Message Decoding UVA - 213

 Some message encoding schemes require that an encoded message be sent in two parts. The fifirst part,called the header, contains the characters of the message. The second part contains a pattern thatrepresents 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 fifirst 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 fromthe fifirst 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 fifirst key (0) is mapped to the fifirst character in the header, the second key (00) to the second character in the header, the kthkey 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 fifirst 3 digits of a segment give the binary representation of the length of the keys in the segment. For example, if the fifirst 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 fifile 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
##*\$

HINT

  这道题相比前面的题的难度要大一些。需要解决的问题是如何来区分各种各个编码，以及如何来输出结果判断输出字符。具体看下面代码

Accepted

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

int main()
{
	char code[200];
	while (gets(code))							//每一个样例录入
	{
		char s[4];								//用来存储三位编码头
		while (1)
		{
			char temp;
			for (int i = 0;i < 3;i++)
			{
				while ((temp = getchar()) == '\n')		//输入如果出现回车换行跳过重新输入
					continue;
				s[i] = temp;					//存储
			}
			s[3] = '\0';
			int len = 0;
			len = (((len * 2 + s[0] - '0') * 2 + s[1] - '0') * 2 + s[2] - '0');//将三位编码头转化为十进制
			if (!len)break;												//判断程序是否终止

			while (1)							//用来录入编码输出编码
			{
				int num = 0;
				for (int i = 0;i < len;i++)		//用编码头获得的长度来录入编码
				{
					while ((temp = getchar()) == '\n')
						continue;
					num = num * 2 + temp - '0';		//边录入边转化为10进制
				}
				if (num == (1 << len) - 1)break;	//这里用到了左移运算符，加入len=3,那么1向左移动三位编程1000（二进制）然后减去一变为111（二进制）刚好是结束条件
				else printf("%c", code[(int)pow(2, len) - len + num-1]);//(int)pow(2, len) - len + num-1；这里是位置计算的公式，看着上面的编码好好推一下，就是每一个等比数列减去一然后加上再本编码段的位置
			}

		}
		putchar('\n');
		getchar();
	}
}