Solution -「51nod 1355」斐波那契的最小公倍数

$\mathcal{Description}$

  Link.

  令 $f$ 为 $\text{Fibonacci}$ 数列，给定 $\{a_n\}$，求：

$$\operatorname{lcm}\{f_{a_1},f_{a_2},\cdots,f_{a_n}\}\bmod(10^9+7)$$

  $n\le5\times10^4$，$a_i\le10^6$。

$\mathcal{Solution}$

  你得知道：

$$\gcd(f_i,f_j)=f_{\gcd(i,j)}\tag1$$

$$\operatorname{lcm}(S)=\prod_{T\subseteq S\land T\not=\varnothing}\gcd(T)^{(-1)^{|T|+1}}\tag2$$

  $(1)$ 老经典的结论了；$(2)$ 本质上是一个 $\text{Min-Max}$ 反演。

  记 $F=\{f_{a_n}\},S=\{a_n\},m=\max(S)$，开始推导：

$$\begin{aligned} \operatorname{lcm}(F)&=\prod_{T\subseteq F\land T\not=\varnothing}\gcd(T)^{(-1)^{|T|+1}}\\ &=\prod_{T\subseteq S\land T\not=\varnothing}f_{\gcd(T)}^{(-1)^{|T|+1}}\\ &=\prod_{d=1}^mf_d^{\sum_{T\subseteq S\land T\not=\varnothing\land\gcd(T)=d}(-1)^{|T|+1}} \end{aligned}$$

  记 $f_d$ 的指数为 $g(d)$，令 $h(d)=\sum_{T\subseteq S\land T\not=\varnothing\land d|\gcd(T)}(-1)^{|T|+1}=1-\sum_{T\subseteq S\land d|\gcd(T)}(-1)^{|T|}$。设有 $c_d$ 个 $a_x$ 是 $d$ 的倍数，那么：

$$\sum_{T\subseteq S\land d|\gcd(T)}(-1)^{|T|}=\sum_{s=0}^{c_d}\binom{c_d}s(-1)^s$$

  二项式展开逆用，后式为 $(1-1)^{c_d}=[c_d=0]$，所以 $h(d)=[c_d\not=0]$。最后利用 $h$ 反演出 $g$：

$$g(d)=\sum_{d|n}h(n)\mu(\frac{n}d)$$

  $\mathcal O(n\ln n)$ 把 $g$ 求出来就好。

$\mathcal{Code}$

/* Clearink */

#include <cstdio>

inline int rint () {
	int x = 0; int f = 1; char s = getchar ();
	for ( ; s < '0' || '9' < s; s = getchar () ) f = s == '-' ? -f : f;
	for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
	return x * f;
}

inline void wint ( int x ) {
	if ( 9 < x ) wint ( x / 10 );
	putchar ( x % 10 ^ '0' );
}

const int MAXN = 5e4, MAXA = 1e6, MOD = 1e9 + 7;
int pn, pr[MAXA + 5], mu[MAXA + 5];
int n, a[MAXN + 5], fib[MAXA + 5], indx[MAXA + 5];
bool buc[MAXA + 5], vis[MAXA + 5];

inline int qkpow ( int a, int b, const int p = MOD ) {
	int ret = 1; b = ( b % ( p - 1 ) + ( p - 1 ) ) % ( p - 1 );
	for ( ; b; a = 1ll * a * a % p, b >>= 1 ) ret = 1ll * ret * ( b & 1 ? a : 1 ) % p;
	return ret;
}

inline void init ( const int n ) {
	mu[1] = 1;
	for ( int i = 2; i <= n; ++ i ) {
		if ( !vis[i] ) mu[pr[++ pn] = i] = -1;
		for ( int j = 1, t; j <= pn && ( t = i * pr[j] ) <= n; ++ j ) {
			vis[t] = true;
			if ( !( i % pr[j] ) ) break;
			mu[t] = -mu[i];
		}
	}
	fib[1] = 1;
	for ( int i = 1; i <= n; ++ i ) {
		if ( i > 1 ) fib[i] = ( fib[i - 1] + fib[i - 2] ) % MOD;
		for ( int j = i; j <= n; j += i ) {
			buc[i] |= buc[j];
		}
	}
	for ( int i = 1; i <= n; ++ i ) {
		for ( int j = 1, t = n / i; j <= t; ++ j ) {
			indx[i] += mu[j] * buc[i * j];
		}
	}
}

int main () {
	n = rint ();
	int mxa = 0;
	for ( int i = 1; i <= n; ++ i ) {
		buc[a[i] = rint ()] = true;
		if ( mxa < a[i] ) mxa = a[i];
	}
	init ( mxa );
	int ans = 1;
	for ( int i = 1; i <= mxa; ++ i ) {
		ans = 1ll * ans * qkpow ( fib[i], indx[i] ) % MOD;
	}
	wint ( ans ), putchar ( '\n' );
	return 0;
}

$\mathcal{Details}$

  对 $\text{Min-Max}$ 要敏感一点呐……