 微软开发 .mpg .mpeg .mov 苹果公司开发 .rm .ram 允许低带宽的视频流 .swf .fl ...
- C 字节对齐.我的算法学习之路
C/C++基础笔试题1.0(字节对齐) http://blog.csdn.net/dengyaolongacmblog/article/details/37559687 我的算法学习之路 http:/ ...