Hints: A Trivial String Problem

$f(t)$ means: split $t$ into the maximum number of non-empty contiguous pieces, where each piece must be a prefix of $t$.

Before thinking about all queries, solve the easiest subtask: for one fixed string $T$, how would you compute $f(T)$ by brute force?

Answer to Hint 1: Use DP on positions. Let dp[i] be the maximum number of pieces to build prefix $T[0..i-1]$.

Transition: try every end i and every start j < i; if substring $T[j..i-1]$ is a prefix of $T$, then dp[i] = max(dp[i], dp[j] + 1).

This is already $O(m^3)$ if prefix-check is naive (m = |T|), or $O(m^2)$ with faster substring-prefix checks.

Answer to Hint 2: To optimize checks of “is $T[j..i-1]$ a prefix of $T$?”, build the Z-function on $T$.

Recall: z[p] is the longest match length between $T[0..]$ and $T[p..]$. Then $T[p..p+len-1]$ is a prefix iff z[p] >= len.

So you can test transitions in $O(1)$ after one $O(m)$ preprocessing.

Answer to Hint 3: Good. But you can do better than “DP + all previous j”.

Now define F[x] = f(T[0..x-1]) for all x = 1..m.
For each position p >= 1, z[p] creates an interval [p, p+z[p]] of prefix lengths where a jump from boundary p can still be valid.

Can you maintain only the currently useful interval(s) while scanning x from left to right?

Answer to Hint 4: Maintain a stack of pairs (start = p, end = p + z[p]) while increasing x.

• add interval for p = x - 1 (except p = 0)
• remove intervals with end < x (they cannot help current length anymore)
• let pre be start of the last remaining interval; if none, pre = 0

Then:

F[x] = F[pre] + 1

This computes all F[x] in linear time for fixed T.

Answer to Hint 5: Also keep prefix sums of F:

pref[x] = pref[x-1] + F[x]

For a fixed left boundary l in the original string, set T = s[l..n] (reindexed).
Then for any r >= l:

$$ \sum_{j=l}^{r} f(s[l..j]) = \sum_{x=1}^{r-l+1} F[x] = pref[r-l+1]. $$

So one run on T answers all queries with this same l.

Answer to Hint 6: Now optimize across many queries.

If you rebuild from scratch per query, it is too much. Instead, group queries by left endpoint l.

Inside one group:

• let maxR be the largest r
• build T = s[l..maxR]
• compute one Z, one F, one pref
• answer each query (l, r) by pref[r-l+1]

Answer to Hint 7: Complexity:

• per distinct l: O(maxR-l+1) (linear Z + linear scan)
• total over one test: O(n * (#distinct l among queries))

Since q <= 100, this is safe, and the global bound sum n <= 1e6 is comfortable.

Answer to Hint 8: Final implementation checklist:

1. Read all queries and bucket by l.
2. For each bucket, compute up to its maxR.
3. Build z for T.
4. Scan x = 1..m with the interval stack to fill F and pref.
5. Fill ans[idx] = pref[r-l+1] for each query in that bucket.
6. Print in original query order.

That is exactly the easy-to-hard path: naive fixed-string idea -> Z optimization -> subarray sum via pref -> grouped-query optimization.