矩阵快速幂2 3*n铺方格

 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <queue>
 #include <cstdio>
 #include <algorithm>
 #include <map>
 #include <time.h>
 #include <ext/pb_ds/assoc_container.hpp>
 #include <ext/pb_ds/tree_policy.hpp>
 #define LL long long

 using namespace std;
 using namespace __gnu_pbds;

 //U[n]表示3×N棋盘的摆放数，
 //
 //V[n]表示缺了1×1边角的3×N棋盘的摆放数
 //
 //则
 //
 //U[n] = 2*V[n-1] + U[n-2]
 //
 //V[n] = U[n-1] + V[n-2]
 //
 //其中，
 //
 //U[0] = 1, U[2] = 3;
 //
 //V[0] = 1, V[1] =1;

 //[U(n)                       [ 0 2 1 0           [ U(n-1)
 // V(n)                         1 0 0 1             V(n-1)
 // U(n-1)        =              1 0 0 0      *      U(n-2)
 // V(n-1)]                      0 1 0 0 ]           V(n-2) ]
 //
 //                   [ 0 2 1 0                    [ U(1)
 //               =     1 0 0 1                      V(1)
 //                     1 0 0 0     ^  (n-1)         U(0)
 //                     0 1 0 0 ]                    V(0) ]

 const int MOD = ;

 struct Martix
 {
     LL martix[][];
     int row,col;
     Martix(int _row,int _col)
     {
         memset(martix,,sizeof(martix));
         row = _row;
         col = _col;
     }
     Martix operator *(const Martix &A)const
     {
         Martix C(row, A.col);
         for(int i = ; i < row; i++)
             for(int j = ; j < A.col; j++)
                 for(int k = ; k < col; k++)
                     C.martix[i][j] =  (C.martix[i][j] + martix[i][k] * A.martix[k][j])%MOD;
         return C;
     }
 };

 void solve()
 {
     Martix A(,), F(,), S(,);
     A.martix[][] = A.martix[][] = A.martix[][] = A.martix[][] = A.martix[][] = ;
     F.martix[][] = F.martix[][] = F.martix[][] = F.martix[][] = ;
     A.martix[][] = ;
     S.martix[][] = S.martix[][] = S.martix[][] = ;
     int n;
     scanf("%d",&n);
     if(n&)
     {
         printf("%d\n",);
         return ;
     }
     n--;
     while(n > )
     {
         if(n & )
             F = F*A;
         A = A*A;
         n >>= ;
     }
     S = F*S;
     printf("%lld\n",S.martix[][]);
 }

 int main(void)
 {
     solve();
     return ;
 }