题目大意
给你一棵\(n\)个点的树，每个点有权值\(a_i\)，\(a\)为一个排列，求
\[\frac{1}{n(n-1)}\sum_{i=1}^n\sum_{j=1}^n \varphi(a_ia_j)dist_{i,j}
\]
\]
\(n\leq 200000\)
题解
\[\begin{align}
ans&=\frac{1}{n(n-1)}\sum_{i=1}^n\sum_{j=1}^n \varphi(a_ia_j)dist_{i,j}\\
&=\frac{1}{n(n-1)}\sum_{i=1}^n\sum_{j=1}^n\sum_{d=(a_i,a_j)} \frac{\varphi(a_i)\varphi(a_j)d}{\varphi(d)}dist_{i,j}\\
&=\frac{1}{n(n-1)}\sum_{d=1}^n\frac{d}{\mu(d)}\sum_{d=(a_i,a_j)}\varphi(a_i)\varphi(a_j)dist_{i,j}\\
f(d)&=\sum_{d=(a_i,a_j)}\varphi(a_i)\varphi(a_j)dist_{i,j}\\
F(d)&=\sum_{d|a_i,d|a_j}\varphi(a_i)\varphi(a_j)dist_{i,j}\\
F(d)&=\sum_{d|n}f(n)\\
f(d)&=F(d)-\sum_{d|n,d\neq n}f(n)
\end{align}
\]
ans&=\frac{1}{n(n-1)}\sum_{i=1}^n\sum_{j=1}^n \varphi(a_ia_j)dist_{i,j}\\
&=\frac{1}{n(n-1)}\sum_{i=1}^n\sum_{j=1}^n\sum_{d=(a_i,a_j)} \frac{\varphi(a_i)\varphi(a_j)d}{\varphi(d)}dist_{i,j}\\
&=\frac{1}{n(n-1)}\sum_{d=1}^n\frac{d}{\mu(d)}\sum_{d=(a_i,a_j)}\varphi(a_i)\varphi(a_j)dist_{i,j}\\
f(d)&=\sum_{d=(a_i,a_j)}\varphi(a_i)\varphi(a_j)dist_{i,j}\\
F(d)&=\sum_{d|a_i,d|a_j}\varphi(a_i)\varphi(a_j)dist_{i,j}\\
F(d)&=\sum_{d|n}f(n)\\
f(d)&=F(d)-\sum_{d|n,d\neq n}f(n)
\end{align}
\]
\(F(d)\)可以直接建虚树DP求。
然后直接反演统计就可以得到答案。
总的点数是\(\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor=O(n\log n)\)
所以总的时间复杂度是\(O(n\log^2 n)\)
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
ll p=1000000007;
struct graph
{
int h[200010];
int v[400010];
int w[400010];
int t[400010];
int n;
graph()
{
memset(h,0,sizeof h);
n=0;
}
void add(int x,int y,int z)
{
n++;
v[n]=y;
w[n]=z;
t[n]=h[x];
h[x]=n;
}
};
graph g,g2;
int f[200010][20];
int d[200010];
int st[200010];
int ti;
void dfs(int x,int fa,int dep)
{
f[x][0]=fa;
d[x]=dep;
st[x]=++ti;
int i;
for(i=1;i<=19;i++)
f[x][i]=f[f[x][i-1]][i-1];
for(i=g.h[x];i;i=g.t[i])
if(g.v[i]!=fa)
dfs(g.v[i],x,dep+1);
}
int getlca(int x,int y)
{
if(d[x]<d[y])
swap(x,y);
int i;
for(i=19;i>=0;i--)
if(d[f[x][i]]>=d[y])
x=f[x][i];
if(x==y)
return x;
for(i=19;i>=0;i--)
if(f[x][i]!=f[y][i])
{
x=f[x][i];
y=f[y][i];
}
return f[x][0];
}
ll phi[200010];
int b[200010];
int pri[100010];
int cnt;
ll inv[200010];
void init(int n)
{
int i,j;
inv[0]=inv[1]=1;
for(i=2;i<=n;i++)
inv[i]=-(p/i)*inv[p%i]%p;
phi[1]=1;
cnt=0;
for(i=2;i<=n;i++)
{
if(!b[i])
{
pri[++cnt]=i;
phi[i]=i-1;
}
for(j=1;j<=cnt&&i*pri[j]<=n;j++)
{
b[i*pri[j]]=1;
if(i%pri[j]==0)
{
phi[i*pri[j]]=phi[i]*pri[j];
break;
}
phi[i*pri[j]]=phi[i]*phi[pri[j]];
}
}
}
ll a[200010];
ll s[200010];
int c[200010];
int c1[200010];
int ct;
int n;
int stack[200010];
int top;
int cmp(int a,int b)
{
return st[a]<st[b];
}
ll s1[200010];
ll s2[200010];
ll sum;
void add(int x,int y)//f[x]=y
{
ll s3=(s1[x]+(d[x]-d[y])*s2[x])%p;
sum=(sum+s3*s2[y]+s1[y]*s2[x])%p;
s1[y]=(s1[y]+s3)%p;
s2[y]=(s2[y]+s2[x])%p;
}
ll solve(int x)
{
sum=0;
ct=top=0;
int i;
for(i=x;i<=n;i+=x)
c1[++ct]=c[i];
sort(c1+1,c1+ct+1,cmp);
int rt=getlca(c1[1],c1[ct]);
if(rt!=c1[1])
{
stack[++top]=rt;
s1[rt]=s2[rt]=0;
}
for(i=1;i<=ct;i++)
{
if(i>=2)
{
int lca=getlca(c1[i],c1[i-1]);
while(d[stack[top]]>d[lca])
if(d[stack[top-1]]<d[lca])
{
s1[lca]=s2[lca]=0;
add(stack[top],lca);
stack[top]=lca;
}
else
{
add(stack[top],stack[top-1]);
top--;
}
}
stack[++top]=c1[i];
s1[c1[i]]=0;
s2[c1[i]]=phi[a[c1[i]]];
}
while(top>1)
{
add(stack[top],stack[top-1]);
top--;
}
return sum*2%p;
}
int main()
{
scanf("%d",&n);
init(n);
int i,x,y,j;
for(i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
c[a[i]]=i;
}
for(i=1;i<n;i++)
{
scanf("%d%d",&x,&y);
g.add(x,y,0);
g.add(y,x,0);
}
dfs(1,0,1);
for(i=1;i<=n;i++)
s[i]=solve(i);
ll ans=0;
for(i=n;i>=1;i--)
{
for(j=i+i;j<=n;j+=i)
s[i]-=s[j];
ans=(ans+s[i]*i%p*inv[phi[i]]%p)%p;
}
ans=ans*inv[n]%p*inv[n-1]%p;
ans=(ans+p)%p;
printf("%lld\n",ans);
return 0;
}