hdu 3306 Another kind of Fibonacci 矩阵快速幂

参考了某大佬的

我们可以根据(s[n-2], a[n-1]^2, a[n-1]*a[n-2], a[n-2]^2) * A = (s[n-1], a[n]^2, a[n]*a[n-1], a[n-1]^2)

能够求出关系矩阵

|1     0      0     0 |
A =   |1   x^2    x     1 |
        |0  2*x*y  y     0 |
        |0   y^2    0     0 |

这样就A了！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const ll Mod = 10007;
const int N = 5;
int msize;

struct Mat
{
    ll mat[N][N];
};

Mat operator *(Mat a, Mat b)
{
    Mat c;
    memset(c.mat, 0, sizeof(c.mat));
    for(int k = 0; k < msize; ++k)
        for(int i = 0; i < msize; ++i)
            if(a.mat[i][k])
                for(int j = 0; j < msize; ++j)
                    if(b.mat[k][j])
                        c.mat[i][j] = (a.mat[i][k] * b.mat[k][j] + c.mat[i][j])%Mod;
    return c;
}

Mat operator ^(Mat a, ll k)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    for(int i = 0; i < msize; ++i)
        c.mat[i][i]=1;
    for(; k; k >>= 1)
    {
        if(k&1) c = c*a;
        a = a*a;
    }
    return c;
}

int main()
{
    ll n,x,y;
    msize = 4;
    while(~scanf("%I64d%I64d%I64d",&n,&x,&y))
    {
        Mat A;
        A.mat[0][0] = 1, A.mat[0][1] = 1, A.mat[0][2] = 0, A.mat[0][3] = 0;
        A.mat[1][0] = 0, A.mat[1][1] = x*x%Mod, A.mat[1][2] = 2*x*y%Mod, A.mat[1][3] = y*y%Mod;
        A.mat[2][0] = 0, A.mat[2][1] = x, A.mat[2][2] = y, A.mat[2][3] = 0;
        A.mat[3][0] = 0, A.mat[3][1] = 1, A.mat[3][2] = 0, A.mat[3][3] = 0;
        A = A^n;
        printf("%I64d\n", (A.mat[0][0] + A.mat[0][1] + A.mat[0][2] + A.mat[0][3])%Mod);
    }
    return 0;
}