紫书第一章训练1 D -Message Decoding（UVA213） by 16黄睿博

Posted 2020-09-14 台州学院ACM集训队

来源：http://m.blog.csdn.net/article/details?id=70766751

Some message encoding schemes require that an encoded message be sent in two parts. The ?rst 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 ?rst 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 representation of the length of the keys in the segment. For example, if the ?rst 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 ?le 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\n##*\$

这个题是真的有毒，怪不得大家没人做，其实就是字符串的处理。给你你要发的消息的文字，然后再按照每行2的n次方减一个进行转换。然后就是解决01串是干嘛的。”That is, the message segments begin with the 3-digit length sequence and end with the appropriate sequence of 1’s. “这句就是讲你接下来要扫描几位进行匹配，认识3就够了，三个字符的值代表行数，这几位的值代表列数，全是1就退出，段结束。扫三个字符确定新的行，因为不存在0行，000就退出这个字符串的处理，下一个字符串。 
这个题01串会出现换行符，getchar会扫换行符，要处理下，最后这个要结束了肯定是一行的结束，继续getchar，要不读取不了下一位，调了好久好久才发现是这个错误。 
我的代码仅供参考 
<<是左移的意思，算关于2的很方便的，另外有时卡二分的时候，你用移位要比除法的速度快，移位还是强的，同理&1也是妙用的，判断其二进制位最后一位是不是1

 

 1 #include<bits/stdc++.h>
 2 char s[10][1025];
 3 char t[520];
 4 using namespace std;
 5 void la(){
 6    int n=1,m=0;
 7    for(int  i=0;t[i];i++){
 8     s[n][m++]=t[i];
 9     if(m+1==(1<<n)){
10         n++;
11         m=0;}
12    }
13 }
14 int main()
15 {while(gets(t)!=NULL){
16     la();
17     int sum=0;
18     for(int i=0;;i++){
19         char c;
20         c=getchar();
21         if(c==‘\n‘)
22         c=getchar();
23         sum=sum*2+c-‘0‘;
24         if(i%3==2&&sum==0)break;
25         if(i%3==2){
26         for(;;){
27         int s1=0;
28         for(int j=0;j<sum;j++){
29         char c;
30         c=getchar();
31         if(c==‘\n‘)
32         c=getchar();
33         s1=2*s1+c-‘0‘;}
34         if((s1+1)==1<<sum)
35         break;
36         else
37         printf("%c",s[sum][s1]);}
38         sum=0;}
39        }
40        printf("\n");
41        getchar();}
42        return 0;
43 }