uvaoj-213:字符串

 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:

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 AEIOU0010101100011101000100111011001111000$#**\0100000101101100011100101000

Sample output

TAN ME##*\$

题解：模拟题，关键是利用函数来把复杂的问题进行简化，分别操作，这样可以降低错误率，虽然作为小白的我依然调试了很长时间；

下边的代码和lrj书上的代码基本相同，我所做的仅仅是做了更加详细的注释而已；估计十五号就能看到数据结构了吧，lrj的白书作为程序猿算法的入门书，里面的代码的确很棒，值得二周目；

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int s[8][1<<8];//code[len][num];
int readchar()//用来跳过回车的，避免读回车；（回车现在一般是\t\n一同进行的，\t是将光标移到行首，\n是将光标移到下一行）；
{
    for(;;)
    {
        int ch=getchar();
        if(ch!='\t'&&ch!='\n')
        return ch;
    }
}

int readints(int c)//read integers;
{
    int ans=0;
    while(c--)
    {
        ans=ans*2+readchar()-'0'; //function fitting;（函数嵌套）；
    }
    return ans;
}
int readcodes() //read string;
{
    memset(s,0 ,sizeof(s));
    s[1][0]=readchar();//因为字符串的读取是在一行就进行完成的，所以用readchar来读到第一个不是回车的字符为止，才开始继续用getchar读；
    for(int i=2; i<=7; i++)
    {
        for(int j=0; j<(1<<i)-1; j++)
        {
            int temp=getchar();

            if(temp==EOF) return 0;
            if(temp=='\n'||temp=='\r')
            return 1;//readchar的返回值为零是跳出，为一是继续循环，这种方式既可以避免在回车出跳出函数，同时可以避免将回车读入；
            s[i][j]=temp;
        }
    }
    return 1;
}

int main()
{
    while(readcodes())
    {
        for(;;)
        {
            int len=readints(3);//整数数；
            if(len == 0)  break;
            for(;;)
            {
                int v=readints(len);
                if(v==(1<<len)-1) break;
                printf("%c", s[len][v]);
            }
        }
        printf("\n");
    }
    return 0;
}

笔记：

函数名称的定义不要放在循环里，之前的一个错误就是把ans的函数再循环的时候不断地进行定义，导致输出的结果全都是零；