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213 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:
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.
Universidad de Valladolid OJ: 213 – Message Decoding 2/2
Sample input
TNM AEIOU
0010101100011
1010001001110110011
11000
$#**\
0100000101101100011100101000
Sample output
TAN ME
##*\$
题意：
题目要求是根据二进制密码输出其对应的信息。编码方式是这样的：长度为1到7的二进制码都有可能对应一个字符，但是各长度的二进制密码都不包括仅包含‘1’的那些二进制码（比如‘1’、‘11’、‘111’）；首先输入一列不包含换行的字符串，我们从左至右依次读取字符串的每一个字符并对其进行编码，从左至右第1个编码为‘0’、第2个为‘00’、第3个为‘01’、第4个为‘000’、第5个为‘001’，依次类推，即当编码长度相同的时候，后一个编码总是前一个编码在二进制下对应加1的那个数（with leading zeros），二进制码长度不会超过7。下面一行输入的就是由‘0’和‘1’组成的字符串。开始的三个‘0’、‘1’串在二进制下代表0~7的一个整数，意味着它之后的那些二进制编码将会以该整数长度进行依次读取并输出相应编码对应的字符，如果该整数为零，则代表该组输入结束。或者读取到全为‘1’的编码，也代表该组数据输入结束。
输入：
多组数据，每组数据第一行为一字符串，用于编码。之后的一行或者数行是二进制密码。
输出：
每组数据输出二进制密码对应的字符串。
分析：
模拟题。书写几个函数将有利于模拟这个过程。用code[len][val]来存储长度为len、二进制对应的十进制值为val的二进制编码对应的字符。readChar()函数将读取下一个非换行字符。readInt(int c)函数将用于读取长度为c的二进制并将其转化为十进制整数（中间忽略换行）。readCodes函数用于编码。
1 #include <stdio.h> 2 #include <string.h> 3 4 // 编码数组 5 // [len][value] 6 int code[8][1<<8]; 7 int readChar() { 8 while(true) { 9 int ch = getchar(); 10 // 读到非换行符为止 11 if(ch != ‘\n‘ && ch != ‘\r‘) { 12 return ch; 13 } 14 } 15 } 16 17 int readInt(int c) { 18 int v = 0; 19 // 获取十进制整数 20 while(c--) { 21 v = v * 2 + readChar() - ‘0‘; 22 } 23 return v; 24 } 25 26 int readCodes() { 27 memset(code, 0, sizeof(code)); 28 // 读取编码头的第一个字符 29 code[1][0] = readChar(); 30 // 从第二个字符开始循环编码 31 for(int len = 2; len <= 7; len++) { 32 for(int i = 0; i < (1 << len) - 1; i++) { 33 int ch = getchar(); 34 // 文件结束，终止程序 35 if(ch == EOF) { 36 return 0; 37 } 38 // 读一行 39 if(ch == ‘\n‘ || ch == ‘\r‘) { 40 return 1; 41 } 42 code[len][i] = ch; 43 } 44 } 45 return 1; 46 } 47 48 int main() { 49 while(readCodes()) { 50 while(true) { 51 // 读入编码长度 52 int len = readInt(3); 53 // 长度为0退出当前编码循环 54 if(!len) break; 55 while(true) { 56 int v = readInt(len); 57 // 全为1退出当前小节 58 if(v == (1 << len) - 1) break; 59 putchar(code[len][v]); 60 } 61 } 62 putchar(‘\n‘); 63 } 64 return 0; 65 }
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原文地址：http://www.cnblogs.com/cyb123456/p/5782195.html
