【HDU 5950】Recursive sequence

1.题目链接。题意：a[n]=a[n-1]+2*a[n-2]+pow(n,4).求an。

2.十分裸的矩阵快速幂，构造个7*7的矩阵就完事了，在对n^4狗杂矩阵时其实很简单，就按照他们的二项展开式的系数填就行了。

#include<bits/stdc++.h>
using  namespace std;
const int N = 1e6 + 100;
#define ll long long
const ll mod = 2147493647;
const int LEN = 7;
struct Sarray {
	int len;
	ll data[LEN][LEN];
	
	
	Sarray(int len, int flag) :len(len)
	{
		for (int i = 0; i < len; i++) {
			for (int j = 0; j < len; j++)data[i][j] = 0;
			data[i][i] = flag;
		}
	}

	Sarray operator *(const Sarray& a) {
		Sarray tem(a.len, 0);
		for (int i = 0; i < len; i++) {
			for (int j = 0; j < len; j++) {
				for (int k = 0; k < len; k++) {
					tem.data[i][j] = (tem.data[i][j] + data[i][k] * a.data[k][j]) % mod;
				}
			}
		}
		return tem;
	}

	Sarray operator +(const Sarray& a) {
		Sarray tem(a.len, 0);
		for (int i = 0; i < len; i++) {
			for (int j = 0; j < len; j++) {
				tem.data[i][j] = (data[i][j] + a.data[i][j]) % mod;
			}
		}
		return tem;
	}

};

Sarray qpow(Sarray a, int b) {//会更改a，不能按引用传递
	Sarray tem(a.len, 1);
	while (b) {
		if (b & 1)tem = a * tem;
		a = a * a;
		b >>= 1;
	}
	return tem;
}

ll tr[7][7] =
{
	1,2,1,4,6,4,1,
	1,0,0,0,0,0,0,
	0,0,1,4,6,4,1,
	0,0,0,1,3,3,1,
	0,0,0,0,1,2,1,
	0,0,0,0,0,1,1,
	0,0,0,0,0,0,1
};
ll init[7][7]
{

	0,0,0,0,0,0,0,
	0,0,0,0,0,0,0,
	16,0,0,0,0,0,0,
	8,0,0,0,0,0,0,
	4,0,0,0,0,0,0,
	2,0,0,0,0,0,0,
	1,0,0,0,0,0,0
};
int main()
{
	int T;
	scanf("%d", &T);
	Sarray trans(7, 0);
	memcpy(trans.data, tr, sizeof(tr));
	Sarray ini(7, 0);
	memcpy(ini.data, init, sizeof(init));
	while (T--)
	{
		int n, a, b;
		scanf("%d%d%d", &n, &a, &b);
		if (n == 1)
		{
			printf("%d\n", a);
			continue;
		}
		if (n == 2)
		{
			printf("%d\n", b);
			continue;
		}


		ini.data[0][0] = b;
		ini.data[1][0] = a;



		int ans = (qpow(trans, n - 2) * ini).data[0][0];
		ans %= mod;
		printf("%d\n", ans);
	}
	return 0;
}