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

Editorial for DMPG '19 S2 - Code Cracking Crisis

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: 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 $e[i + 1]$ $e [i + 1]$ .

Comments

There are no comments at the moment.