hdu5950

hdu5950

题意

(给出 f_1 ， f_2 ，以及递推式 f_n = 2 * f_{n-2} + f_{n-1} + n^4 ，求 f_n (mod=2147493647))

推导一下。

[egin{Bmatrix} f_n\ f_{n-1}\ f_{n-2}\ (n+1)^4\ (n+1)^3\ (n+1)^2\ (n+1)\ 1 end{Bmatrix} = egin{Bmatrix} 1 & 2 & 0 & 1 & 0 & 0 & 0 & 0\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\ 0 & 0 & 0 & 1 & 4 & 6 & 4 & 1\ 0 & 0 & 0 & 0 & 1 & 3 & 3 & 1\ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 1\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 end{Bmatrix} * egin{Bmatrix} f_{n-1}\ f_{n-2}\ f_{n-3}\ n^4\ n^3\ n^2\ n\ 1 end{Bmatrix}]

矩阵快速幂即可。

code

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iostream>
#define ll long long
using namespace std;

const ll MOD = 2147493647;
const int SIZE = 11;
ll n;
//定义结构体
struct Matrix
{
    ll mat[SIZE][SIZE];
    Matrix()
    {
        memset(mat, 0, sizeof mat);
    }
};
//矩阵乘法 重载 * 操作符
Matrix operator * (Matrix a, Matrix b)
{
    Matrix c;
    memset(c.mat, 0, sizeof(c.mat));
    for(int i = 0; i < SIZE; i++)
    {
        for(int j = 0; j < SIZE; j++)
        {
            for(int k = 0; k < SIZE; k++)
                c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j]) % MOD;
        }
    }
    return c;
}
//矩阵快速幂 重载 ^ 操作符
Matrix operator ^ (Matrix a, ll k)
{
    Matrix t;
    memset(t.mat, 0, sizeof(t.mat));
    for(int i = 0; i < SIZE; i++) // 单位矩阵
        t.mat[i][i] = 1;
    while(k)
    {
        if(k & 1)
            t = t * a;
        a = a * a;
        k >>= 1;
    }
    return t;
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        Matrix mt;
        ll a, b;
        scanf("%lld%lld%lld", &n, &a, &b);
        ll c = 2 * a + b + 81;
        c %= MOD;
        if(n == 1) printf("%lld
", a % MOD);
        else if(n == 2) printf("%lld
", b % MOD);
        else if(n == 3) printf("%lld
", c % MOD);
        else
        {
            mt.mat[0][0] = 1;
            mt.mat[0][1] = 2;
            mt.mat[0][3] = 1;
            mt.mat[1][0] = 1;
            mt.mat[2][1] = 1;
            mt.mat[3][3] = 1;
            mt.mat[3][4] = 4;
            mt.mat[3][5] = 6;
            mt.mat[3][6] = 4;
            mt.mat[3][7] = 1;
            mt.mat[4][4] = 1;
            mt.mat[4][5] = 3;
            mt.mat[4][6] = 3;
            mt.mat[4][7] = 1;
            mt.mat[5][5] = 1;
            mt.mat[5][6] = 2;
            mt.mat[5][7] = 1;
            mt.mat[6][6] = 1;
            mt.mat[6][7] = 1;
            mt.mat[7][7] = 1;
            mt = mt ^ (n - 3);
            ll ans = mt.mat[0][0] * c + mt.mat[0][1] * b + mt.mat[0][2] * a + mt.mat[0][3] * 256
                    + mt.mat[0][4] * 64 + mt.mat[0][5] * 16 + mt.mat[0][6] * 4 + mt.mat[0][7];
            printf("%lld
", ans % MOD);
        }
    }
    return 0;
}