洛谷 P1939 矩阵加速（数列）

题意简述

(a[1]=a[2]=a[3]=1)
(a[x]=a[x−3]+a[x−1](x>3))
求a数列的第n项对1000000007取余的值。

题解思路

矩阵加速
设$$ F=egin{bmatrix} 0&0&11&0&0&1&1end{bmatrix}, G=egin{bmatrix} 111end{bmatrix}$$
则$$ egin{bmatrix} a[n-3]a[n-2]a[n-1]end{bmatrix} = G * F ^ {n-3} (n > 3)$$

代码

#include <cstdio>
typedef long long ll;
const int mod = 1000000007;
int n, T, N;
struct Matrix
{
	int a[4][4];
	Matrix& operator =(const Matrix& x)
	{
		for (register int i = 1; i <= N; ++i)
			for (register int j = 1; j <= N; ++j)
				a[i][j] = x.a[i][j];
		return *this;
	}
};
Matrix a, b, c;
Matrix Mul(const Matrix& x, const Matrix& y)
{
	Matrix s;
	for (register int i = 1; i <= N; ++i)
		for (register int j = 1; j <= N; ++j)
			s.a[i][j] = 0;
	for (register int i = 1; i <= N; ++i)
		for (register int j = 1; j <= N; ++j)
			for (register int k = 1; k <= N; ++k)
				s.a[i][j] = (s.a[i][j] + (ll)x.a[i][k] * y.a[k][j] % mod) % mod;
	return s;
}
Matrix _pow(Matrix x, int y)
{
	Matrix s;
	for (register int i = 1; i <= N; ++i)
		for (register int j = 1; j <= N; ++j)
			s.a[i][j] = (i == j);
	for (; y; y >>= 1, x = Mul(x, x)) if (y & 1) s = Mul(s, x);
	return s;
}
int main()
{
	N = 3;
	scanf("%d", &T);
	while (T--)
	{
		c.a[1][1] = c.a[1][2] = c.a[1][3] = 1;
		a.a[1][3] = a.a[2][1] = a.a[3][2] = a.a[3][3] = 1;
		scanf("%d", &n);
		if (n <= 3) {printf("1
"); continue; }
		b = Mul(c, _pow(a, n - 3));
		printf("%d
", b.a[1][3]);
	}
}