Editorial for DMOPC '23 Contest 1 P2 - Knights on Chessboard

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Author: yzhao123

There are many solutions to this problem. The intended solution will be described here.

Notice that placing a full row of knights covers five rows: itself, and the two rows above and below it. Thus, starting from the third row, and every five rows, we place a full row of knights. To fit within the bound of knights, casework is required for handling $N = 1 \pmod 5$ ~N = 1 \pmod 5~ and $N = 2 \pmod 5$ ~N = 2 \pmod 5~.

Comments

-15

volcano commented on Dec. 6, 2023, 1:40 a.m.

If you always fill the third-last row with knights (similar to the editorial) you only have one edge case: $N = 1 \pmod{5}$ ~N = 1 \pmod{5}~ because the second one is always handled.

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