Editorial for DMPG '19 S2 - Code Cracking Crisis - Olympiads Online Judge

Editorial for DMPG '19 S2 - Code Cracking Crisis

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Author: AvaLovelace

First, we find the inverse $k^{-1}$ ~k^{-1}~ of the encryption keys— $k^{-1}[i][j]$ ~k^{-1}[i][j]~ is equal to the position of digit $j$ ~j~ in the $i$ ~i~-th key.

Then we can decrypt the code from left to right. As we obtain each $d[i]$ ~d[i]~, we can find $d[i + 1]$ ~d[i + 1]~ by seeing which digit in the $d[i]$ ~d[i]~-th key would result in an encrypted digit of $e[i + 1]$ ~e[i + 1]~.

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