Recursive sequence HDU - 5950

Posted 2021-08-04 Jozky86

Recursive sequence HDU - 5950

题意：

给你一个式子：f[n]=2f[n-2]+f[n-1]+n⁴
给你f[1]和f[2]，给你一个n，求f[n]
f[1],f[2],n<=2³¹

题解：

很明显，矩阵快速幂，但是太久没做这种题，我都忘了怎么推导矩阵的了

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
 
using namespace std;
 
typedef long long ll;
int t, n, a, b;
ll mod = 2147493647;
struct Matrix
{
	ll mat[15][15];
	Matrix()
	{
		memset(mat, 0, sizeof(mat));
	}
	friend Matrix operator * (Matrix A, Matrix B)
	{
		Matrix ans;
		for(int i = 1; i <= 7; ++ i)
		{
			for(int j = 1; j <= 7; ++ j)
			{
				for(int k = 1; k <= 7; ++ k)
				{
					ans.mat[i][j] += (A.mat[i][k] * B.mat[k][j]) % mod;
					ans.mat[i][j] %= mod;
				}
			}
		}
		return ans;
	}
};
 
 
Matrix quick_matrix(Matrix A, int b)
{
	Matrix ans;
	for(int i = 1; i <= 7; ++ i)
	{
		ans.mat[i][i] = 1;
	}
	while(b)
	{
		if(b & 1)
		{
			ans = ans * A;
		}
		A = A * A;
		b >>= 1;
	}
	return ans;
}
 
int main()
{
	cin >> t;
	while(t--)
	{
		cin >> n >> a >> b;
		if(n == 1 || n == 2)
		{
			if(n == 1)
				cout << a << endl;
			else 
				cout << b << endl;
			continue;
		}
		Matrix A, B;
		A.mat[1][1] = b;
		A.mat[2][1] = a;
		A.mat[3][1] = 3 * 3 * 3 * 3;
		A.mat[4][1] = 3 * 3 * 3;
		A.mat[5][1] = 3 * 3;
		A.mat[6][1] = 3;
		A.mat[7][1] = 1;
 
		B.mat[1][1] = 1;
		B.mat[1][2] = 2;
		B.mat[1][3] = 1;
 
		B.mat[2][1] = 1;
 
		B.mat[3][3] = 1;
		B.mat[3][4] = 4;
		B.mat[3][5] = 6;
		B.mat[3][6] = 4;
		B.mat[3][7] = 1;
 
		B.mat[4][4] = 1;
		B.mat[4][5] = 3;
		B.mat[4][6] = 3;
		B.mat[4][7] = 1;
 
		B.mat[5][5] = 1;
		B.mat[5][6] = 2;
		B.mat[5][7] = 1;
 
		B.mat[6][6] = 1;
		B.mat[6][7] = 1;
 
		B.mat[7][7] = 1;
		B = quick_matrix(B, n - 2);
		A = B * A;
		cout << A.mat[1][1] << endl;
	}
	return 0;
}