C - The Hardest Problem Ever
Description
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
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
Sample Input
START
NS BFW, JAJSYX TK NRUTWYFSHJ FWJ YMJ WJXZQY TK YWNANFQ HFZXJX
END
START
N BTZQI WFYMJW GJ KNWXY NS F QNYYQJ NGJWNFS ANQQFLJ YMFS XJHTSI NS WTRJ
END
START
IFSLJW PSTBX KZQQ BJQQ YMFY HFJXFW NX RTWJ IFSLJWTZX YMFS MJ
END
ENDOFINPUT
Sample Output
IN WAR, EVENTS OF IMPORTANCE ARE THE RESULT OF TRIVIAL CAUSES
I WOULD RATHER BE FIRST IN A LITTLE IBERIAN VILLAGE THAN SECOND IN ROME
DANGER KNOWS FULL WELL THAT CAESAR IS MORE DANGEROUS THAN HE
解码,左移五位
#include <iostream>
#include <string.h>
#include <string>
using namespace std;
int main()
{
string a,b;
while(cin>>a){
if(a=="ENDOFINPUT")break;
if(a=="START"){
string c;
char s[];
int p=;
while(cin>>c){
if(c=="END")break;
int n;
n=c.length();
for(int i=;i<n;i++){
if(c[i]>='A'&&c[i]<='Z'){
if(c[i]-<'A')c[i]=c[i]-+'Z'-'A';
else c[i]-=;
}
s[p]=c[i];
p++;
}
s[p]=' ';
p++;
}
for(int i=;i<p-;i++)cout<<s[i];
cout<<endl;
}
}
//system("pause");
return ;
}
输入输出格式!
C - The Hardest Problem Ever的更多相关文章
- HDU1048The Hardest Problem Ever
The Hardest Problem Ever Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & ...
- The Hardest Problem Ever(字符串)
The Hardest Problem Ever Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 24039 Accept ...
- (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 ...
- 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 ...
- 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 ...
- 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 ...
随机推荐
- ConcurrentQueue对列的基本使用方式
队列(Queue)代表了一个先进先出的对象集合.当您需要对各项进行先进先出的访问时,则使用队列.当您在列表中添加一项,称为入队,当您从列表中移除一项时,称为出队. ConcurrentQueue< ...
- djangoPOST请求403 forbidden
处理过程 网上搜索修改setting.py,在MIDDLEWARE_CLASSES增加django.middleware.csrf.CsrfResponseMiddleware 没能解决问题 有说在 ...
- oc语言--内存管理
一.基本原理 1.什么是内存管理 1> 移动设备的内存及其有限,每个app所能占用的内存是有限制的 2> 当app所占用的内存较多时,系统就会发出内存警告,这是需要回收一些不需要的内存空间 ...
- RHEL 7.0 修改防火墙配置
RHEL 7.0默认使用的是firewall作为防火墙,这里改为iptables防火墙. 关闭firewall: systemctl stop firewalld.service #停止firewal ...
- python学习资料
http://woodpecker.org.cn/diveintopython/ http://www.cnblogs.com/txw1958/archive/2012/12/10/A_Byte_of ...
- HttpWebResponse类
HttpWebResponse类的作用用于在客户端获取返回的响应的信息,还记得HttpResponse类吗?你是否在写B/S程序的时候,经常用到Response.Write()呢? HttpRespo ...
- FileAccess枚举
FileAccess用于控制对文件的读访问.写访问或读/写访问的常熟.从源代码可以看到FileAccess是一个简单枚举. 枚举成员 成员值 描述 Read 1 对文件的读访问,拥有读取权限. Wri ...
- Vitamio视频播放
activity代码 package com.hck.player.ui; import io.vov.utils.StringUtils; import io.vov.vitamio.LibsChe ...
- 绘图工具graphviz学习使用
画图工具: http://www.tuicool.com/articles/r2iAfa http://www.tuicool.com/articles/RjQfey 绘图工具graphviz学习使用 ...
- jQuery 局部div刷新和全局刷新方法
div的局部刷新 $(".dl").load(location.href+".dl"); 全页面的刷新方法 window.location.reload( ) ...