Editorial for CPC '21 Contest 1 P2 - AQT and Multiset
Submitting an official solution before solving the problem yourself is a bannable offence.
For the first subtask, it suffices to try possible values of less than . The resulting multisets can be compared with sorting in .
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 : .
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.