This is one of the problems in SPOJ (Sphere Online Judge (SPOJ))

The solution consists of constructing the suffix array and then finding the number of distinct substrings based on the Longest Common Prefixes.

One key observation here is that:

If you look through the prefixes of each suffix of a string, you have covered all substrings of that string.

Let us take an example: BANANA

Suffixes are:

0) BANANA
1) ANANA
2) NANA
3) ANA
4) NA
5) A

It would be a lot easier to go through the prefixes if we sort the above set of suffixes, as we can skip the repeated prefixes easily.

Sorted set of suffixes:

5) A
3) ANA
1) ANANA
0) BANANA
4) NA
2) NANA

From now on,

LCP = Longest Common Prefix of 2 strings.

Initialize

ans = length(first suffix) = length("A") = 1.

Now consider the consecutive pairs of suffixes, i.e, [A, ANA], [ANA, ANANA], [ANANA, BANANA], etc. from the above set of sorted suffixes.

We can see that,
LCP("A", "ANA") = "A".

All characters that are not part of the common prefix contribute to a distinct substring. In the above case, they are 'N' and 'A'. So they should be added to ans.

So we have,

ans += length("ANA") - LCP("A", "ANA")  
ans = ans + 3 - 1 = ans + 2 = 3 

Do the same for the next pair of consecutive suffixes: ["ANA", "ANANA"]

LCP("ANA", "ANANA") = "ANA". 
ans += length("ANANA") - length(LCP) 
=> ans = ans + 5 - 3 
=> ans = 3 + 2 = 5. 

Similarly, we have:

LCP("ANANA", "BANANA") = 0 
ans = ans + length("BANANA") - 0 = 11 

LCP("BANANA", "NA") = 0 
ans = ans + length("NA") - 0 = 13 

LCP("NA", "NANA") = 2 
ans = ans + length("NANA") - 2 = 15 

Hence the number of distinct substrings for the string "BANANA" = 15.