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此法好牛!In Reply To:Re:是用公式吗? Posted by:LIANGLIANG at 2004-07-17 22:09:09 > 也许算吧，基于你的求对数的思路，然后 > lg1+lg2+...lgn = [lg(1/n)+lg(2/n)+...lg(n/n)] + nlg(n) > 前部可以近似求积分。。。 等于n/ln10 当然会不精确，调整误差搞了很久的说。。。 Followed by:
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