NYOJ 1075 （递推 + 矩阵快速幂）

标签：矩阵   dp   

#include<cstdio>
#include<cstring>

#define mod 10007

struct Matrix {
    int mat[4][4];
    Matrix() {
        memset(mat, 0, sizeof(mat));
        for(int i = 0; i < 4; i++)
            mat[i][i] = 1;
    }
};

Matrix Multi(Matrix a, Matrix b) {
    Matrix res;
    for(int i = 0; i < 4; i++) {
        for(int j = 0; j < 4; j++) {
            res.mat[i][j] = 0;
            for(int k = 0; k < 4; k++) {
                res.mat[i][j] += a.mat[i][k] * b.mat[k][j];
                res.mat[i][j] %= mod;
            }
        }
    }
    return res;
}

Matrix Pow(Matrix a, int x) {
    Matrix res;
    while(x) {
        if(x&1) res = Multi(res, a);
        a = Multi(a, a);
        x >>= 1;
    }
    return res;
}

int main() {
    int T, n;

    Matrix A;  //系数矩阵
    A.mat[0][0] = 2; A.mat[0][1] = 1; A.mat[0][2] = 1; A.mat[0][3] = 0;
    A.mat[1][0] = 1; A.mat[1][1] = 2; A.mat[1][2] = 0; A.mat[1][3] = 1;
    A.mat[2][0] = 1; A.mat[2][1] = 0; A.mat[2][2] = 2; A.mat[2][3] = 1;
    A.mat[3][0] = 0; A.mat[3][1] = 1; A.mat[3][2] = 1; A.mat[3][3] = 2;

    scanf("%d",&T);
    while(T--) {
        scanf("%d", &n);
        Matrix ans = Pow(A, n);
        printf("%d\n", ans.mat[0][0]);
    }

    return 0;
}

NYOJ 1075 （递推 + 矩阵快速幂）

标签：矩阵   dp   

原文地址：http://blog.csdn.net/lyhvoyage/article/details/39525785