Hints: Amanojaku and Sequence (Easy Version)

Before subarrays: suppose the whole array is $b=(-1,-1,\ldots,-1)$ of length $n$ and you need $g(b,m)$. This is exactly the warmup function $G(n,m)$ from E0 (sum of $f(c)$ over all compositions of $m$ into $n$ non-negative parts). What four DP tables did E0 use?

Answer to Hint 1: For each total $s$ on free slots processed so far: $\mathrm{cnt}[s]$, $\mathrm{pref}[s]$ (sum of current prefix), $\mathrm{pref2}[s]$ (sum of prefix squared), $\mathrm{sq}[s]$ (sum of earlier prefix squares). Start $\mathrm{cnt}[0]=1$; answer $G(n,m)=\mathrm{sq}[m]$ after $n$ free layers.

On one free position, naively looping $v=0,1,\ldots,R$ for each $s$ costs $O(R^2)$ per layer. Can you update all $s \to s+v$ transitions in $O(R)$ using prefix sums over $s$?

Answer to Hint 3: Yes. For example $\mathrm{cnt}'[k]=\sum_{s\le k}\mathrm{cnt}[s]$. The $\mathrm{pref}$, $\mathrm{pref2}$, and $\mathrm{sq}$ updates are similar prefix sums plus weighted sums $\sum \mathrm{cnt}[s]\cdot s$ and $\sum \mathrm{cnt}[s]\cdot s^2$. One free layer becomes $O(R)$ with $R=m$ when every entry is $-1$.

E1 asks for $g([a_l,\ldots,a_r],m)$. If every $a_i=-1$ on the whole array, does the answer for $[a_l,\ldots,a_r]$ depend on $l$ and $r$ individually, or only on $L=r-l+1$ and $m$?

Answer to Hint 5: Only on $L$ and $m$. The subarray is still $L$ copies of $-1$, so

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

The seven sample tests with $a=(-1,-1,-1,-1)$ and lengths $1,2,3,4$ are exactly $G(1,4)$, $G(2,5)$, $G(3,6)$, $G(4,7)$.

You may need $G(L,m)$ with $L$ up to $3\cdot 10^5$ and $m$ up to $10^6$. Repeating one $O(R)$ layer $L$ times is $O(L\cdot R)$ — too slow. How do you apply the same free-layer map $T$ many times using binary exponentiation?

Answer to Hint 7: Treat $T$ as “one free position” on the four tables. Build $T^{2^k}$ by doubling: $T^{2k}=T^k\circ T^k$. If composing two maps costs $O(R)$ (two applications), then $T^L(\mathrm{init})$ costs $O(R\log L)$. The answer is still $\mathrm{sq}[m]$ in the final state.

Now $b=[a_l,\ldots,a_r]$ may contain values $b_i\ge 0$ (forced $c_i=b_i$) and $b_i=-1$ (free). With $F=\sum_{b_i\ge 0}b_i$ and $R=m-F$, when is the answer $0$?

Answer to Hint 9: If $R<0$, fixed values already exceed $m$. Otherwise run the same E0 DP on the subarray $b$, but only index $s=0,\ldots,R$ (free mass left to distribute). Fixed steps keep $s$; free steps use the $O(R)$ layer above.

For a fixed $b_i=v\ge 0$, every partial assignment at sum $s$ adds $v$ to its prefix. Write the bulk update to $\mathrm{sq}[s]$, $\mathrm{pref}[s]$, $\mathrm{pref2}[s]$ using $(p+v)^2=p^2+2vp+v^2$ summed over assignments — same as E0.

Answer to Hint 11: With $c=\mathrm{cnt}[s]$,

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

then update $\mathrm{pref}$ and $\mathrm{pref2}$. Cost $O(R)$ per fixed index.

E1 algorithm per query: build $b$ from $[a_l,\ldots,a_r]$. If all $b_i=-1$, use $G(L,m)$ in $O(R\log L)$. Otherwise scan $b$ left to right with fixed/free steps. What is the final answer?

Answer to Hint 13: $\mathrm{sq}[R]$ after processing all of $b$, modulo $998\,244\,353$. Read one query line ($\mathrm{op}=2$, $l$, $r$, $m$) per test case.

Complexity: Let $L=r-l+1$, $R=m-F$. All-free subarray: $O(R\log L)$. General subarray: $O(L\cdot R)$ with $O(R)$ per position (often fast when $R$ is small because fixed values ate most of $m$). E2 adds many queries and point updates; E1 is one subarray DP per test.