## Editorial for An Animal Contest 5 P2 - Permutations & Products

Remember to use this editorial

**only**when stuck, and**not to copy-paste code from it**. Please be respectful to the problem author and editorialist.**Submitting an official solution before solving the problem yourself is a bannable offence.**Author:

#### Subtask 1

Since , there are only possible permutations. Casework can be done for each of them.

**Time Complexity:**

#### Subtask 2

This editorial will highlight one of several solutions to this problem. If you have a different solution, feel free to share it in the comments!

We will query with , storing the result in an array (let's call it ). Let be the smallest element in and let be the largest element in . If , this means that . Otherwise, which implies that . Thus, we have found .

Now that we have found , we can divide every element in by it, and at this point, we have found every element in .

**Time Complexity:**

## Comments

There's an alternative solution that extends into the hard version quite nicely.

First, find three values using three queries:

With the value of , use your remaining queries to get .

Finally, since is a permutation of the first integers, . For the hard version, you can simply select different indices to find your first three values and query accordingly.