计算三个不重叠的子字符串，将它们连接起来形成一个回文串

String s = “abbacab”      
Acceptable substrings of length 3 are: “abb”, “bac”, and “bba”.

string_name,size();  

#include <bits/stdc++.h>
using namespace std;
 
// user defined function to check formed substrings are palindrome or not
bool isStringPalin(int a, int b, int c, int d, int x, int y, string st){
   int begin = a, stop = y;
   while (begin < stop) {
      if (st[begin] != st[stop])
         return false;
 
      begin++;
      if (begin == b + 1)
         begin = c;
      stop--;
      if (stop == x - 1)
         stop = d;
   }
   return true;
}
 
// User defined function to count the number of useful substrings
int countSubString(string st){
   //Counting variable to count and return the number of substrings
   int ct = 0;
   int l = st.size();
 
   //It is to select the first substring
   for (int a = 0; a < l - 2; a++) {
      for (int b = a; b < l - 2; b++){
 
         // This loop selects the second useful substring
         for (int c = b + 1; c < l - 1; c++) {
            for (int d = c; d < l - 1; d++) {
 
               // this for loop will select the third substring
               for (int x = d + 1; x < l; x++) {
                  for (int y = x; y < l; y++) {
 
                     // If condition to check the selected substrings are forming palindrome or not
                     if (isStringPalin(a, b, c, d, x, y, st)) {
                        ct++;
                     }
                  }
               }
            }
         }
      }
   }
   // returning the count variable that stores the number of useful substrings
   return ct;
}
 
// Controlling code
int main(){
   string st = "abcab";
   cout << "The possible number of substrings are: "<< countSubString(st);
 
   return 0;
}

The possible number of substrings are: 4