Editorial for DMOPC '23 Contest 1 P1 - Peak Grid

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Author: yzhao123

Note that there are no impossible cases. For $K = 1$ ~K = 1~, it suffices to place the largest $N$ ~N~ numbers on the bottom row. For $K = N$ ~K = N~, it suffices to place the largest $N$ ~N~ numbers on the main diagonal of the grid. Starting from the $K = 1$ ~K = 1~ grid, we can achieve each $K$ ~K~ from $2$ ~2~ to $N$ ~N~ by swapping the element on the diagonal with the element on the bottom row.

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