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大开方便之门,这题解法

Posted by Base at 2008-11-20 13:55:31 on Problem 2651
假设所求为f(n) , 即表示还有n题的情况下,可以得到多少奖金

显然有 f(0) = 2^n
对于f(k) , 有两种选择,
	(1) 放弃,拿走2^{n-k}的奖金,
	(2) 继续,则有p的概率得到f(k-1)的奖金,期望为p*f(k-1)

由于p是一个[t,1)均匀分布的随机量,所以它的概率密度为1/(1-t)

所以 f(k) = 1/(1-t) * int_t^1 max(2^{n-k},p*f(k-1))dp

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