题意
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≤200000)
答案对 (10^9+7) 取模
Sol
莫比乌斯反演
然后要求
[sum_{d=1}^{n}frac{d}{varphi(d)}sum_{i=1}^{lfloorfrac{n}{d}
floor}mu(i)F(i*d)
]
其中
(F(d)=sum_{d|a_i}sum_{d|a_j}varphi(a_i)varphi(a_j)dist(i,j))
直接把(F(d))求出来，把满足要求的点拿出来建虚树(DP)就好了
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
const int maxn(2e5 + 5);
const int mod(1e9 + 7);
typedef int Arr[maxn];
int n, cnt, idx, tot, len, ret;
Arr dfn, size, fa, first1, first2, s, deep, id;
int lg[maxn << 1], st[20][maxn << 1];
Arr a, prime, isprime, phi, mu, f, mark, sum;
vector <int> p[maxn], fac[maxn];
struct Edge{
int to, next;
} e1[maxn << 1], e2[maxn];
IL int Pow(RG ll x, RG ll y){
RG ll ret = 1;
for(; y; y >>= 1, x = x * x % mod)
if(y & 1) ret = ret * x % mod;
return ret;
}
IL void Add1(RG int u, RG int v){
e1[cnt] = (Edge){v, first1[u]}, first1[u] = cnt++;
}
IL void Add2(RG int u, RG int v){
e2[cnt] = (Edge){v, first2[u]}, first2[u] = cnt++;
}
IL void Dfs(RG int u){
size[u] = 1, dfn[u] = ++idx, st[0][id[u] = ++len] = u;
for(RG int i = 0, j = fac[a[u]].size(); i < j; ++i)
p[fac[a[u]][i]].push_back(u);
for(RG int e = first1[u]; e != -1; e = e1[e].next){
RG int v = e1[e].to;
if(!size[v]){
deep[v] = deep[u] + 1;
Dfs(v), size[u] += size[v];
fa[v] = u, st[0][++len] = u;
}
}
}
IL int Min(RG int x, RG int y){
return deep[x] < deep[y] ? x : y;
}
IL int LCA(RG int u, RG int v){
u = id[u], v = id[v];
if(u > v) swap(u, v);
RG int lgn = lg[v - u + 1];
return Min(st[lgn][u], st[lgn][v - (1 << lgn) + 1]);
}
IL int Dis(RG int u, RG int v){
RG int lca = LCA(u, v);
return deep[u] + deep[v] - 2 * deep[lca];
}
IL void Upd(RG int &x, RG int y){
x += y;
if(x >= mod) x -= mod;
}
IL void DP(RG int u){
sum[u] = mark[u] * phi[a[u]];
Upd(ret, mod - 1LL * sum[u] * sum[u] % mod * deep[u] % mod);
for(RG int e = first2[u]; e != -1; e = e2[e].next){
RG int v = e2[e].to;
DP(v);
Upd(ret, mod - 2LL * sum[u] * sum[v] % mod * deep[u] % mod);
Upd(sum[u], sum[v]);
}
}
IL int Calc(RG int x){
RG int l = p[x].size(), ret1 = 0, ret2 = 0, t = 0; cnt = 0;
for(RG int i = 0; i < l; ++i) first2[p[x][i]] = -1, mark[p[x][i]] = 1;
for(RG int i = 0; i < l; ++i){
RG int lca = s[t] ? LCA(s[t], p[x][i]) : 0;
if(lca == s[t]) s[++t] = p[x][i];
else{
while(t > 1 && deep[lca] <= deep[s[t - 1]]) Add2(s[t - 1], s[t]), --t;
if(s[t] != lca) first2[lca] = -1, Add2(lca, s[t]), s[t] = lca;
s[++t] = p[x][i];
}
}
while(t > 1) Add2(s[t - 1], s[t]), --t;
for(RG int i = 0; i < l; ++i){
Upd(ret1, phi[a[p[x][i]]]);
Upd(ret2, 1LL * deep[p[x][i]] * phi[a[p[x][i]]] % mod);
}
ret = 0;
DP(s[1]), Upd(ret, ret);
Upd(ret, 2LL * ret1 * ret2 % mod);
for(RG int i = 0; i < l; ++i) mark[p[x][i]] = 0;
return ret;
}
IL void Sieve(){
phi[1] = mu[1] = 1;
for(RG int i = 2; i <= n; ++i){
if(!isprime[i]) prime[++tot] = i, phi[i] = i - 1, mu[i] = -1;
for(RG int j = 1; j <= tot && i * prime[j] <= n; ++j){
isprime[i * prime[j]] = 1;
if(i % prime[j]){
phi[i * prime[j]] = phi[i] * phi[prime[j]];
mu[i * prime[j]] = -mu[i];
}
else{
phi[i * prime[j]] = phi[i] * prime[j];
mu[i * prime[j]] = 0;
break;
}
}
}
for(RG int i = 1; i <= n; ++i)
for(RG int j = 1; j * i <= n; ++j) fac[j * i].push_back(i);
}
int main(){
n = Input(), Sieve();
for(RG int i = 1; i <= n; ++i) first1[i] = first2[i] = -1, a[i] = Input();
for(RG int i = 1; i < n; ++i){
RG int u = Input(), v = Input();
Add1(u, v), Add1(v, u);
}
Dfs(1);
for(RG int i = 2; i <= len; ++i) lg[i] = lg[i >> 1] + 1;
for(RG int j = 1; j <= lg[len]; ++j)
for(RG int i = 1; i + (1 << j) - 1 <= len; ++i)
st[j][i] = Min(st[j - 1][i], st[j - 1][i + (1 << (j - 1))]);
for(RG int i = 1; i <= n; ++i) f[i] = Calc(i);
RG int ans = 0;
for(RG int i = 1; i <= n; ++i){
RG int ret = 0;
for(RG int j = 1, t = n / i; j <= t; ++j)
Upd(ret, 1LL * (mu[j] + mod) % mod * f[i * j] % mod);
ret = 1LL * ret * i % mod;
Upd(ans, 1LL * ret * Pow(phi[i], mod - 2) % mod);
}
ans = 1LL * ans * Pow(1LL * n * (n - 1) % mod, mod - 2) % mod;
printf("%d
", ans);
return 0;
}