POJ 3734 Blocks

Matrix Fast power.

Set AI to indicate that the red and green are even-numbered schemes after finishing the block

Set BI to show the number of programs with even numbers of red after I have finished painting

Set CI to indicate that the green is an even number of schemes after I have finished painting

Setting di means that after I have been finished, it is not even number of scenarios

Ai+1=ai+ai+bi+ci

Bi+1=ai+2*bi+di

Ci+1=ai+2*ci+di

Di+1=bi+ci+2*di

Then it is easy to construct the matrix.

#include <cstdio>#include<cstring>#include<cmath>#include<vector>#include<algorithm>using namespacestd;intN;Long LongMod=10007;structmatrix{Long Longa[5][5]; intR, C; Matrixoperator*(Matrix b);}; Matrix X, Y, Z; Matrix Matrix::operator*(Matrix B) {matrix C; memset (C.A,0,sizeof(C.A)); intI, j, K;  for(i =1; I <= R; i++)         for(j =1; J <= B.C; J + +)             for(k =1; K <= C; k++) C.a[i][j]= (C.a[i][j] + (a[i][k] * b.a[k][j])% MOD)%MOD; C.R= R; C.C =B.C; returnC;}voidinit () {n=n-1; memset (X.A,0,sizeofx.a); memset (Y.A,0,sizeofy.a); memset (Z.A,0,sizeofz.a); Y.R=4; Y.C =4;  for(inti =1; I <=4; i++) Y.a[i][i] =1; X.R=4; X.C =4; x.a[1][1]=2; x.a[1][2]=1; x.a[1][3]=1; x.a[1][4]=0; x.a[2][1]=1; x.a[2][2]=2; x.a[2][3]=0; x.a[2][4]=1; x.a[3][1]=1; x.a[3][2]=0; x.a[3][3]=2; x.a[3][4]=1; x.a[4][1]=0; x.a[4][2]=1; x.a[4][3]=1; x.a[4][4]=2; Z.R=1; Z.C =4; z.a[1][1]=2; z.a[1][2]=1; z.a[1][3]=1; z.a[1][4]=0;}voidRead () {scanf ("%d", &n);}voidWork () { while(n) {if(n%2==1) Y = y*y; N= N >>1; X= x*X; } Z= z*Y; printf ("%lld\n", z.a[1][1]);}intMain () {intCase=1; intT; scanf ("%d", &T);  while(t--) {read ();        Init ();    Work (); }    return 0;}

POJ 3734 Blocks