HDU 4332 state compression matrix acceleration DP thank sue God

Now there are countless 1*1*2 bricks to build a chimney with a length of N. A brick can be erected and put in a horizontal manner. The problem is that when the four bricks are flat and F, there are two situations in the question.

Practice: use 0 to indicate null, and 1 to indicate overwriting. After simulating the placement of bricks in each section, we can find that the number of 1 is always an even number, which can be used to simplify the process. Due to the special nature of placement, the section can be abstracted into an eight-bit binary number that can affect each other at the beginning and end. In this way, all situations are considered for judgment.

# Include <cstdio> # include <cstring> # define mod 1000000007 # define LMT 130 # define ll long //~ The priority is lower than the displacement. // The memory will burst when defined in the structure .. // less defined in main // (x> I) & 1 is used to determine. // The using namespace STD for matrix multiplication; Class matrix {public: ll mat [LMT] [LMT]; int n, m; void Init (void) {memset (MAT, 0, sizeof mat);} friend Matrix Operator * (Matrix &, matrix &) ;}; matrix Matt, Trans; int State [260], Lim = 1 <8; Matrix Operator * (Matrix & A, Matrix & B) {matrix TEM; TEM. N =. n; TEM. M = B. m; TEM. init (); For (int K = 0; k <. m; k ++) for (INT I = 0; I <. n; I ++) if (. mat [I] [k]) for (Int J = 0; j <B. m; j ++) TEM. mat [I] [J] = (TEM. mat [I] [J] +. mat [I] [k] * B. mat [k] [J]) % MOD; return TEM;} bool Yes (INT code) {int S = 0; while (CODE) {S + = Code & 1; code >>=1;} return S % 2 = 0;} ll can (int A, int B) {int St, I, j; bool end = false; if (A = 0) return B = 255? 1ll: 0; for (I = 0; I <8; I ++) {J = (I + 7) % 8; if (A & (1 <I) & (A & (1 <j) = 0) {ST = I; break ;}} for (I = sT; I! = ST | End = false; I = (I + 1) % 8) {end = true; j = (I + 1) % 8; if (0 = (A & (1 <I) {If (0 = (B & (1 <I) return 0 ;} else {If (A & (1 <j) & (B & (1 <I) & (B & (1 <j ))) I ++; else if (B & (1 <I) = 0) continue; else return 0 ;}} return 1ll ;} void Init (void) {Lim = 0; memset (State,-1, sizeof (State); For (INT I = 0; I <1 <8; I ++) if (Yes (I) State [I] = lim ++; trans. init (); Trans. M = trans. N = lim; For (INT I = 0; I <1 <8; I ++) for (Int J = I + 1; j <1 <8; j ++) If (State [I]! =-1 & State [J]! =-1) trans. mat [State [J] [State [I] = trans. mat [State [I] [State [J] = Can (I, j); Trans. mat [lim-1] [lim-1] = 2;} matrix POW (Matrix T, int N) {matrix TEM; TEM. M = TEM. N = lim; TEM. init (); For (INT I = 0; I <Lim; I ++) TEM. mat [I] [I] = 1; while (n) {If (N & 1) TEM = TEM * t; t = T * t; n >>= 1 ;} return TEM;} int main (void) {int T, I, n; Init (); scanf ("% d", & T); for (I = 1; I <= T; I ++) {scanf ("% d", & N); Matt = POW (trans, n); printf ("case % d: % i64d \ n ", I, Matt. mat [lim-1] [lim-1]);} return 0 ;}