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    @ 2025-3-16 11:36:42

    记事件 AAf(a)=1f(a) = 1;事件 BBa1ka_1 \geq k。所求即为 P(AB)P(A \mid B)。显然有 P(AB)=P(AB)P(B)=P(BA)P(A)P(AB) = P(A \mid B)P(B) = P(B \mid A)P(A)

    P(B)P(B) 容易用组合数求。P(BA)P(B \mid A) 即最大值大于等于 kk,同样容斥即可。P(A)=1nP(A) = \frac 1n,证明可以考虑将所有循环移位相等的数组配对。

    时间复杂度 O(n+m)\mathcal O(n + m)

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    ID
    1148
    Time
    2000ms
    Memory
    512MiB
    Difficulty
    10
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