Message Decoding  

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:
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
##*\$

제목은 디코딩 프로그램을 작성하여 숫자열을 디코딩하는 것이다
첫 번째 줄을 입력하면 디코딩 키 키가 왼쪽에서 오른쪽으로 각각 0,00,01,10000000111001100000001,11100000001,...,1101,1110,00000,.......
len 길이의 문자 인코딩은 2^n-1개이며, 마침 2진법으로 0에서 2^n-2로 점차적으로 증가하였으며, 문자 인코딩의 최대 길이는 7로 2^7-1=127자 가능합니다
우리는 단지 키 [len] [val] 수조에 len 길이의 val+1 문자를 인코딩하고 디코딩할 때 모든 문자를 출력하면 된다
예를 들어 2 디코딩 키는 "S#**\"입니다.
key[1][0] 해당 인코딩'0'에 저장된 문자는'$'key[2][0] 해당 인코딩'00'에 저장된 문자는'#'이다
key[2][1] 해당 인코딩'01'에 저장된 문자는'*'key[2][2] 해당 인코딩'10'에 저장된 문자는'*'
key[3][0] 해당 인코딩 "000"에 저장된 문자는 "\"
디코딩해야 할 텍스트는 여러 개의 소절로 구성되어 있으며, 각 소절의 앞의 세 숫자는 소절의 인코딩 길이를 대표한다(이진법으로 표시한다. 예를 들어 010으로 표시한다). 그리고 이 길이의 부정확한 문자 인코딩은 전체 1로 끝난다. 길이가 2시'11'이면 묶음 길이가 000일 때 인코딩해야 할 텍스트의 끝을 나타낸다.
이 문제는 알고리즘 경연 입문 고전 2판의 예제 83페이지에 구체적인 해설이 담겨 있다.

#include<cstdio>
#include<cstring>
#include<cctype>
using namespace std;
const int N = 1000;
char s[N], key[8][1 << 8], c;
int val, len, l;

void read()
{
    len = 1; l = strlen (s);
    for (int i = val = 0; i < l; ++i)
        if (val < ( (1 << len) - 1)) 
            key[len][val++] = s[i];
        else
        {
            val = 0;
            key[++len][val++] = s[i];
        }
}

void print ()
{
    for (int i = val = 0; i < len; ++i)
    {
        do scanf ("%c", &c); while (!isdigit (c));
        val = val * 2 + (c - '0');
    }
    if (val >= ( (1 << len) - 1))  return;
    else
    {
        printf ("%c", key[len][val]);
        print();
    }
}

int main()
{
    int l1, l2, l3;
    while (gets (s) != NULL)
    {
        if (!s[0]) continue;
        memset (key, 0, sizeof (key));
        read();
        while (scanf ("%1d%1d%1d", &l1, &l2, &l3), len = 4 * l1 + 2 * l2 + l3)
            print ();
        printf ("
");
    }
    return 0;
}

판권 성명: 본문 블로그 오리지널 글.블로그는 동의 없이 전재할 수 없다.