## Editorial for CPC '21 Contest 1 P2 - AQT and Multiset

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**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:

For the first subtask, it suffices to try possible values of less than . The resulting multisets can be compared with sorting in .

**Time complexity:**

For the second subtask, we observe that if a value for exists, there is some such that , meaning . With this, we have possible candidates for : .

**Time complexity:**

For the final subtask, we will use the fact that is odd. Let's begin by assuming that we have found a value of which makes the multisets equal. This implies that:

Let's XOR both sides by :

The left hand side has an odd number of copies of , so we have:

We have proven if a solution exists, then it is unique and given by the expression above. All that remains is to check if this value of works.

**Time complexity:**

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