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其实很水的，直接排个序就可以O(n)解决了。原因：a1<=a2<=a3<=a4<=a5<=.....an； 对于a1，从序列an开始递减查找，找到第一个aj+a1<=S的话，那么ans+=j-1,跳出。 对于a2,则只要从序列aj开始递减查找就可以了。 ...... 当j>i的情况下，按照上面的方式查找，每次ans+=j-i即可。 Followed by:
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