设逆序对个数为 $I$，显然其数量一直在减少，最终变为 $0$。注意到翻转 $\texttt{10}$ 会让 $I$ 减少 $1$，其余操作会让 $I$ 减少 $2$，且只要 $I\ge 2$，就一定存在一种操作让 $I$ 减少 $2$。因此先手必胜当且仅当 $I\not\equiv 0\mod 3$。
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Discussion #316 for Problem #1644. Reverse Game
Type: Editorial
Status: Open
Posted by: jiangly
Posted at: 2025-12-14 07:03:23
Last updated: 2025-12-14 07:03:28
题解
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