星月轮转
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算法导论书上有个练习，让证明n个元素的第2个最小的元素可以用n + ceil(lg n) – 2次比较找出来，给的提示只是：把最小的也找出来。
我想了好久，终于出来了。思路是：n-1次比较找出最小，然后再用ceil(lg n)-1次比较就找到第2个最小。具体见这一页。
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