bzoj5015 [Snoi2017]礼物 矩阵快速幂+二项式展开

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=5015

题解

设 \(f_i\) 表示第 \(i\) 个朋友的礼物，\(s_i\) 表示从 \(1\) 到 \(i\) 的 \(f_i\) 的和。

考虑用矩阵维护转移，但是这个 \(i^k\) 不太方便转移。

发现 \(k \leq 10\)，可以考虑使用二项式展开。

所以可以用矩阵维护一下 \(i^j(0 \leq j \leq k)\) 转移就可以了。

时间复杂度 \(O(k^3\log n)\)。

#include

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair pii;

template inline void read(I &x) {
	int f = 0, c;
	while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
	x = c & 15;
	while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
	f ? x = -x : 0;
}

const int N = 13 + 3;
const int P = 1e9 + 7;

ll T;
int k, n;

inline int smod(int x) { return x >= P ? x - P : x; }
inline void sadd(int &x, const int &y) { x += y; x >= P ? x -= P : x; }
inline int fpow(int x, int y) {
	int ans = 1;
	for (; y; y >>= 1, x = (ll)x * x % P) if (y & 1) ans = (ll)ans * x % P;
	return ans;
}

struct Matrix {
	int a[N][N];
	
	inline Matrix() { memset(a, 0, sizeof(a)); }
	inline Matrix(const int &x) {
		memset(a, 0, sizeof(a));
		for (int i = 1; i <= n; ++i) a[i][i] = x;
	}
	
	inline Matrix operator * (const Matrix &b) {
		Matrix c;
		for (int k = 1; k <= n; ++k)
			for (int i = 1; i <= n; ++i)
				for (int j = 1; j <= n; ++j) sadd(c.a[i][j], (ll)a[i][k] * b.a[k][j] % P);
		return c;
	}
} A, B;

inline Matrix fpow(Matrix x, ll y) {
	Matrix ans(1);
	for (; y; y >>= 1, x = x * x) if (y & 1) ans = ans * x;
	return ans;
}

inline void work() {
	B.a[3][1] = 1;
	A.a[1][1] = 0, A.a[1][2] = 1;
	A.a[2][1] = 0, A.a[2][2] = 2;
	A.a[3][3] = 1;
	for (int i = 4; i <= k + 3; ++i)
		for (int j = 3; j <= i; ++j) A.a[i][j] = smod(A.a[i - 1][j - 1] + A.a[i - 1][j]);
	for (int i = 1; i <= n; ++i) sadd(A.a[1][i], A.a[n][i]), sadd(A.a[2][i], A.a[n][i]);
	printf("%d\n", B.a[1][1]);
}

inline void init() {
	read(T), read(k);
	n = k + 3;
}

int main() {
#ifdef hzhkk
	freopen("hkk.in", "r", stdin);
#endif
	init();
	work();
	fclose(stdin), fclose(stdout);
	return 0;
}