Mapping Cipher

题目： 输入字母表中所映射的密文， 再将你输入的原文按照对应的密文所转换后输出。（加密）

        Mapping Cipher

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!

Input

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

described above. Following are plain text enclosed by two lines of string "***".

Your task is to output the cypher text encrypted using the given mapping string. 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 the test case is no more than 1000 characters. Each

line in test cases is no more than 1000 characters.

Output

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

sample output. Place a blank line between cases.

Sample Input

incpfvztsmbexowhlgkqjyrdau

***

There are 2 dogs and 3 cats.

It rains dogs & cats!

***

xowhlgkqjyrdauincpfvztsmbe

***

There are 2 dogs and 3 cats.

It rains dogs & cats!

***

###

Sample Output

Cypher 1:

Qtfgf igf 2 pwzk iop 3 ciqk.

Sq gisok pwzk & ciqk!

Cypher 2:

Vqlpl xpl 2 hikf xuh 3 wxvf.

Jv pxjuf hikf & wxvf!

#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}; //存取输入的原文
		
		gets(arr);
		if(arr[0]=='#' && arr[1]=='#' && arr[2]=='#') break;
		
		gets(z);
		
		if(flag) flag=0;
		else printf("\n");
		
		printf("Cypher %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') printf("%c",arr[str[i]-'a']);
				else if(str[i]>='A' && str[i]<='Z') printf("%c",arr[str[i]-'A']+'A'-'a');
				else printf("%c",str[i]);
			}
			
			printf("\n");
		}
		
		scanf("\n");
			
	}
    
    return 0;
}