Here you will find solutions of many problems on spoj. If you want solution of some problem which is not listed in blog or have doubt regarding any spoj problem (which i have solved) or any programming concept (data structure) you can mail me @ raj.nishant360@gmail.com
And my humble request to you all that don't copy the code only try to understand the logic and algorithm behind the code. I have started this because if you tried as hard as you can and still can't find any solution to the problem then you can refer to this.
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Tuesday, September 15, 2015

PLD-Palindromes

Palindromes

Given below code is for pld spoj or palindromes spoj.

Hint :- Simple implementation of Rolling Hash.




/*
===================================================
Name :- Nishant Raj
Email :- raj.nishant360@gmail.com
College :- Indian School of Mines
Branch :- Computer Science and Engineering
Time :- 15 September 2015 (Tuesday) 00:19
===================================================*/
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair < int , int >
#define pb push_back
#define mp make_pair
#define mod 1000000009
#define base 1223
ll prime[30009] , fh[30009] , bh[30009];
void prime_power(int n){
    prime[0] = 1;
    for(int i = 1 ; i <= n + 5 ; i++)
        prime[i] = (prime[i-1]*base)%mod;
}
void Hash(string &s , int n){
    for(int i = 1, j = n ; i<= n ;j--, i++){
        fh[i] = (fh[i-1] + s[i-1]*prime[i])%mod;
        bh[j] = (bh[j+1] + s[j-1]*prime[i])%mod;
    }
}
int count_palindromic_substring(string &s , int k){
    int n = s.size();
    int count = 0;
    int flag = 1;
    for(int i = k , j = 0 ,p_pow = n-k ; i<= n ;p_pow-=2 , j++,i++){
        ll hs1 , hs2;
        if(p_pow>=0){
            hs1 = (fh[i] - fh[j] + mod)%mod;
            hs1 = (hs1*prime[p_pow])%mod;
            hs2 = (bh[j+1] - bh[i+1] + mod)%mod;
        } else{
            hs2 = (bh[j+1] - bh[i+1] + mod)%mod;
            hs2 = (hs2*prime[abs(p_pow)])%mod;
            hs1 = (fh[i] - fh[j] + mod)%mod;
        }
        if(hs1 == hs2)
            count++;
    }
    return count;
}
int main(){
    
    string s ;
    int k;
    cin>>k>>s;
    prime_power(s.size());
    Hash(s , s.size());
    cout<<count_palindromic_substring(s , k)<<endl;
    return 0;
}