Hints: Learning Binary Search

$f(a,k,1,n)$ counts comparisons (plus the final success check) in a standard binary search for $k$ in the sorted array $a$ on indices $[1,n]$. If $k$ is missing, the function returns $0$ immediately.

Over all nondecreasing arrays with values in $[1,m]$ of length $n$, you must sum these costs for every $k \in [1,m]$. What linearity-of-expectation or double-counting viewpoint helps?

Answer to Hint 1: Sum over arrays first, then over $k$: equivalently, for each array add $\sum_{k=1}^m f(a,k,1,n)$. You can also think of summing, for each comparison depth in the implicit binary-search tree on indices $[1,n]$, how many pairs $(a,k)$ provoke that comparison.

The recursion splits $[l,r]$ at $\text{mid} = \lfloor(l+r)/2\rfloor$.

Answer to Hint 2: Let $F(l,r,\ell,\rho,d)$ be a DFS that mirrors the comparison tree: $\ell$ and $\rho$ are the last pivot indices from the left and right child calls (or sentinels if none), and $d$ is the current depth contribution.

At node $(l,r)$ with midpoint $\text{mid}$, the stars-and-bars count of nondecreasing arrays consistent with “still searching this interval” is needed. Precompute $C(i) = \binom{i+m-1}{i}$, the number of weakly increasing sequences of length $i$ with values in $[1,m]$.

Answer to Hint 3: When the search interval is all of $[1,n]$ (both sides “empty”), every good array contributes $d \cdot C(n)$ to the total for that subtree’s weighting—here $d$ tracks accumulated comparison cost along the path.

When one side is missing (e.g. target entirely to the right of the last pivot), subtract sequences that would have terminated earlier: counts like $C(n - (\text{mid}-\ell))$ appear.

Answer to Hint 4: When both left and right pivots exist, inclusion–exclusion prevents double subtraction:

$$ C(n) - C(n - (\text{mid}-\ell)) - C(n - (\rho-\text{mid})) + C(n - (\rho-\ell)). $$

Each term is multiplied by the current depth parameter $d$ (starting at $1$ for the root call) and accumulated modulo $676,767,677$.

Answer to Hint 5: Recurse into $[l,\text{mid}-1]$ and $[\text{mid}+1,r]$, passing updated pivot markers $(\ell,\text{mid})$ or $(\text{mid},\rho)$ and incrementing $d$ by $1$ for child calls.

Precompute factorials inverses modulo the prime for binomials up to needed $n+m$.

Answer to Hint 6: The recursion depth is $O(\log n)$; each test case runs in time proportional to the number of recursive nodes (roughly $O(n)$ for the comparison tree). The answer is the accumulated ans from all DFS visits.