【UVA 213 --- Message Decoding】

题目来源：点击进入【UVA 213 — Message Decoding】

Description

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
##*$

解题思路

由题意可知可以通过一个s[i][j]来记录对应的字符。
i代表编码长度。
j代表编码对应的10进制。
这样s[i][j]就直接对应字符。我们可以通过输入的数值直接得到对应的字符。

AC代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char s[8][1 << 8];

char readchar() 
{
	while (true) 
	{
		char ch = getchar();
		if (ch != '\n' && ch != '\r')
			return ch; 
	}
} 

int read()
{
    s[1][0]=readchar();
    if(s[1][0]==-1) return 0;
    for(int i=2;i<8;i++)
    {
        for(int j=0;j<(1<<i)-1;j++)
        {
            char ch=getchar();
            if (ch == '\n' || ch == '\r') return 1;
            s[i][j]=ch;
        }
    }
    return 1;
}

int readint(int c) 
{
	int v = 0;
	while (c--) v = v * 2 + readchar() - '0';
	return v;
}

int main()
{
    while(read())
    {
        while (true) 
		{
			int len = readint(3);
			if (len == 0) break;
			while(true)
			{
				int v = readint(len);
				if (v == (1 << len) - 1) 
					break;
				putchar(s[len][v]);
			}
		} 
		printf("\n");
    }
    return 0;
}