HDU 1588 Gauss Fibonacci

Hdu_1588

Assume that F is the same matrix containing the Fibonacci items, and the final result is the matrix S (n-1) = f ^ (B) + f ^ (K + B) +... + F ^ (K * (n-1) + B) elements in the upper left corner.

For the matrix S and F, the recursive formula S (n) = S (n-1) + f ^ (K * n + B), f ^ (K * (n + 1) can be obtained) + B) = a ^ (k) * f ^ (K * n + B). matrix A is the matrix shown in. Therefore, we can construct the two Recursive formulas into a matrix of Recursive Relations, and then use the binary matrix method to quickly obtain S (n-1.

 

# Include <stdio. h> # Include < String . H> # Define Maxd 8 Int  K, B, n, m; Struct  Matrix {  Int  A [maxd] [maxd]; matrix () {memset (,  0 , Sizeof  (A) ;}}; matrix multiply (Matrix & X, Matrix & Y ){  Int  I, J, K; Matrix Z;  For (K = 0 ; K < 8 ; K ++ ) For (I = 0 ; I < 8 ; I ++ )  If  (X. A [I] [k]) {  For (J = 0 ; J < 8 ; J ++ )  If  (Y. A [k] [J]) Z. A [I] [J] = (Z. A [I] [J] + ( Long  Long ) X. A [I] [k] * Y. A [k] [J]) % M ;}  Return  Z;} matrix powmod (Matrix & Unit, Matrix & mat, Int  N ){  While  (N ){  If (N & 1  ) Unit = Multiply (MAT, Unit); n >>= 1 ; MAT = Multiply (MAT, mat );}  Return  Unit;} matrix getf (  Int  N) {matrix unit, mat; Unit. [  0 ] [ 0 ] = 1  ; Mat. [  0 ] [ 0 ] = Mat. [ 0 ] [ 1 ] = Mat. [1 ] [ 0 ] = 1  ; Powmod (unit, mat, n - 1  );  Return  Unit;} matrix geta (  Int  N) {matrix unit, mat; Unit. [  0 ] [ 0 ] = Unit. [ 1 ] [ 1 ] =1  ; Mat. [  0 ] [ 0 ] = Mat. [ 0 ] [ 1 ] = Mat. [ 1 ] [ 0 ] = 1  ; Unit = Powmod (unit, mat, N );  Return  Unit ;}  Void  Solve (){ Int  I, J; matrix unit, mat, AK, AB, AKB;  If  (B) {AB = Getf (B );  For (I = 0 ; I < 4 ; I ++ )  For (J = 0 ; J < 4 ; J ++ ) Unit. A [I] [J] =AB. A [I] [J];}  If (K + B) {AKB = Getf (B + K );  For (I = 0 ; I < 4 ; I ++ )  For (J = 0 ; J < 4 ; J ++ ) Unit. A [I + 4 ] [J] =AKB. A [I] [J];}  For (I = 0 ; I < 4 ; I ++ ) Mat. A [I] [I] = Mat. A [I] [I + 4 ] = 1  ; AK = Geta (k );  For (I = 0 ; I < 4 ; I ++ )  For (J = 0 ; J < 4 ; J ++ ) Mat. A [I + 4 ] [J + 4 ] = AK. A [I] [J]; Unit = Powmod (unit, mat, N- 1  ); Printf (  "  % D \ n  " , Unit. [ 0 ] [ 0  ]);} Int  Main (){  While (Scanf ( "  % D  " , & K, & B, & N, & M) = 4  ) {Solve ();}  Return   0  ;}