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
##*\$
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <vector>
using namespace std;
const double EXP=1e-;
const double PI=acos(-1.0);
const int INF=0x7fffffff;
const int MS=; char code[][<<];
int readchar()
{
while()
{
int ch=getchar();
if(ch!='\n'&&ch!='\r')
return ch;
}
}
int readint(int cnt)
{
int v=;
while(cnt--)
v=v*+readchar()-'';
return v;
}
int readcode()
{
memset(code,,sizeof(code));
code[][]=readchar(); //特别注意,因为读到长度为0的时候就break了,后面还有个换行符
for(int len=;len<=;len++)
{
for(int i=;i<(<<len)-;i++)
{
int ch=getchar();
if(ch==EOF)
return ;
if(ch=='\n'||ch=='\r')
return ;
code[len][i]=ch;
}
}
return ;
}
int main()
{
while(readcode())
{
//printcode(); for debug
while()
{
int len=readint();
if(len==)
break;
//printf("len=%d\n",len);
while()
{
int v=readint(len);
//printf("v==%d\n",v);
if(v==(<<len)-)
break;
putchar(code[len][v]);
}
}
putchar('\n');
}
return ;
}

 
 

Message Decoding的更多相关文章

  1. UVa 213 Message Decoding(World Finals1991,串)

     Message Decoding  Some message encoding schemes require that an encoded message be sent in two part ...

  2. [算法竞赛入门经典]Message Decoding,ACM/ICPC World Finals 1991,UVa213

    Description Some message encoding schemes require that an encoded message be sent in two parts. The ...

  3. 紫书第一章训练1 D -Message Decoding

    Some message encoding schemes require that an encoded message be sent in two parts. The first part, c ...

  4. Message Decoding UVA - 213

     Some message encoding schemes require that an encoded message be sent in two parts. The fifirst par ...

  5. uva 213 Message Decoding

    思路来自紫书...开始时的思路估计100行+,果断放弃!关键:1.正确提取出函数!   initmap():初始化字母与整数的映射.   returnint(x):向后读取x位,并转换为十进制数返回. ...

  6. UVA - 213 Message Decoding (输入字符串并对单个字符进行操作的输入输出)

    POINT: 关于表示一个编码:利用code字符数组表示一个编码字符,其中code[len][val]表示长度为len,二进制值为val的字符: 主程序如下: #include <iostrea ...

  7. 【每日一题】 UVA - 213 Message Decoding 模拟解码+读入函数+阅读题

    题意:阅读理解难度一道比一道难orz.手摸了好久样例 题解: 读入:大循环用getline读入header顺便处理一下,  里面再写两重循环,外层一次读三个串,内层一次读num个串. 之后就查表,线性 ...

  8. uva 213 - Message Decoding (我认为我的方法要比书上少非常多代码,不保证好……)

    #include<stdio.h> #include<math.h> #include<string.h> char s[250]; char a[10][250] ...

  9. uvaoj 213 - Message Decoding(二进制,输入技巧)

    https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...

随机推荐

  1. Python 变量 对象 引用

    1.变量 变量第一次赋值时被创建,变量在使用前必须赋值 变量本身没有类型,变量类型为它引用的对象类型: 变量在使用时被替换成它引用的对象 2.对象 对象本身具有计数和类型,变量引用对象,当对象的引用变 ...

  2. Python 核心数据类型

    1.Python中一切皆对象 2.Python中不需要申明对象类型,对象的类型由运行的表达式决定 3.创建了对象意味着绑定了对象的操作到此对象,也就是在固有的对象上只能调用该对象特有的操作.比如只能将 ...

  3. LeetCode 232 Implement Queue using Stacks

    Implement the following operations of a queue using stacks. push(x) -- Push element x to the back of ...

  4. IIS中使用PUT方法錯誤記錄

    在IIS7.5中使用PUT,DELETE方法時會遇到404,405錯誤,特記錄解決辦法:404: 405: 在web.config的system.webServer節點中加入 <modules ...

  5. Java缓存学习之三:CDN缓存机制

    CDN是什么? 关于CDN是什么,此前网友详细介绍过. CDN是Content Delivery Network的简称,即"内容分发网络"的意思.一般我们所说的CDN加速,一般是指 ...

  6. 修复Debian(Ubuntu)Grub2 引导

    重装win7, 之前的系统debian 的引导就没有了. 而debian 的盘似乎没有ubuntu的livecd模式,于是用ultraISO将ubuntu的ios文件写入到u盘中. boot时选择启动 ...

  7. DB2 递归查询

    上一篇中讲解了ORACLE中的递归查询,下面我们看一下DB2中如何使用递归查询: 同样的我们先新建一个表来存储以上信息,并插入测试数据: --建表 create table FAMILY ( pers ...

  8. CodeForces 682E Alyona and Triangles (计算几何)

    Alyona and Triangles 题目连接: http://acm.hust.edu.cn/vjudge/contest/121333#problem/J Description You ar ...

  9. Azure 公网及内网ip绑定方法

    此文章为我的云服务绑定情况,仅供参考,适用于已经创建vm,但开始未绑定vip,后期进行vip的绑定,注意:绑定ip会造成虚拟机暂时离线. -Location(vm所在地) 注意区分 北部和东部: Ch ...

  10. [iOS基础控件 - 6.10.3] DatePicker & UIToolBar

    A.需求 1. 学习DatePicker的基本配置 2.使用TextField召唤指定类型的输入键盘View,这里使用DatePicker 3.给输入键盘上方加上一个UIToolBar,实现如关闭键盘 ...