Message dicoding, ACM/ICPC World Finals 1991, 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:

displaymath26

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

##*\$

用自己最笨的方法搞了几天,做个辣鸡的笔记,还没有AC,确定是跑偏了路线了,决定重新写一遍了,已经能通过大部分测试数据,但是潜在的bug太多,算法设计的也有问题,很多漏洞,还是重新看看rujia的代码重新写一遍比较好,现在这里记下自己的渣代码(注释方便自己以后查看).

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

using namespace std;

char codes[1000001];
char* binary[100] = { "0", "00", "01", "10", "000", "001", "010", "011", "100",
"101", "110" };
//char decodes[1000001] = "011001111010010111011010100111000";
char decodes[1000001] = "010101101100010111100101000";

void getCodes_2()
{
int index = 0;
char temp;
while (true)
{
temp = getchar();
if (temp == '\n')
{
break;
}
codes[index] = temp;
index++;
}
}

int BtoD(int p, int len)
{
int res = 0;
while (len--)
{
res += (decodes[p] - '0') * pow(2, len);
p++;
}
if (res == 0)
{
cout << "三个数全都是000" << endl;
}
return res;
}

bool allOnes(int p, int len)
{
while (len--)
{
if (decodes[p] == '0')
{
return false;
}
p++;
}
//  cout << "小节执行完毕" << endl;
return true;
}

int getCodes(int p, int len)
{
//  cout << "*****" << endl;
char step_code[10<
4000
/span>] = { '\0' };
int i = 0;
while (len--)
{
step_code[i] = decodes[p];
p++;
i++;
}
//  cout<<"step: "<<step_code<<endl;
for (int i = 0; i <= 9; i++)
{
//      cout << "i: " << i << endl;
//      cout << "binary step: " << binary[i] << endl;
int ok = 1;
int step_len = strlen(step_code), binary_len = strlen(binary[i]);
if (step_len == binary_len)
{
//          cout<<"step_len===================: "<<step_len<<endl;
for (int j = 0; j < step_len; j++)
{
//              cout << "binary i j : " << binary[i][j] << endl;
//              cout << "step_code: " << step_code[j] << endl;
if (binary[i][j] != step_code[j])
{
//                  cout << "not ok i: " << i << endl;
ok = 0;
}
}
if (ok)
{
//              cout << "ok i: " << i << endl;
return i;
break;
}
}
}
//  cout << "*****" << endl;
}

int main()
{
int length = 3, p = 0;
getCodes_2();
while (BtoD(p, length) != 0)
{
int len = length;
int len_2 = BtoD(p, length);
len = BtoD(p, len);
p += length;
int test = 0;
while (!allOnes(p, len))
{
int index = getCodes(p, len);
cout<<codes[index];
p += len;
//          cout << "p: " << p << endl;
//          cout << "len: " << len << endl;
}
p += len_2;
//      cout << "小节结束后的位置: " << p << endl;
}
return 0;
}