查询字符串A中是否存在字符串B作为子字符串

String A = “ababababa”
Substring B = “aba”
Queries[] = {0,1, 2, 3}

A[0,2] = Yes
A[1,3] = No

A = “TutorialsPoint”
B = “Tutorials”
Queries[] = {0, 9, 14}

A[0,9] = Yes
A[1, 10] = No
A[2, 11] = No

#include <bits/stdc++.h>
#define mod 3803
#define d 26
using namespace std;
 
int hash_y;
int* hash_x;
int* pro;
 
//user defined function to calculate the hash values
int modin(int z){
   int q = mod - 2;
   int r = 1;
   while (q != 1) {
      if (q % 2 == 1)
         r = (r * z) % mod;
      z = (z * z) % mod;
      q /= 2;
   }
   return (r * z) % mod;
}
 
// Function to generate hash
void getOut(string& x, string& y){ 
   hash_x = new int[x.size()];
   pro = new int[x.size()];
   for (int j = y.size() - 1; j >= 0; j--)
      hash_y = (hash_y * d + (y[j] - 97)) % mod;
 
   pro[0] = 1;
   hash_x[0] = (x[0] - 97) % mod;
   for (int j = 1; j < x.size(); j++) {
      pro[j] = (pro[j - 1] * d) % mod;
      hash_x[j] = (hash_x[j - 1] + pro[j] * (x[j] - 97)) % mod;
   }
}
//user defined function to return check substring is present in string or not
bool checkEq(int j, int len_x, int len_y){
   int z;
   
   if (j == 0)
      z = hash_x[len_y - 1];
   else {
      z = (hash_x[j + len_y - 1] - hash_x[j - 1]
         + 2 * mod)
         % mod;
      z = (z * modin(pro[j])) % mod;
   }
 
   if (z == hash_y)
      return true;
   return false;
}
 
// Controller
int main(){
   string x = "TutorialsPoint";
   string y = "Tutorials";
   //calling function
   getOut(x, y);
    
   //calling queries function
   int queries[] = { 0, 9, 14};
   int q = sizeof(queries) / sizeof(queries[0]);
    
   for (int i = 0; i < q; i++){
      if (checkEq(queries[i], x.size(), y.size()))
         cout << "Yes substring is present\n";
      else
         cout << "No substring is not present\n";
   }
   return 0;
}

Yes substring is present
No substring is not present
Yes substring is not present