Alice 必败当且仅当前 $p_1$ 个数都大于等于 $p_1$。不论 Alice 怎么样操作，Bob 都能重新把 $p_1$ 放回第一个。同理，如果条件不满足，Alice 先把前 $p_1$ 个数排序，也有同样的必胜策略。计数只需要枚举 $p_1$ 即可。时间复杂度 $O(n)$。
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Discussion #154 for Problem #6736. Alice and Bob
Type: Editorial
Status: Open
Posted by: jiangly
Posted at: 2025-12-12 23:37:27
Last updated: 2025-12-12 23:37:31
题解
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