BZOJ2173: 整数的lqp拆分

传送门

Sol

构造 (fib) 数列的母函数 (F(x))
那么答案就是

[[x^n]sum_{i=1}^{infty}F^i(x)=[x^n]frac{F(x)}{1-F(x)} ]

而

[F(x)=xF(x)+x^2F(x)+x,F(x)=frac{x}{1-x-x^2} ]

所以答案就是

[[x^n]frac{x}{1-2x-x^2} ]

递推式出来了直接 (Theta(n)) 递推
或者求出通项

[frac{(sqrt{2}+1)^n-(1-sqrt{2})^n}{2sqrt{2}} ]

然后 (Theta(logn)) 出解

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int mod(1e9 + 7);

inline int Inc(int x, int y) {
	return x + y >= mod ? x + y - mod : x + y;
}

inline int Dec(int x, int y) {
	return x - y < 0 ? x - y + mod : x - y;
}

inline int Pow(ll x, int y) {
	register ll ret = 1;
	for (x = (x % mod + mod) % mod; y; y >>= 1, x = x * x % mod)
		if (y & 1) ret = ret * x % mod;
	return ret;
}

struct Pair {
	int a, b;

	inline Pair() {
		a = b = 0;
	}

	inline Pair(int _a, int _b) {
		a = _a, b = _b;
	}

	inline Pair operator +(Pair x) const {
		return Pair(Inc(a, x.a), Inc(b, x.a));
	}

	inline Pair operator -(Pair x) const {
		return Pair(Dec(a, x.a), Dec(b, x.b));
	}

	inline Pair operator *(Pair x) const {
		return Pair(((ll)a * x.a + (ll)2 * b * x.b) % mod, ((ll)a * x.b + (ll)x.a * b) % mod);
	}

	inline Pair operator /(Pair x) const {
		register Pair t1(a, b), t2(x.a, mod - x.b), t3(Pow((ll)x.a * x.a - (ll)2 * x.b * x.b, mod - 2), 0);
		return t1 * t2 * t3;
	}
} v1, v2, v3;

int n, x;

int main() {
    scanf("%d", &n), v1 = Pair(1, 1), v2 = Pair(1, mod - 1);
	for (v3 = Pair(1, 0), x = n; x; x >>= 1, v1 = v1 * v1) if (x & 1) v3 = v3 * v1;
	for (v1 = Pair(1, 0), x = n; x; x >>= 1, v2 = v2 * v2) if (x & 1) v1 = v1 * v2;
	v1 = (v3 - v1) / Pair(0, 2), printf("%d
", v1.a);
    return 0;
}