Some message encoding schemes require that an encoded message be sent in two parts. The first part, called the header, contains the characters of the message. The second part contains a pattern that represents the message. You must write a program that can decode messages under such a scheme. The heart of the encoding scheme for your program is a sequence of “key” strings of 0’s and 1’s as follows: 0, 00, 01, 10, 000, 001, 010, 011, 100, 101, 110, 0000, 0001, . . . , 1011, 1110, 00000, . . . The first key in the sequence is of length 1, the next 3 are of length 2, the next 7 of length 3, the next 15 of length 4, etc. If two adjacent keys have the same length, the second can be obtained from the first by adding 1 (base 2). Notice that there are no keys in the sequence that consist only of 1’s. The keys are mapped to the characters in the header in order. That is, the first key (0) is mapped to the first character in the header, the second key (00) to the second character in the header, the kth key is mapped to the kth character in the header. For example, suppose the header is: AB#TANCnrtXc Then 0 is mapped to A, 00 to B, 01 to #, 10 to T, 000 to A, ..., 110 to X, and 0000 to c. The encoded message contains only 0’s and 1’s and possibly carriage returns, which are to be ignored. The message is divided into segments. The first 3 digits of a segment give the binary representation of the length of the keys in the segment. For example, if the first 3 digits are 010, then the remainder of the segment consists of keys of length 2 (00, 01, or 10). The end of the segment is a string of 1’s which is the same length as the length of the keys in the segment. So a segment of keys of length 2 is terminated by 11. The entire encoded message is terminated by 000 (which would signify a segment in which the keys have length 0). The message is decoded by translating the keys in the segments one-at-a-time into the header characters to which they have been mapped.
Input
The input file contains several data sets. Each data set consists of a header, which is on a single line by itself, and a message, which may extend over several lines. The length of the header is limited only by the fact that key strings have a maximum length of 7 (111 in binary). If there are multiple copies of a character in a header, then several keys will map to that character. The encoded message contains only 0’s and 1’s, and it is a legitimate encoding according to the described scheme. That is, the message segments begin with the 3-digit length sequence and end with the appropriate sequence of 1’s. The keys in any given segment are all of the same length, and they all correspond to characters in the header. The message is terminated by 000. Carriage returns may appear anywhere within the message part. They are not to be considered as part of the message.
Output
For each data set, your program must write its decoded message on a separate line. There should not be blank lines between messages.
Sample input
TNM AEIOU
0010101100011
1010001001110110011
11000
$#**\
0100000101101100011100101000
Sample output
TAN ME
##*\$
弄了一上午，中午去吃个饭回来电脑竟然关机了，TMD，只能从新打一遍了
首先我们先输入编码头，而编码头会对应01串；例如：
A & * M $ N（输入时没有空格）
对应的01串 0 00 01 10 000 001
然后我们开始输入01编码串，规则我们都知道，全为1时小结结束，小结结束后接着全为0表示01编码输入完毕
例如：
0100000101101100011100101000
如上面所示加粗表示小节开始，下划线代表小节结束，斜体表示输入结束
我们首先输入编码头，然后再输入编码串；
#include<stdio.h>
#include<string.h>
int readchar()//读取字符，当遇到'\n'&&\r'时返回输入的字符
{
for(;;)
{
int ch=getchar();
if(ch!='\n'&&ch!='\r')
return ch;
}
}
int readint(int c)//者读取c位二进制字符（即0和1），并转化为十进制整数。
{
int v=0;
while(c--)
{
v=v*2+readchar()-'0';
}
return v;
}
int code[8][1<<8];
int readcodes()
{
memset(code,0,sizeof(code));
code[1][0] = readchar(); //直接调到下一行开始读取。如果输入已经结束，会读到EOF
for(int len = 2; len <=7; len++)
{
for(int i=0;i<(1<<len)-1;i++)
{
int ch = getchar();
if(ch==EOF)
return 0;
if(ch=='\n'||ch=='\r')
return 1;
code[len][i] = ch;
}
}
return 1;
}
int main()
{
while(readcodes())
{
for(;;)
{
int len = readint(3);
if(len == 0)
break;
for(;;)
{
int v=readint(len);
if(v==(1<<len)-1)
break;
putchar(code[len][v]);
}
}
putchar('\n');
}
return 0;
}