[反演 数论] 51Nod 1355 斐波那契的最小公倍数

我好菜啊 出过一万遍的原题 我怎么第一次看见啊
某乎链接
按照zyz的做法 orzz

lcm(fS)==∏T⊆S,T≠∅gcd(fT)(−1)|T|+1∏T⊆S,T≠∅f(−1)|T|+1gcd{T}

令fn=∑d|ngd

lcm(fS)==∏T⊆S,T≠∅(∏d|gcd{T}gd)(−1)|T|+1∏dg∑T⊆S,T≠∅,d|gcd{T}(−1)|T|+1d

由于

∑T⊆S,T≠∅,d|gcd{T}(−1)|T|+1={1,0,∃a∈S,d|aotherwise

所以

lcm(fS)=∏∃a∈S,d|agd

实际上g是不需要的，可以直接枚举所有倍数容斥掉fi的指数
O(nlogn)次乘法是我跑的慢的原因

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int maxn=1000000;const int P=1e9+7;inline ll Pow(ll a,int b){  ll ret=1; for (;b;b>>=1,a=a*a%P) if (b&1) ret=ret*a%P; return ret;}inline ll Inv(ll a){  return Pow(a,P-2);}ll f[maxn+5],g[maxn+5];inline void Pre(int n){  f[0]=0; f[1]=1; for (int i=2;i<=n;i++) f[i]=(f[i-1]+f[i-2])%P;  for (int i=1;i<=n;i++) g[i]=f[i];  for (int i=2;i<=n;i++){    ll inv=Inv(g[i]);    for (int j=i+i;j<=n;j+=i)      (g[j]*=inv)%=P;  }}const int N=50005;int n,a[maxn+5];int main(){  int x;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  Pre(maxn);  read(n); for (int i=1;i<=n;i++) read(x),a[x]=1;  ll ans=1;  for (int i=1;i<=maxn;i++){    int t=0;    for (int j=i;j<=maxn && !t;j+=i) t|=a[j];    if (t) ans=ans*g[i]%P;  }  printf("%d\n",ans);  return 0;}