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如果直接计算A+A^3+A^5+...+A^(N-(!(N&0x1))) 这样会TLE么。。?In Reply To:针对1002的进一步说明 Posted by:lyg at 2008-10-20 11:57:15 > S是初始向量,R是答案向量,A转换矩阵,那么有 > > R=S*A+S*A^3+S*A^5+...+S*A^(2k+1) > > 因为矩阵运算满足结合律和分配律 > 所以有 > > R=S*A*(I+A^2+A^4+A^6+...+A^(2k)) > > 两边右乘(A^2-I) > > 得到 > > R*(A^2-I) = S*A*(I+A^2+A^4+A^6...+A^(2k))*(A^2-I) > > 根据结合律 和分配律 > > 得到 > > R*(A^2-I) = S*A*( A^(2k+2) -I) > > 所以 > > R= S*A*(A^(2k+2)-I)* (A^2-I)^(-1) > > :-) Followed by: Post your reply here: |
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