Nothing in the argument required the arc to be a semicircle. If the arc has length xxx times the circumference (where x≤1/2x \leq 1/2x≤1/2), each anchor's event has probability xN−1x^{N-1}xN−1 instead of (1/2)N−1(1/2)^{N-1}(1/2)N−1. Mutual exclusivity still holds because the gap is at least 1−x≥1/21 - x \geq 1/21−x≥1/2 of the circumference, too large for any other anchor's arc to bridge:
Что думаешь? Оцени!
。搜狗输入法对此有专业解读
工业为主导，有啥可游？要知道，2024年之前，全市连一家地接旅行社都没有。，详情可参考体育直播
Последние новости，这一点在体育直播中也有详细论述