HDU 3306 矩阵快速幂

来源：互联网发布：linux 重启oracle 编辑：程序博客网时间：2024/05/21 07:36

题解: S(n) = S(n - 1) + A(n)^2 = S(n-1) + x^2*A(n-1)^2 + 2*x*y*A(n-1)A(n-2) +y^2*A(n-2)^2构造矩阵

code :

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int p = 10007;int n, x, y;struct Matrix{    int mat[4][4];    void clr(){memset(mat, 0, sizeof mat);}    void E(){        clr();        for(int i = 0; i < 4; ++i) mat[i][i] = 1;    }    Matrix operator * (const Matrix &rhs) const{        Matrix res;        res.clr();        for(int i = 0; i < 4; ++i)            for(int j = 0; j < 4; ++j)                for(int k = 0; k < 4; ++k)                       res.mat[i][j] = (res.mat[i][j] + mat[i][k] * rhs.mat[k][j]) % p;        return res;    }    Matrix operator ^ (int b){        Matrix res, tmp;        res.E();        memcpy(tmp.mat, mat, sizeof mat);        while(b){            if(b & 1) res = res * tmp;            tmp = tmp * tmp;            b >>= 1;        }        return res;    }};int main(){    //freopen("in.txt", "r", stdin);    while(cin >> n >> x >> y){        x %= p;        y %= p;        Matrix base;        base.clr();        base.mat[0][0] = 1;         /**row 1*/        base.mat[1][0] = base.mat[1][1] = x * x % p;        base.mat[1][2] = 1;        base.mat[1][3] = x;        /**row 2*/        base.mat[2][0] = base.mat[2][1] = y * y % p;        /**row 3*/        base.mat[3][0] = base.mat[3][1] = 2 * x * y % p;        base.mat[3][3] = y;        /**row 4*/        Matrix data;        data.clr();        data.mat[0][0] = 2;        data.mat[0][1] = data.mat[0][2] = data.mat[0][3] = 1;        Matrix res = data * (base ^ (n - 1));        cout << res.mat[0][0] << endl;    }    return 0;}

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