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

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Author: Tzak

For the first subtask, it suffices to try possible values of c less than 2^{15}. The resulting multisets can be compared with sorting in \mathcal O(N \log N).

Time complexity: \mathcal O(\max A \cdot N \log N)

For the second subtask, we observe that if a value for c exists, there is some A_i such that A_i \oplus c = B_1, meaning c = A_i \oplus B_1. With this, we have 2N+1 possible candidates for c: A_1 \oplus B_1, A_2 \oplus B_1, \dots, A_{2N+1} \oplus B_1.

Time complexity: \mathcal O(N^2 \log N)

For the final subtask, we will use the fact that 2N+1 is odd. Let's begin by assuming that we have found a value of c which makes the multisets equal. This implies that:

\displaystyle A_1 \oplus c \oplus A_2 \oplus c \oplus \dots \oplus A_{2N+1} \oplus c = B_1 \oplus B_2 \oplus \dots \oplus B_{2N+1}

Let's XOR both sides by A_1 \oplus A_2 \oplus \dots \oplus A_{2N+1}:

\displaystyle c \oplus c \oplus \dots \oplus c = B_1 \oplus B_2 \oplus \dots \oplus B_{2N+1} \oplus A_1 \oplus A_2 \oplus \dots \oplus A_{2N+1}

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

\displaystyle c = B_1 \oplus B_2 \oplus \dots \oplus B_{2N+1} \oplus A_1 \oplus A_2 \oplus \dots \oplus A_{2N+1}

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 c works.

Time complexity: \mathcal O(N \log N)


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