题目描述：输入一个大小为\(n\)的集合\(S\)，求\(\text{lcm}_{k\in S}f_k\)，其中\(f_k\)是第$$个Fibonacci数。
数据范围：\(n\le 5\times 10^4,u\le 10^6\)
数论经典题？
首先你要想到min-max容斥。
\[\text{lcm}(f_S)=\prod_{\varnothing\ne T\subseteq S}\gcd(f_T)^{(-1)^{|T|-1}}
\]
然后你知道\(\gcd(f_a,f_b)=f_\gcd(a,b)\)，所以。
\[\text{lcm}(f_S)=\prod_{\varnothing\ne T\subseteq S}f_{\gcd(T)}^{(-1)^{|T|-1}}
\]
不知道为什么你开始反演，设\(f_n=\prod\limits_{d|n}g_d\)，则\(g_n=\prod\limits_{d|n}f_{d}^{\mu(\frac{n}{d})}\)。
\[\begin{aligned}
\text{lcm}(f_S)&=\prod_{\varnothing\ne T\subseteq S}(\prod_{d|\gcd(T)}g_d)^{(-1)^{|T|-1}} \\
&=\prod_{d}g_d^{\sum\limits_{\varnothing\ne T\subseteq S,d|T}(-1)^{|T|-1}}
\end{aligned}
\]
我们看看指数是啥。设\(S_d=\{n|n\in S\and d|n\}\)。
\[\sum_{\varnothing\ne T\subseteq S_d}(-1)^{|T|-1}=[|S_d|>0]
\]
所以
\[\text{lcm}(f_S)=\prod_{\exist a\in S,d|a}g_d
\]
直接做，时间复杂度\(O(k\log k)\)
code
```cpp
#include
#define Rint register int
using namespace std;
typedef long long LL;
const int mod = 1e9 + 7, N = 1000003;
int n, a[N], mx, f[N], g[N], ans = 1;
bool vis[N];
inline int add(int a, int b){return (a + b >= mod) ? (a + b - mod) : (a + b);}
inline int kasumi(int a, int b){
int res = 1;
while(b){
if(b & 1) res = (LL) res * a % mod;
a = (LL) a * a % mod; b >>= 1;
}
return res;
}
int main(){
scanf("%d", &n);
for(Rint i = 1;i <= n;i ++) scanf("%d", a + i), vis[a[i]] = true, mx = max(mx, a[i]);
for(Rint i = 1;i <= mx;i ++)
for(Rint j = (i << 1);j <= mx;j += i) vis[i] |= vis[j];
f[0] = 0; f[1] = g[1] = 1;
for(Rint i = 2;i <= mx;i ++) f[i] = add(f[i - 1], f[i - 2]), g[i] = 1;
for(Rint i = 1;i <= mx;i ++){
int tmp = kasumi(g[i] = (LL) g[i] * f[i] % mod, mod - 2);
for(Rint j = (i << 1);j <= mx;j += i) g[j] = (LL) g[j] * tmp % mod;
}
for(Rint i = 1;i <= mx;i ++) if(vis[i]) ans = (LL) ans * g[i] % mod;
printf("%d", ans);
}
```
