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如果直接计算A+A^3+A^5+...+A^(N-(!(N&0x1))) 这样会TLE么。。?

Posted by O___O at 2008-10-20 13:02:29 and last updated at 2008-10-20 13:02:35
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)
> 
> :-)

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