(字符串 枚举)The Hardest Problem Ever hdu1048
The Hardest Problem Ever
链接:http://acm.hdu.edu.cn/showproblem.php?pid=1048
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31307 Accepted Submission(s):
14491
The hardest situation Caesar ever faced was keeping himself alive. In order for
him to survive, he decided to create one of the first ciphers. This cipher was
so incredibly 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.
series of up to 100 data sets. Each data set will be formatted according to the
following description, and there will be no blank lines separating data sets.
All characters will 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. This is the original message by Caesar.
N BTZQI WFYMJW GJ KNWXY NS F QNYYQJ NGJWNFS ANQQFLJ YMFS XJHTSI NS WTRJ
IFSLJW PSTBX KZQQ BJQQ YMFY HFJXFW NX RTWJ IFSLJWTZX YMFS MJ
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
@SuppressWarnings("resource")
Scanner inScanner = new Scanner(System.in);
while(inScanner.hasNext()) {
String string = inScanner.nextLine();
if(string.equals("ENDOFINPUT")) {
break;
}
else if(string.equals("START") || string.equals("END")) {
continue;
}
else {
for(int i = ;i<string.length();i++) {
if(string.charAt(i)>='A' && string.charAt(i)<='E') {
System.out.printf("%c",string.charAt(i)+-);
}
else if(string.charAt(i)>='F'&&string.charAt(i)<='Z'){
System.out.printf("%c", string.charAt(i)-);
}
else {
System.out.printf("%c",string.charAt(i));
}
}
}
System.out.println();
}
}
}
(字符串 枚举)The Hardest Problem Ever hdu1048的更多相关文章
- The Hardest Problem Ever(字符串)
The Hardest Problem Ever Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 24039 Accept ...
- HDUOJ-------The Hardest Problem Ever
The Hardest Problem Ever Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java ...
- hdu_1048_The Hardest Problem Ever_201311052052
The Hardest Problem Ever Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java ...
- HDU1048The Hardest Problem Ever
The Hardest Problem Ever Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & ...
- POJ 1298 The Hardest Problem Ever【字符串】
Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar ever faced was ke ...
- Poj1298_The Hardest Problem Ever(水题)
一.Description Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar eve ...
- 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 ...
- HDOJ 1048 The Hardest Problem Ever(加密解密类)
Problem Description Julius Caesar lived in a time of danger and intrigue. The hardest situation Caes ...
- C - The Hardest Problem Ever
Description Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar ever ...
随机推荐
- SQL Server 2008 开启远程连接
除了 IP1.IP2 外,也要把 IPALL 的端口也设置为 1433 参考:SQL Server开启1433端口,彻底解决方案
- moogodb 安装及简单介绍
1,安装Moogodb 因为是windows 64位操作系统,直接到官网上下载.msi文件,下载完成后点击安装,点击同意协议之后,出现下面的对话框, Choose Setup Type, 就是选择安装 ...
- codeforces401C
Team CodeForces - 401C Now it's time of Olympiads. Vanya and Egor decided to make his own team to ta ...
- 3、springframe常用注解
1.@controller 控制器(注入服务) 2.@service 服务(注入dao) 3.@repository dao(实现dao访问) 4.@component (把普通pojo实例化到spr ...
- CF558E-A Simple Task-线段树+计数排序
计数排序的原理,只要知道了有几个数比i小,就可以知道i的位置 这道题只有26个字母,搞26颗线段树,然后区间更新 #include <cstdio> #include <cstrin ...
- Codeforces1101F Trucks and Cities 【滑动窗口】【区间DP】
题目分析: 2500的题目为什么我想了这么久... 考虑答案是什么.对于一辆从$s$到$t$的车,它有$k$次加油的机会.可以发现实际上是将$s$到$t$的路径以城市为端点最多划分为最大长度最小的$k ...
- 线性基求第k小异或值
题目链接 题意:给由 n 个数组成的一个可重集 S,每次给定一个数 k,求一个集合 \(T \subseteq S\), 使得集合 T 在 S 的所有非空子集的不同的异或和中, 其异或和 \(T_1 ...
- HNOI2019 游记
HNOI2019 游记 Day 0 其实考前几天,心里还是挺慌的.结果最后 Day 0 的时候,因为种种原因反而释然了.也许是觉得,在这一步退役,也没有什么好害怕的吧. OI 本身就是一项偶然性太大的 ...
- Android工程图片资源命名禁忌
Android工程中,res\drawable\ 文件夹下所有的图片资源文件命名,不允许: 1. 大写字母 从Eclipse的这个报错可以知道资源文件的命名规则. Invalid file name: ...
- Android ViewSwitcher 的功能与用法
ViewSwitcher 代表了视图切换组件, 本身继承了FrameLayout ,可以将多个View叠在一起 ,每次只显示一个组件.当程序控制从一个View切换到另个View时,ViewSwitch ...