(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
- linux 安装python 和pip
下载文件 python官网:https://www.python.org/downloads/ 百度网盘http://pan.baidu.com/s/1mixGB12 密码 9nzu [r ...
- 实体类注解错误:Could not determine type for: java.util.List
今天配置实体类注解时,出现以下错误: Caused by: org.hibernate.MappingException: Could not determine type for: java.uti ...
- struts2 核心过滤器的配置
<!-- struts2 过滤器核心配置--> <filter> <filter-name>struts2</filter-name> <filt ...
- M3U8文件
M3U本质上说不是音频文件,它是音频文件的列表文件,是纯文本文件.你下载下来打开它,播放软件并不是播放它,而是根据它的记录找到网络地址进行在线播放. M3U文件的大小很小,也就是因为它里面没有任何音频 ...
- 洛谷P1373小a和uim大逃离题解
题目 这个题好坑啊,首先是他会卡空间,然后我们就只能把一种比较好理解的状态给舍弃,因为空间开不下,然而采用一种难理解的状态就是\(dp[i][j][l][0/1]\)表示\(i\),\(j\)位置,两 ...
- [POJ2976] Dropping tests
传送门:>Here< 题意:给出长度相等的数组a和b,定义他们的和为$\dfrac{a_1+a_2+...+a_n}{b_1+b_2+...+b_n}$.现在可以舍弃k对元素(一对即$a[ ...
- YC的基本创业建议
原文出处:https://blog.ycombinator.com/ycs-essential-startup-advice/ 我们给初创公司的许多建议都是战术性的; 意味着在日常或周到周的基础上有所 ...
- 【XSY2727】Remove Dilworth定理 堆 树状数组 DP
题目描述 一个二维平面上有\(n\)个梯形,满足: 所有梯形的下底边在直线\(y=0\)上. 所有梯形的上底边在直线\(y=1\)上. 没有两个点的坐标相同. 你一次可以选择任意多个梯形,必须满足这些 ...
- Leetcode 167. 两数之和 II - 输入有序数组 By Python
给定一个已按照升序排列 的有序数组,找到两个数使得它们相加之和等于目标数. 函数应该返回这两个下标值 index1 和 index2,其中 index1 必须小于 index2. 说明: 返回的下标值 ...
- 【CF487E】Tourists(圆方树)
[CF487E]Tourists(圆方树) 题面 UOJ 题解 首先我们不考虑修改,再来想想这道题目. 我们既然要求的是最小值,那么,在经过一个点双的时候,走的一定是具有较小权值的那一侧. 所以说,我 ...