形如: { f ( 0 ) = 0 f ( 1 ) = 1 f ( n ) = a f ( n − 2 ) + b f ( n − 1 ) ( a , b ) = 1 \small \begin{cases} \ f(0)=0\\ \ f(1)=1\\ \ f(n)=af(n-2)+bf(n-1)~~~~~~~~~~~~(a,b)=1\\ \end{cases} ⎩⎨⎧ f(0)=0 f(1)=1 f(n)=af(n−2)+bf(n−1) (a,b)=1
这样的数列的前缀最小公倍数可以在 O ( n l o g n ) \small O(nlogn) 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;
}