Decrypting Mapping Cipher

题目： 先输入字母映射的密文， 再将输入的加密后的句子中的字母， 得其所对应密文中相同字母的位置， 将其位置的位数所对应字母表中的字母输出。（解密）

    Decrypting Mapping Cipher

Recall the Mapping Cipher laboratory work. You are given an alphabetic mapping

table to create cipher text. The table is defined by a string using 26 lower-case

alphabets, such as

incpfvztsmbexowhlgkqjyrdau

In the string all characters are different. Using this table to generate cipher text, it

performs the following mapping scheme:

a -> i, A->I

b -> n, B->N

c -> c, C->C

d -> p, D->P

. . . . . . .

y -> a, Y->A

z -> u, Z->U

Any non-alphabetic character are kept unchanged.

Apply the scheme to plain text:

There are 2 dogs and 3 cats.

It rains dogs & cats!

Obtain the following cipher text:

Qtfgf igf 2 pwzk iop 3 ciqk.

Sq gisok pwzk & ciqk!

Now, you are asked to do its inverse task. You are given the same mapping string, but

you will receive cipher text. Your mission is to decrypt the cipher text to a plain one.

Input

There are several test cases. Each test case starts with a line of mapping string used to

create cipher text. Following are cipher text enclosed by two lines of string "***".

Your task is to output the plain text before encryption. There is a blank line after each

case. Finally, you will reach a line of string "###" which signals the end of test.

Each line in test cases is no more than 1000 characters.

Output

The cipher text of each test case is prefixed with a plain text number as shown in the

sample output. Place a blank line between cases.

Sample Input

incpfvztsmbexowhlgkqjyrdau

***

Qtfgf igf 2 pwzk iop 3 ciqk.

Sq gisok pwzk & ciqk!

***

xowhlgkqjyrdauincpfvztsmbe

***

Vqlpl xpl 2 hikf xuh 3 wxvf.

Jv pxjuf hikf & wxvf!

***

###

Sample Output

Plain 1:

There are 2 dogs and 3 cats.

It rains dogs & cats!

Plain 2:

There are 2 dogs and 3 cats.

It rains dogs & cats!

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
	int flag=1, test=1;
	
	while(1)
	{
		char arr[26]={0}; //存取字母映射的密文
		char z[10]={0}; //存取密文后的***
		char str[1001]={0}; //存取输入的加密后的句子
		
		gets(arr);
		if(arr[0]=='#' && arr[1]=='#' && arr[2]=='#') break;
		
		gets(z);
		
		if(flag) flag=0;
		else printf("\n");
		
		printf("Plain %d:\n",test++);
		
		while(gets(str))
		{
			if(str[0]=='*' && str[1]=='*' && str[2]=='*') break;
			
			for(int i=0; i<strlen(str); i++)
			{
				if(str[i]>='a' && str[i]<='z' || str[i]>='A' && str[i]<='Z')
				{
					for(int j=0; j<26; j++)
					{
						if(str[i]==arr[j])
						{
							printf("%c",j+'a');
							break;
						}
						
						else if(str[i]+'a'-'A'==arr[j])
						{
							printf("%c",j+'A');
							break;
						}	
					}
				}
				
				else printf("%c",str[i]);
			}
			
			printf("\n");
		}
		
		scanf("\n");
			
	}
    
    return 0;
}