HDU1048The Hardest Problem Ever
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
Sample Output
#include<bits/stdc++.h>
using namespace std;
int main() {
char s[],s1[],s2[];
while(gets(s1)&&strcmp(s1,"START")==) {
gets(s);
int len=strlen(s);
for(int i=; i<len; i++) {
if(isupper(s[i]))
s[i]=(s[i]-'A'+)%+'A';
}
while(gets(s2)&&strcmp(s2,"END")==) {
puts(s);
break;
}
}
return ;
}
HDU1048The Hardest Problem Ever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 ...
- (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 ...
- C - The Hardest Problem Ever
Description Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar ever ...
- 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
- Java迷题:等于,还是不等于?
等于还是不等于? 看来看下面的一段代码: public static void main(final String[] args) { Integer a = new Integer(100); In ...
- Partitioner没有被调用的情况
map的输出,通过分区函数决定要发往哪个reducer. 有2种情况,我们自定义的Partitioner不会被调用 1) reducer个数为0 这种情况,没有reducer,不需要分区 2) red ...
- 常用的PC/SC接口函数
PC/SC规范是一个基于WINDOWS平台的一个标准用户接口(API),提供了一个从个人电脑(Personal Computer)到智能卡(SmartCard)的整合环境,PC/SC规范建立在工业标准 ...
- linux kernel 0.11 head
head的作用 注意:bootsect和setup汇编采用intel的汇编风格,而在head中,此时已经进入32位保护模式,汇编的采用的AT&T的汇编语言,编译器当然也就变成对应的编译和连接器 ...
- myeclipse激活+Aptana安装配置
一.Myeclipse安装激活. 安装过程一路向下. 1.破解公钥,确保MyEclipse没有开启,否则失败! 用WinRAR或7-zip打开安装目录下Common\plugins\com.genui ...
- ToolBar存档
上图是将本阶段要完成的结果画面做了标示,结合下面的描述希望大家能明白. colorPrimaryDark(状态栏底色):在风格 (styles) 或是主题 (themes) 里进行设定. App ba ...
- Swift基础小结_2
import Foundation // MARK: - ?和!的区别// ?代表可选类型,实质上是枚举类型,里面有None和Some两种类型,其实nil相当于OPtional.None,如果非nil ...
- mongoDB 3.0 安全权限访问控制
MongoDB3.0权限,啥都不说了,谷歌百度出来的全是错的.先安装好盲沟,简单的没法说. 首先,不使用 —auth 参数,启动 mongoDB: mongodb-linux-i686-3.0.0/b ...
- shell 基本结构
就像其他的编程语言一样,shell也有三种基本的结构:顺序结构.分支结构.循环结构.顺序结构就是按照命令的出现顺序依次执行,比较简单.如下分别介绍分支结构和循环结构. 分支结构 格式1: if com ...
- Entity Framework学习笔记(三)----CRUD(2)
请注明转载地址:http://www.cnblogs.com/arhat 昨天晚上老魏配的机器终于到了,可是拿回来之后什么都组装好了,唯独差一个非常重要的组件"电源线",老魏那个汗 ...