poj 3734 Blocks

从左边开始染色，到第$i$个方块为止，红绿都是偶数的方案数为$a_i$，红绿恰有一个是偶数的方案为$b_i$，红绿都是奇数的方案为$c_i$，从而有如下状态转移方程：

$a_{i+1} = 2 imes a_i + b_i$

$b_{i+1} = 2 imes  a_i + 2 imes b_i + 2 imes c_i$

$c_{i+1} = b_i + 2 imes c_i$

从而可以用矩阵表示：

egin{equation} left( egin{array}{c} a_i \ b_i \ c_i end{array} ight) = left( egin{array}{ccc} 2 & 1 & 0 \ 2 & 2 & 2 \ 0 & 1 & 2 end{array} ight)^i left( egin{array}{c} 1\ 0\ 0\ end{array} ight)  end{equation}

 最终使用矩阵的快速幂即可求解。

#define     MOD 10007
#define     M 3
#include  <cstdio>
#include  <cstdlib>
#include  <iostream>
#include  <cstring>
int N;
using namespace std;
class Matrix
{
    public:
        int a[M][M];
        Matrix(bool init = 0)
        {
            memset(a, 0, sizeof(a));
            if( init == 1 )
            {
                for( int i = 0 ; i < M ; i++ )
                {
                    a[i][i] = 1;
                }
            }
        }
        int * operator [](int i)
        {
            return a[i];
        }
        const int * operator [](int i)const
        {
            return a[i];
        }
};
Matrix A;
void add(int &a, int b)
{
    a += b;
    while( a >= MOD )
    {
        a -= MOD;
    }
}
Matrix operator * (const Matrix &a, const Matrix &b)
{
    Matrix c;
    for( int i = 0 ; i < M ; i++ )
    {
        for( int j = 0 ; j < M ; j++ )
        {
            for( int k = 0 ; k < M ; k++ )
            {
                add(c[i][k], 1ll * a[i][j] * b[j][k] %MOD);
            }
        }
    }
    return c;
}
Matrix quick_pow(Matrix m, int n)
{
    Matrix ans(1);
    while( n )
    {
        if( n & 1 )
        {
            ans = ans * m;
        }
        n = n>>1;
        m = m * m;
    }
    return ans;
}
void solve()
{
    Matrix ans;
    ans = quick_pow(A, N);
    printf ( "%d
", ans[0][0]);
}
int main(int argc, char *argv[])
{
    int T;
    scanf ( "%d", &T );
    A[0][0] = 2, A[0][1] = 1;
    A[1][0] = 2, A[1][1] = 2, A[1][2] = 2;
    A[2][1] = 1, A[2][2] = 2;
    while( T-- )
    {
        scanf ( "%d", &N );
        solve();
    }
}