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我的也是TA的算法，数据比较弱吧。In Reply To:O(n*a*t)----->O(a*t) Posted by:library2 at 2009-10-25 04:55:07 > 把O(n*t^20）方法中f[j]与f[j+1] 的方程列出来后做差，O（n*t）方程便出炉。 > 注意：由于过程中mod10^6 所以算出的数可能为负数，用10^6加上即可 Followed by: Post your reply here: |
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