> int A[logk][n][n], B[logk][n][n];
> 
> // A[x][i][j] = A^(2^x);
> // B[x][i][j] = A^1 + A^2 + ... + A^(2^x);
> 
> 一下是递推公式：
> 
> A[x+1] = A[x] * A[x];
> B[x+1] = A[x] * B[x] + B[x];
> 
> S = A^1 + A^2 + .... + A^k
>    
> 将 k 分解为 2^bi + 2^bi-1 + …… + 2^bj；
> 则 
> S = (A^1 + A^2 + ... + A^(2^bi) ) + A^(2^bi) * (A^1 + A^2 + ... + A^(2^bi-1)) + A^(2^bi + 2^bi-1) * (A^1 + A^2 + ... + A^(2^bi-2) )  + ... + A^(2^bi + 2^bi-1 + ... + 2^bj+1) * (A^1 + A^2 + ... + A^(2^bj) ) 
> 
> = B[bi] + A[bi] * B[bi-1] + A[bi] * A[bi-1] * B[bi-2] + ... + A[bi] * A[bi-1] * ... * A[bj+1] * B[bj]  ;
> 
> Sample:
> 
> S = A + A^2 + ... + A^14
> 
> 14 = 2^1 + 2^2 + 2^3
> 
> S = (A^1 + A^2) + A^2 * (A^1 + A^2 + A^3 + A^4) + A^2 * A^4 * (A^1 + A^2 + ... + A^(2^3))
> 
> = B[1] + A[1] * B[2] + A[1] * A[2] * B[3]
> 
> 复杂度为 O(n^3 * logk)