计算长度为N的二进制字符串，它们是子字符串的重复拼接

Let us take the Input, N = 3

Output: 2

"000":The substring "0" is repeatedly concatenated to form this string.
"111":The substring "1" is repeatedly concatenated to form this string.

Let us take the Input, N = 8

Output: 16

“00000000”: The substring "0" is repeatedly concatenated to form this string.
“11111111”: The substring "1" is repeatedly concatenated to form this string.
“01010101”: The substring "01" is repeatedly concatenated to form this string.
“10101010”: The substring "10" is repeatedly concatenated to form this string.
"00110011”: The substring "0011" is repeatedly concatenated to form this string.
"11001100”: The substring "1100" is repeatedly concatenated to form this string.
"11011101”: The substring "1101" is repeatedly concatenated to form this string.
"00100010”: The substring "0010" is repeatedly concatenated to form this string.
"10111011”: The substring "1011" is repeatedly concatenated to form this string.
"01000100”: The substring "0100" is repeatedly concatenated to form this string.
"10001000”: The substring "1000" is repeatedly concatenated to form this string.
"00010001”: The substring "0001" is repeatedly concatenated to form this string.
"11101110”: The substring "1110" is repeatedly concatenated to form this string.
"01110111”: The substring "0111" is repeatedly concatenated to form this string.
"01100110”: The substring "0110" is repeatedly concatenated to form this string.
"10011001”: The substring "1001" is repeatedly concatenated to form this string.

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;

// Storing all the states of recurring recursive 
map<int, int> dp;

// Function for counting the number of strings of length n wherein thatstring is a  concatenation of its substrings
int countTheStrings(int n){

   //the single character cannot be repeated
   if (n == 1)
      return 0;
      
   // Checking whether the state is calculated already or not
   if (dp.find(n) != dp.end())
      return dp[n];
      
      // Storing those value of the result or the count for the present recursive call
   int res = 0;
   
   // Iterate through all of the divisors
   for(int d= 1; d <= sqrt(n); d++){
      if (n % d== 0){
         res += (1 << d) -  countTheStrings(d);
         int div1 = n/d;
         if (div1 != d and d!= 1)
         
            // Non-Rep = Total - Rep
            res += (1 << div1) -  countTheStrings(div1);
      }
   }
   
   // Storing the result of the above calculations
   dp[n] = res; 
   
   // Returning the obtained result
   return res;
}
int main(){
   int n = 8;
   cout<< "Count of 8-length binary strings that are repeated concatenation of a substring: "<< endl;
   cout << countTheStrings(n) << endl;
}

Count of 8-length binary strings that are repeated concatenation of a substring −
16