The key observation here is how many combinations a given output has. Consider the sample output grouped by type: 2 9 9 9. The counts of the elements look like this:

2 : 1
9 : 3

Now for each element, recall that two sets with the same amount of all elements are the same. Assuming the rest of our elements have $K$ combinations, and our last element has $3$ total, we have $4K$ total sets. Why? We can choose to have any number from $0,1,2,3$ of our last element, and no matter which we choose, we have $K$ combinations for the rest of our elements, for a total of $4K$ combinations.

Let us have $M$ distinct elements, each with counts $C_i$. Then the total number of combinations is:

$${\displaystyle \prod _{i=1}^M(C_i+1)-1}$$

We subtract the empty combination here.

As an example suppose we are given $K=7$, as in the sample input. Then we need:

$${\displaystyle \prod _{i=1}^M(C_i+1)-1=7}$$ $${\displaystyle \prod _{i=1}^M(C_i+1)=8}$$

Well, since we get to choose $M$, what if we just factor $8$?

And indeed, the sample input has $C=\{1,3\}$. Another valid output for $K=7$ is 1 2 3, which has $C=\{1,1,1\}$.

This should give us the understanding to solve the first subtask. We can simply output $K$ of the same integer, since $K$ is so small.

For the remaining subtasks, we note that we can find the sum of $C$ to be minimal in order to check if it fits under the constraint $N\le {10}^5$. In other words, we must find a set of numbers $C_i$ such that

$${\displaystyle \prod _{i=1}^M(C_i+1)=K+1}$$

and

$${\displaystyle \sum _{i=1}^M{C}_i}$$

is minimum. By making a change of variable of $C_i+1=D_i$, it follows that

$${\displaystyle \prod _{i=1}^M{D}_i=K+1}$$

and we are trying to minimize

$${\displaystyle \sum _{i=1}^M{D}_i-1}$$

Now consider that there exists a $D_i$ such that there exist positive integers $a,b$ where $ab=D_i$. We can try to immediately minimize the sum by replacing $D_i$ with $a$ and $b$. In other words, we will replace $D_i$ with $a$ and $b$ when

$${\displaystyle \begin{array}{rl}ab-1& >a-1+b-1\\ ab-a-b+1& >0\\ (a-1)(b-1)& >0\end{array}}$$

The only positive integer solutions to the inequality are when $a,b\ge 2$, so it is always better to replace $D_i$ with $a$ and $b$. Thus, the optimal solution should have all of $D_i$ not factorable into $a,b\ge 2$ meaning that all $D_i$ should be prime. So, after factoring $K$ into its prime factors, we can compare the sum against ${10}^5$.

The second subtask rewards contestants who made these observations but implemented an inefficient solution, potentially by factoring in linear time.

The final subtask can be solved by factoring in $\mathcal{O}(\sqrt{K})$ time or other efficient methods. Implementation details are left as an exercise to the reader.

Time Complexity: $\mathcal{O}(\sqrt{K})$