Assuming you have read the editorial for Problem F, we present the case analysis for \(M = 5\).

Since \(5\) is prime, it suffices to handle the six cases where the set of elements in shortage is one of \(\lbrace \rbrace,\lbrace 4 \rbrace,\lbrace 1,4 \rbrace,\lbrace 3,4 \rbrace,\lbrace 2,3,4 \rbrace,\lbrace 1,2,3,4 \rbrace\), as all \(2^4 = 16\) cases can be reduced to one of these by multiplying all elements by a constant.

• When $\lbrace \rbrace$ is in shortage

Possible.

• When $\lbrace 4 \rbrace$ is in shortage

There are five options: $\lbrace 1,3 \rbrace,\lbrace 2,2 \rbrace,\lbrace 1,1,1,1 \rbrace,\lbrace 1,1,2 \rbrace,\lbrace 3,3,3 \rbrace$. When at least one each of $1$ and $3$ is available, we may use $\lbrace 1,3 \rbrace$. We prove this by assuming a solution that does not contain $\lbrace 1,3 \rbrace$ and case-splitting based on whether $1$ and $3$ are leftover in the end.

• When both $1$ and $3$ are leftover

Produce one copy using $\lbrace 1,3 \rbrace$ and remove one other set.

• When only $1$ is leftover

Since $\lbrace 3,3,3 \rbrace$ is the only way to consume $3$, at least one $\lbrace 3,3,3 \rbrace$ is being used. Remove one of them and produce $\lbrace 1,3 \rbrace$.

• When only $3$ is leftover

$1$ must be consumed by $\lbrace 1,1,1,1 \rbrace$ or $\lbrace 1,1,2 \rbrace$, so remove one of them and produce $\lbrace 1,3 \rbrace$.

• When neither $1$ nor $3$ is leftover

At least one of $\lbrace 1,1,1,1 \rbrace$ or $\lbrace 1,1,2 \rbrace$ and at least one $\lbrace 3,3,3 \rbrace$ are being used, so remove one of each and produce two copies of $\lbrace 1,3 \rbrace$.

This completes the argument. We repeat this until at least one of $1$ and $3$ is exhausted. Then we case-split based on which is leftover, $1$ or $3$.

• When $1$ is exhausted

Produce as many as possible independently using $\lbrace 2,2 \rbrace$ and $\lbrace 3,3,3 \rbrace$.

• When $3$ is exhausted

The candidates are $\lbrace 2,2 \rbrace,\lbrace 1,1,2 \rbrace,\lbrace 1,1,1,1 \rbrace$. Pair up $1$'s as much as possible, then pair up $\lbrace 1,1 \rbrace$'s and $\lbrace 2 \rbrace$'s as much as possible.

Thus, this case is resolved. Summarizing the above argument, the greedy strategy of producing as many as possible in the order $\lbrace 1,3 \rbrace,\lbrace 2,2 \rbrace,\lbrace 1,1,1,1 \rbrace,\lbrace 1,1,2 \rbrace,\lbrace 3,3,3 \rbrace$ is in fact correct.

• When $\lbrace 1,4 \rbrace$ is in shortage

We produce $1$ using $\lbrace 2,2,2 \rbrace$ or $\lbrace 3,3 \rbrace$, and $4$ using $\lbrace 2,2 \rbrace$ or $\lbrace 3,3,3 \rbrace$. When $1$ is in shortage and at least two copies of $3$ are available, we may use $\lbrace 3,3 \rbrace$ to produce $1$. Assuming there is a solution that does not use $\lbrace 3,3 \rbrace$, we can construct one that uses at least one $\lbrace 3,3 \rbrace$ by the following case analysis.

• When two or more copies of $3$ are leftover

Produce $1$ using $\lbrace 3,3 \rbrace$ and remove one $\lbrace 2,2,2 \rbrace$.

• When at most one copy of $3$ remains

At least one each of $\lbrace 2,2,2 \rbrace$ and $\lbrace 3,3,3 \rbrace$ are being used to produce $1$ and $4$ respectively, so remove one of each and replace them with $\lbrace 2,2 \rbrace$ and $\lbrace 3,3 \rbrace$.

Thus, we may keep using $\lbrace 3,3 \rbrace$ to produce $1$ until either $3$ is reduced to at most one copy or all $1$'s have been produced. Then we case-split based on which occurred.

• When $3$ is reduced to at most one copy

We must produce $1$ using $\lbrace 2,2,2 \rbrace$ and $4$ using $\lbrace 2,2 \rbrace$.

• When all $1$'s have been produced

Produce as many $4$'s as possible using $\lbrace 2,2 \rbrace$ and $\lbrace 3,3,3 \rbrace$.

Thus, this case is resolved. Summarizing the above argument, the greedy strategy of producing as many as possible in the order $\lbrace 2,2 \rbrace,\lbrace 3,3 \rbrace,\lbrace 2,2,2 \rbrace,\lbrace 3,3,3 \rbrace$ is in fact correct.

• When $\lbrace 3,4 \rbrace$ is in shortage

We produce $3$ using $\lbrace 1,1,1 \rbrace,\lbrace 1,2 \rbrace,\lbrace 2,2,2,2 \rbrace$ and $4$ using $\lbrace 1,1,1,1 \rbrace,\lbrace 1,1,2 \rbrace,\lbrace 2,2 \rbrace$. We case-split based on whether $X_1 \ge Y_3$ or not.

• When $X_1 \ge Y_3$

Clearly $X_1 + 2X_2 \ge 3Y_3 + 4Y_4$ is necessary. When this is satisfied, we give one copy of $1$ to each $3$, and then we can produce $Y_3$ copies of $3$ and $Y_4$ copies of $4$ from the remaining $1$'s and $2$'s using $\lbrace 1,1 \rbrace$ and $\lbrace 2 \rbrace$.

• When $X_1 < Y_3$

The $Y_3 - X_1$ copies of $3$ must be produced using $\lbrace 2,2,2,2 \rbrace$, so $X_1 + 2X_2 \ge 3X_1 + 4Y_4 + 8(Y_3-X_1)$ is necessary. Conversely, when this inequality holds, we give one $1$ to each of the $X_1$ copies of $3$, and we need to produce $X_1$ copies of $2$, $Y_4$ copies of $4$, and $Y_3-X_1$ copies of $3$ from $X_2$ copies of $2$. Subtracting $X_1$ from both sides and dividing by $2$ gives $X_2 \ge X_1 + 2Y_4 + 4(Y_3-X_1)$, which holds, so this is achievable.

Thus, this case is resolved. Summarizing the above argument, a greedy strategy such as producing as many as possible in the order $\lbrace 1,2 \rbrace,\lbrace 1,1,1 \rbrace,\lbrace 2,2 \rbrace,\lbrace 1,1,2 \rbrace,\lbrace 1,1,1,1 \rbrace,\lbrace 2,2,2,2 \rbrace$ is correct.

• When $\lbrace 2,3,4 \rbrace$ is in shortage

The only options are to produce $2,3,4$ using $\lbrace 1,1 \rbrace,\lbrace 1,1,1 \rbrace,\lbrace 1,1,1,1 \rbrace$, respectively. The condition is $2Y_2 + 3Y_3 + 4Y_4 \le X_1$.

• When $\lbrace 1,2,3,4 \rbrace$ is in shortage

Impossible.

Thus, all cases have been resolved. The time complexity is \(\mathrm{O}(Q)\) treating \(M\) as a constant.