模板——特殊数列的前缀最小公倍数

形如：$\{\begin{array}{l}f(0)=0\\ f(1)=1\\ f(n)=af(n-2)+bf(n-1)(a,b)=1\end{array}$
这样的数列的前缀最小公倍数可以在$O(nlogn)$的复杂度求出。

#include<iostream>
using namespace std;

typedef long long ll;
const ll mod = 1e9 + 7;
const int maxn = 1e6 + 10;

//初始化数组
ll f[maxn] = { 0,1 };

//求逆元
ll quick_pow(ll a, ll b) {
	ll res = 1;
	while (b) {
		if (b & 1) {
			res = (res * a) % mod;
		}
		a = (a * a) % mod;
		b >>= 1;
	}
	return res;
}

int main() {

	ll a, b, n;
	cin >> a >> b >> n;

	for (ll i = 2; i <= n; i++) {
		f[i] = (a * f[i - 2] + b * f[i - 1]) % mod;
	}

	for (ll i = 2; i <= n; i++) {
		ll inv = quick_pow(f[i], mod - 2);
		for (ll j = 2 * i; j <= n; j += i) {
			f[j] = (f[j] * inv) % mod;
		}
	}

	for (ll i = 2; i <= n; i++) {
		f[i] = (f[i - 1] * f[i]) % mod;
	}

	for (ll i = 1; i <= n; i++) {
		cout << f[i] << " ";
	}
	cout << endl;
	return 0;
}