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 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:

1. Start line - A single line, "START"
2. Cipher message - A single line containing from one to two hundred characters, inclusive, comprising a single message from Caesar
3. 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

【题意】：

你是Caesar军队的一个分队长。你的工作是破译Caesar送来的信息并汇报给你的上级。
密码很简单，每一个字母对应着一个明文，你将明文向右五步来得到安全的信息。（比如，假如那个字母是‘A’，密文就是‘F’）当你将Caesar的信息中找出明文后，你要反过来做：
加密文本
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 
明文文本 
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 
密文中只有字母被切换了，非字母的字符应该保持不变，所有的字母都是大写的。

这个问题的输入包括一系列（非空）最多100个数据。每一个数据的格式会按照以下格式，并且在不同组数据间不会有空行分隔。所有的字符都是大写的。
一个单独的测试数据包括三个部分：
1. 开始行：单独的一行“START” 。
2. 加密的信息：单独的一行，由1~200个字符组成来自Caesar的一行信息。
3. 结束行：单独的一行“END” 。
最后一组测试数据结束会跟着单独的一行“ENDOFINPUT”。

【代码】：

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <set>
#include <map>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long ll;
#pragma comment(linker, "/STACK:102400000,102400000")
#define Abs(x) ((x^(x >> 31))-(x>>31))
#define Swap(a,b) (a^=b,b^=a,a^=b)
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define EPS 1e-8
#define MOD 1000000007
#define max_ 10005
#define maxn 200002

using namespace std;
char a[];
int main()
{

    while(gets(a))
    {
        if(!strcmp(a,"ENDOFINPUT")) break;
        else if(!strcmp(a,"START")||!strcmp(a,"END")) continue;
        for(int i=;a[i];i++)
        {
            if(isalpha(a[i]))
            {
                a[i]=((a[i]-'A')+-)%+'A';
            }
        }
        puts(a);
    }
    return ;
}

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <set>
#include <map>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long ll;
#pragma comment(linker, "/STACK:102400000,102400000")
#define Abs(x) ((x^(x >> 31))-(x>>31))
#define Swap(a,b) (a^=b,b^=a,a^=b)
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define EPS 1e-8
#define MOD 1000000007
#define max_ 10005
#define maxn 200002

using namespace std;
char a[],b[];
int main()
{
    char s[]={"VWXYZABCDEFGHIJKLMNOPQRSTU"};
    while(gets(a))
    {
        int k=;
        if(!strcmp(a,"ENDOFINPUT")) break;
        if(strcmp(a,"START")!=&&strcmp(a,"END")!=)
        {
            for(int i=;a[i];i++)
            {
                if(isalpha(a[i]))
                {
                    printf("%c",s[a[i]-'A']);
                }
                else
                {
                    printf("%c",a[i]);
                }
            }
            printf("\n");
        }
    }
    return ;
}

常量字符数组妙用