Editorial: Amanojaku and Sequence (Easy Version)

E1 is the same combinatorial core as E0, but you are given a subarray $b=[a_l,\ldots,a_r]$ and a single query per test ($q=1$). We build the solution in three steps:

All $-1$, full array — warmup $G(n,m)$.
All $-1$, subarray — answer depends only on length.
Fixed and free — DP on the subarray.

Step 1: All -1 on the full array

If $b=(-1,\ldots,-1)$ has length $n$, every valid $c$ is a composition of $m$ into $n$ non-negative parts. Define

$$G(n,m)=\sum_{c_1+\cdots+c_n=m} f(c),\qquad f(c)=\sum_{i=1}^{n} P_i^2,\quad P_i=\sum_{j=1}^{i}c_j.$$

Then $g(b,m)=G(n,m)$. The four-table DP from E0 applies: for each $s\in[0,m]$, maintain $\mathrm{cnt}$, $\mathrm{pref}$, $\mathrm{pref2}$, $\mathrm{sq}$; one free position is a layer $T$; answer $G(n,m)=\mathrm{sq}[m]$.

See E0 editorial for the full transition formulas.

Step 2: All -1 on a subarray

Only the length matters

If every $a_i=-1$ on the whole array, any subarray $[a_l,\ldots,a_r]$ is $L=r-l+1$ copies of $-1$. Valid $c$ for the subarray are compositions of $m$ into $L$ parts, so

$$g([a_l,\ldots,a_r],m)=G(L,m).$$

The sample block with $a=(-1,-1,-1,-1)$ and queries $(l,r,m)=(1,1,4),(1,2,5),(1,3,6),(1,4,7)$ returns $G(1,4)=16$, $G(2,5)=205$, $G(3,6)=1736$, $G(4,7)=12180$.

One free layer in O(R) time

Let $R=m$ (no fixed values). A naïve free step loops $v$ for each $s$ and costs $O(R^2)$. It can be done in $O(R)$ using prefix sums:

$\mathrm{cnt}'[k]=\sum_{s\le k}\mathrm{cnt}[s]$.
$\mathrm{pref}'[k]=\sum_{s\le k}(\mathrm{pref}[s]+\mathrm{cnt}[s]\cdot(k-s))$.
$\mathrm{pref2}'$ and $\mathrm{sq}'$ use the same prefix structure (see model_sol.cpp).

Many free positions: O(R log L)

When the subarray is all $-1$, you need $T^L(\mathrm{init})$ — apply the free layer $L$ times. Binary exponentiation on the map $T$:

keep maps mul and res (composition of layers);
if $L$ has a bit set, res ← res ∘ mul;
mul ← mul ∘ mul;
answer from res(init).

Each composition is two $O(R)$ applications, so $O(R\log L)$ per all-free query. With $m\le 10^6$, $R\le m$ and this fits E1 limits.

Step 3: Fixed and free on the subarray

Feasibility

For general $b=[a_l,\ldots,a_r]$:

$$F=\sum_{b_i\ge 0} b_i,\qquad R=m-F.$$

If $R<0$, no valid $c$. Otherwise DP only on sums $s=0,\ldots,R$ (mass still to place on $-1$ slots).

Transitions on the subarray

Scan $b$ left to right (same as E0 on the whole array):

Type	Effect on $s$	Cost
$b_i=-1$	free layer $T$	$O(R)$
$b_i=v\ge 0$	add $v$ to prefix, same $s$	$O(R)$

Fixed $v$ at sum $s$ (mass $c=\mathrm{cnt}[s]$):

$$\mathrm{sq}[s]\mathrel{+}=\mathrm{pref2}[s]+2v\,\mathrm{pref}[s]+c\,v^2,$$

then update $\mathrm{pref}[s]$, $\mathrm{pref2}[s]$.

Free slot: use the $O(R)$ layer from Step 2.

Answer: $\mathrm{sq}[R]$ modulo $998\,244\,353$.

Walk-through: $[6,-1]$, $m=8$ (sample test 3)

$F=6$, $R=2$. After fixed $6$: one assignment with prefix $6$, $\mathrm{sq}[0]=36$. One free layer gives $c=[6,2]$, $\mathrm{sq}[2]=100$.

Walk-through: $[-1,-1,3,-1]$, $m=5$ (sample test 10)

$F=3$, $R=2$. Fixed $3$ in the middle, then free/free around it; answer $286$.

E1 algorithm (one query per test)

Read $n$, array $a$, query $2\;l\;r\;m$.
Build $b=a_l,\ldots,a_r$, $L=|b|$.
If every $b_i=-1$: output $G(L,m)$ via $O(R\log L)$ exponentiation ($R=m$).
Else: run the mixed DP on $b$ in $O(L\cdot R)$.
If $R<0$, output $0$.

Complexity

Case	Time
Subarray all $-1$	$O(R\log L)$, $R\le m$
General subarray	$O(L\cdot R)$

Per test case one query; $\sum n\le 3\cdot 10^5$ over the input. All-free queries with large $L$ use logarithmic layering; mixed queries are $O(L\cdot R)$, and $R$ is often small when fixed values use most of $m$.

E2 keeps the same $g(b,m)$ but adds many queries and point updates on $a$ — you need a data structure on top of this DP core.

Reference code

model_sol.cpp implements:

apply_free_layer — $O(R)$;
pow_free_layers — binary exponentiation for all-free subarrays;
apply_fixed and a linear scan for the general case.

Uses AtCoder modint998244353.