Message Decoding （UVA-213）
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 rst 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 rst 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 rst 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 rst 3 digits of a segment give the binary representationof the length of the keys in the segment. For example, if the rst 3 digits are 010, then the remainderof the segment consists of keys of length 2 (00, 01, or 10). The end of the segment is a string of 1’swhich 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 segmentin which the keyshave 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 le contains several data sets. Each data set consists of a header, which is on a single lineby itself, and a message, which may extend over several lines. The length of the header is limitedonly 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 of1’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
##*\$
题意：先给你一个字符串，能有空格，不能有换行，每个字符用0;00;01;10;000;001;010;011;100;101;110;0000;0001;…;1011;1110;00000;…表示，也就是说0表示第一个字符，00表示第二个字符，01表示第三个字符，等等，但是不能全是1；
接着给你一些01的字符串，一开始先读前三位，把前三位转化成十进制，这个代表之后要读的位数，例如一开始是010，也就2，那么之后就读取2位，然后将这2位所代表的字符输出，当这几位全是1的时候跳出，从新设置位数，接着进行，当位数为0的时候停止；
思路：将字符存到二维数组 g[l][v] 里，l代表有几位，v代表十进制数是多少，然后读取01字符串，并按照之前的解码将其输出；
ps：这个题理解起来有点困难，还有就是输入的时候有点麻烦；
一开始输入字符串的时候有空格，所以不能用cin，而且用gets一直显示编译错误，所以只能用 scanf("%[^\n]",s) 了；
还有就是输入01待解码字符串的时候有换行，需要麻烦点；
#include
#include
#include
#define maxn 1<<9
using namespace std;
char g[8][maxn],s[1010];
int len;
int main()
{
while(scanf("%[^\n]",s)!=EOF) //输入带空格的字符串不会停止，输入回车停止；
{
memset(g,0,sizeof(g));
len=strlen(s);
int v,l;
v=0;
l=1;
for(int i=0; i=((1<