Countingstrings.cpp

/*

				Name: Mehul Chaturvedi
				IIT-Guwahati

*/

/*
				PROBLEM STATEMENT
Given a string 's' consisting of upper case alphabets, i.e. from 'A' to 'Z'. Your task is to find how many strings 't' with length equal to that of 's', also consisting of upper case alphabets are there satisfying the following conditions:
-> String 't' is lexicographical larger than string 's'.
-> When you write both 's' and 't' in the reverse order, 't' is still lexicographical larger than 's'.
Find out number of such strings 't'. As the answer could be very large, take modulo 10^9 + 7.
*/

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unordered_map<int, int> umapii;
typedef unordered_map<int, bool> umapib;
typedef unordered_map<string, int> umapsi;
typedef unordered_map<string, string> umapss;
typedef map<string, int> mapsi;
typedef map<pair<int, int>, int> mappiii;
typedef map<int, int> mapii;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef unordered_set<int> useti;

#define uset unordered_set
#define it iterator
#define mp make_pair
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define f first
#define s second
#define MOD 1000000007

#include<bits/stdc++.h>
using namespace std;
long long ans[100001],mod=1e9+7;
int go(string s){
	string str="";
	int n=s.length();
	for(int i=n-1;i>=1;i--)
	str.push_back(s.at(i));
	for(int i=0;i<n-1;i++)
	ans[i]=(ans[i-1]*26%mod+90-str.at(i))%mod;
	long long a=0;
	ans[-1]=0;
	for(int i=0;i<n;i++)
	{
		a=(a+(90-s.at(i))*(ans[n-i-2]+1)%mod)%mod;
	}
	return a;
}

int countStrings(char* s) {
    int i=0;
    string s1="";
    while(s[i]!='\0'){
        s1=s1+s[i];
        i++;
    }
    
    return go(s1);
}

int main( int argc , char ** argv )
{
	ios_base::sync_with_stdio(false) ; 
	cin.tie(NULL) ; 
	
	char s[100005];
    cin>>s;
    int ans = countStrings(s);
    cout<<ans<<endl;


	return 0 ; 



}