poj 1298(水题)
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24241 | Accepted: 13227 |
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
to this problem will consist of a (non-empty) 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
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
Source
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
int main(){
// freopen("in.txt","r",stdin);
string s,s1,s2;
char m[]="VWXYZABCDEFGHIJKLMNOPQRSTU";
int j,k;
while(cin>>s){
if(s=="ENDOFINPUT")
break;
cin.ignore(INT_MAX,'\n');
getline(cin,s1,'\n');
j=s1.size();
for(int i=;i<j;i++){
if(s1[i]>='A'&&s1[i]<='Z')
cout<<m[s1[i]-'A'];
else
cout<<s1[i];
}
cin>>s2;
cout<<endl;
}
return ;
}
poj 1298(水题)的更多相关文章
- poj 1269 水题
题目链接:http://poj.org/problem?id=1269 #include<cstdio> #include<cstring> #include<cmath ...
- poj 2245 水题
求组合数,dfs即可 #include<cstdio> #include<iostream> #include<algorithm> #include<cst ...
- poj 3006水题打素数表
#include<stdio.h> #include<string.h> #define N 1100000 int isprim[N],prime[N]; void ispr ...
- POJ 3672 水题......
5分钟写完 水水更开心 //By SiriusRen #include <cstdio> #include <iostream> #include <algorithm& ...
- 【转】POJ百道水题列表
以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight ...
- POJ 1488 Tex Quotes --- 水题
POJ 1488 题目大意:给定一篇文章,将它的左引号转成 ``(1的左边),右引号转成 ''(两个 ' ) 解题思路:水题,设置一个bool变量标记是左引号还是右引号即可 /* POJ 1488 T ...
- POJ 水题若干
POJ 3176 Cow Bowling 链接: http://poj.org/problem?id=3176 这道题可以算是dp入门吧.可以用一个二维数组从下向上来搜索从而得到最大值. 优化之后可以 ...
- poj 3080 Blue Jeans(水题 暴搜)
题目:http://poj.org/problem?id=3080 水题,暴搜 #include <iostream> #include<cstdio> #include< ...
- POJ 水题(刷题)进阶
转载请注明出处:優YoU http://blog.csdn.net/lyy289065406/article/details/6642573 部分解题报告添加新内容,除了原有的"大致题意&q ...
随机推荐
- Bzoj4870 [SXOI2017]组合数问题
Time Limit: 10 Sec Memory Limit: 512 MBSubmit: 155 Solved: 78 Description Input 第一行有四个整数 n, p, k, ...
- 【洛谷 P3809】 【模板】后缀排序
题目链接 先占个坑,以后再补. \(SA\)的总结肯定是要写的. 等理解地深入一点再补. #include <cstdio> #include <cstring> const ...
- nodejs入门教程之http的get和request简介及应用
nodejs入门教程之http的get和request简介及应用 前言 上一篇文章,我介绍了nodejs的几个常用的模块及简单的案例,今天我们再来重点看一下nodejs的http模块,关于http模块 ...
- Android中的异常情况
1.setText()方法中,如果参数是int类型,Android会把它当做是一个id查找,报以下异常,因此解决办法就是将参数转化为String类型 如:setText(num) è setText( ...
- js删除数组中重复的元素
1.方法一 将数组逐个搬到另一个数组中,当遇到重复元素时,不移动,若元素不重复则移动到新数组中 function unique(arr){ var len = arr.length; var resu ...
- CreateProcess中的部分参数理解
函数原型,这里写Unicode版本 WINBASEAPIBOOLWINAPICreateProcessW( _In_opt_ LPCWSTR lpApplicationName, //可执行文件名字 ...
- Python【模块】importlib,requests
内容概要: 模仿django中间件的加载方式 importlib模块 requests模块 rsplit() 用实际使用的理解来解释两个模块 importlib模块 ...
- SD卡 模拟SPI总线控制流程
SD卡为移动设备提供了安全的,大容量存储解决方法.它本身可以通过两种总线模式和MCU进行数据传输,一种是称为SD BUS的4位串行数据模式,另一种就是大家熟知的4线SPI Bus模式.一些廉价,低端的 ...
- Chrome控制台的妙用之使用XPATH
谷歌浏览器,对于作为程序员的我们来说可以是居家必备了,应该用的相当的熟悉了,我们用的最多的应该是network选项吧,一般用来分析网页加载的请求信息,比如post参数之类的,这些基本的功能基本上够用了 ...
- spark 环境搭建坑
spark的新人会有什么坑 spark是一个以java为基础的,以Scala实现的,所以在你在安装指定版本的spark,需要检查你用的是对应spark使用什么版本的scala,可以通过spark-sh ...