记事件 AAA:f(a)=1f(a) = 1f(a)=1;事件 BBB:a1≥ka_1 \geq ka1≥k。所求即为 P(A∣B)P(A \mid B)P(A∣B)。显然有 P(AB)=P(A∣B)P(B)=P(B∣A)P(A)P(AB) = P(A \mid B)P(B) = P(B \mid A)P(A)P(AB)=P(A∣B)P(B)=P(B∣A)P(A)。
P(B)P(B)P(B) 容易用组合数求。P(B∣A)P(B \mid A)P(B∣A) 即最大值大于等于 kkk,同样容斥即可。P(A)=1nP(A) = \frac 1nP(A)=n1,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)O(n+m)。
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