ZOJ Problem Set - 1392 The Hardest Problem Ever

本文介绍了一种基于凯撒密码的解密算法实现,通过将输入的密文进行特定偏移来还原原始消息。该算法适用于包含大写字母的字符串,并保持非字母字符不变。
The Hardest Problem Ever

Time Limit: 2 Seconds       Memory Limit: 65536 KB

Introduction

Julius Caesar lived in a time of danger and intrigue. 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.


Input

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

For each data set, there will be exactly one line of output. This is the original message by Caesar.


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: South Central USA 2002  





分析:

题意:

输入多组测试数据。每组测试数据包含三行,第一行是 “START”,不处理;第二行是“END”,也不处理;第三行是一行密码(可能包含空格等多种字符)。要求对大写字母做减5运算(字母表上往左移动五位,其中A~E分别对应V~Z),其他字符不做处理,原样输出就可以。



涉及到有空格等字符的处理,首先想到的就是cin.getline()函数。用这个函数就可以轻松解决这个题了。因为时间方面并没有卡紧。



ac代码:

#include <iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
char ss[205];
string s;
int main()
{
    int i,j;
    while(cin.getline(ss,205))//输入为char
    {
        s=ss;//转换为string
        if(s=="ENDOFINPUT") break;
        else if(s=="START")  continue;
        else if(s=="END") continue;
        else
        {
            for(i=0;i<s.size();i++)
            {
                if(s[i]>='F'&&s[i]<='Z')
                cout<<char(s[i]-5);
                else if(s[i]=='A')
                cout<<"V";
                else if(s[i]=='B')
                cout<<"W";
                else if(s[i]=='C')
                cout<<"X";
                else if(s[i]=='D')
                cout<<"Y";
                else if(s[i]=='E')
                cout<<"Z";
                else
                cout<<s[i];
            }
            cout<<endl;
        }
    }
    return 0;
}

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