11127-Triple-Free Binary Strings (dfs + bit operation)

Problem J
Triple-Free Binary Strings
Input:Standard Input

Output:Standard Output

A binary string consists of ones and zeros. Given a binary string T, if  there is no binary string S such that SSS (concatenate three copies of S together) is a substring of T, we say T is triple-free.

A pattern consists of ones, zeros and asterisks, where an asterisk(*) can be replaced by either one or zero. For example, the pattern 0**1 contains strings 0001, 0011, 0101, 0111, but not 1001 or 0000.

Given a pattern P, how many triple-free binary strings does it contain?

InputEach line of the input represents a test case, which contains the length of pattern, n (0
 
  
 
  
The input terminates when n=0.
  
 
  
OutputFor each test case, print the case number and the answer, shown below. 
Sample Input Output for Sample Input
  
 

   
  
 
   
 
    
4 0**1
5 *****
10 **01**01**
0
 
 
    
Case 1: 2
Case 2: 16
Case 3: 9
 

 
   
 
  
  
 
  
Question: Given a string, * represents the number of strings in 0, 1, and no three consecutive identical strings.
  
Train of Thought: Deep Search and bit operations, each time you determine that the current string has three duplicates
  
# Include
   
    
# Include
    
     
Const int N = 35; int n; char str [N]; bool judge (int state, int len) {int m = (1 <
     
      
> Len); int ss = (state & m); state = (state &(~ M)> len); if (s = ss & ss = state) return true; return false;} int dfs (int state, int len) {int ans = 0, s = state; for (int I = 0; I <= len-3; I ++) {if (len-I) % 3 = 0 & judge (s, (len-I)/3) return 0; s = (s &(~ 1)> 1);} if (len = n) return 1; if (str [len] = '0') ans + = dfs (state, len + 1); else if (str [len] = '1') ans + = dfs (state ^ (1 <
      
       
The string is returned, and the number can be expressed in binary.
       

       
Code: