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
$#**
0100000101101100011100101000
Sample output
TAN ME
'##*$'（这个两个单引号是我加的，因为我用的是MarkDown编辑器，##会出事...）
这个题目的题意有点难理解啊......看到这个题目我当时是懵逼的......
题意：
我用样例来解释题意吧。。
例如样例中的字符串 $#** 从第一个字符开始，每个字符对应一个01串
考虑下面的01串序列：0，00,01,10,000,001,010,011,100，101,110,0000,0001，...，1101,1110,00000，...
首先是长度为1的串，然后是长度为2的串，以此类推。相同长度的后一个串等于前一个串加1.注意上述序列中不存在全为1的串。
接下来会有一段01串读入，你的任务是把它分解成许多01子串。
注意：像样例中提到的
“0010101100011
1010001001110110011
11000”
这个样例，我们需要将这些01串连在一起，看做一个01串来进行分解。
编码文本由多个小节组成，每个小节的前3个数字代表小节中每个编码的长度（用二进制表示，such as 010代表长度为2），然后是各个字符的编码，以全1结束（例如，编码长度为2的小节以1结束）。编码文本以长度为000的小节结束。
Attention:上面所说的最开始的字符串所一一对应的01串序列最大长度为7（二进制数）。
题解：这就是个裸的模拟啊（当然是膜你）。。。
代码如下：
#include<bits/stdc++.h>
using namespace std;
int code[8][1<<8];
int read()
{
int x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int readchar()
{
while(1)
{
int ch=getchar();
if(ch!='
'&&ch!='
')return ch;
}
}
int readint(int c)
{
int v=0;
while(c--)v=v*2+readchar()-'0';
return v;
}
bool readcodes()
{
memset(code,0,sizeof(code));
code[1][0]=readchar();
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=='
'||ch=='
')return 1;
code[len][i]=ch;
}
return 1;
}
int main()
{
while(readcodes())
{
while(1)
{
int len=readint(3);
if(!len)break;
while(1)
{
int v=readint(len);
if(v==(1<<len)-1)break;
putchar(code[len][v]);
}
}
putchar('
');
}
return 0;
}