Let Z[i] be the Z-value array of the suffix of S starting at position i. Then, for each position k, compute maxZ[k], the maximum of Z[0][k], Z[1][k-1], Z[2][k-2], ..., Z[k-1][1] (the maximum prefix of S[k..n) that also occurs in some earlier prefix of S). Then, all substrings starting at position k of length up to maxZ[k] have already occurred so you do not want to count those, but you count all longer substrings. Therefore, the answer is sum{k=0 to n} (n — k — maxZ[k]).

for each prefix i of the word, reverse it and do z_function over it, the number of new distinct substrings that end in the prefix i is (the length of the prefix) — (maximum value in the z_function array) the pseudo code look like this:

string s; cin >> s;
int sol = 0
foreach i to s.size()-1
    string x = s.substr( 0 , i+1 );
    reverse( x.begin() , x.end() );
    vector<int> z = z_function( x );
    //this work too
    //vector<int> z = prefix_functionx(x); 
    int mx = 0;
    foreach j to x.size()-1
        mx = max( mx , z[j] );
    sol += (i+1) - mx; 

cout << sol;

The following is the wrong algorithm, sorry for this.

I guess it is better to use trie data structure that will reduce the time complexity to O(26*N) but it effectively use to give number of distinct substrings as you want to calculate.

Code link: https://wtools.io/paste-code/bDox