题目传送门：CF809E。
题意简述：
有一棵 (n) 个点的树，第 (i) 个节点有一个值 (a_i)，保证 (a_i) 是一个从 (1) 到 (n) 的排列。
请求出在树中均匀随机选择两个点 (u,v)（(u
e v)）时 (varphi(a_ua_v)mathrm{dist}(u,v)) 的期望。
其中 (mathrm{dist}(u,v)) 定义为树上 (u,v) 两点间简单路径经过的边数。
题解：
定义 (hat{a}) 是 (a) 的逆置换，推式子：
[egin{aligned}mathrm{Ans}&=frac{1}{n(n-1)}sum_{1le i,jle n}varphi(a_ia_j)mathrm{dist}(i,j)\n(n-1)mathrm{Ans}&=sum_{1le i,jle n}frac{varphi(a_i)varphi(a_j)gcd(a_i,a_j)}{varphi(gcd(a_i,a_j))}mathrm{dist}(i,j)\&=sum_{d=1}^{n}frac{d}{varphi(d)}sum_{1le i,jle n}varphi(a_i)varphi(a_j)mathrm{dist}(i,j)[gcd(a_i,a_j)=d]\&=sum_{d=1}^{n}frac{d}{varphi(d)}sum_{1le i,jle n}varphi(i)varphi(j)mathrm{dist}(hat{a}_i,hat{a}_j)[gcd(i,j)=d]\&=sum_{d=1}^{n}frac{d}{varphi(d)}sum_{1le i,jle ndiv d}varphi(id)varphi(jd)mathrm{dist}(hat{a}_{id},hat{a}_{jd})sum_{e|i,e|j}mu(e)\&=sum_{t=1}^{n}f(t)sum_{i=1}^{n}sum_{j=1}^{n}[t|a_i][t|a_j]varphi(a_i)varphi(a_j)mathrm{dist}(i,j)end{aligned}
]
其中 (displaystyle f(n)=sum_{d|n}frac{d}{varphi(d)}mu!left(frac{n}{d} ight))，仅有 (displaystyle f(p)=frac{1}{p-1})，可以线筛。
令后面的部分为 (g(t))，则有：
[egin{aligned}g(t)&=sum_{i=1}^{n}sum_{j=1}^{n}[t|a_i][t|a_j]varphi(a_i)varphi(a_j)mathrm{dist}(i,j)\&=sum_{i=1}^{n}sum_{j=1}^{n}[t|a_i][t|a_j]varphi(a_i)varphi(a_j)(mathrm{dep}[i]+mathrm{dep}[j]-2mathrm{dep}[mathrm{lca}(i,j)])\frac{1}{2}g(t)&=sum_{i=1}^{n}[t|a_i]varphi(a_i)sum_{i=1}^{n}[t|a_i]varphi(a_i)mathrm{dep}[i]-sum_{i=1}^{n}sum_{j=1}^{n}[t|a_i][t|a_j]varphi(a_i)varphi(a_j)mathrm{dep}[mathrm{lca}(i,j)]end{aligned}
]
即 (displaystylemathrm{Ans}=frac{2}{n(n-1)}sum_{t=1}^{n}f(t)!left(sum_{i=1}^{n}[t|a_i]varphi(a_i)sum_{i=1}^{n}[t|a_i]varphi(a_i)mathrm{dep}[i]-sum_{i=1}^{n}sum_{j=1}^{n}[t|a_i][t|a_j]varphi(a_i)varphi(a_j)mathrm{dep}[mathrm{lca}(i,j)] ight))。
每次建虚树，树上统计一下即可。
下面是代码，时间复杂度 (mathcal{O}(nlog n))：
#include <cstdio>
#include <vector>
#define csi const int
typedef long long LL;
csi Mod = 1000000007;
csi MN = 200005, MP = 17995;
inline int qPow(int b, int e) {
int a = 1;
for (; e; e >>= 1, b = (LL)b * b % Mod)
if (e & 1) a = (LL)a * b % Mod;
return a;
}
bool ip[MN];
int p[MP], pc;
int Inv[MN], phi[MN], f[MN];
std::vector<int> F[MN];
inline void Init(csi &N) {
Inv[0] = 1;
for (int i = 1; i <= N; ++i)
Inv[i] = (LL)Inv[i - 1] * i % Mod;
int iFac = qPow(Inv[N], Mod - 2);
for (int i = N; i >= 1; --i)
Inv[i] = (LL)iFac * Inv[i - 1] % Mod,
iFac = (LL)iFac * i % Mod;
phi[1] = f[1] = 1;
for (int i = 2; i <= N; ++i) {
if (!ip[i]) p[++pc] = i, phi[i] = i - 1, f[i] = Inv[i - 1];
for (int j = 1; j <= pc; ++j) {
csi &k = p[j] * i;
if (k > N) break;
ip[k] = 1;
if (i % p[j])
phi[k] = phi[i] * (p[j] - 1),
f[k] = (LL)f[i] * f[p[j]] % Mod;
else { phi[k] = phi[i] * p[j]; break; }
}
}
for (int i = 1; i <= N; ++i) if (f[i])
for (int j = i; j <= N; j += i)
F[j].push_back(i);
}
int N, A[MN], Ans;
std::vector<int> G[MN];
int dep[MN], dfn[MN], idf[MN], dfc;
int ieu[MN * 2], leu[MN], reu[MN], euc;
void DFS(csi &u, csi &fz) {
dep[u] = dep[fz] + 1;
idf[dfn[u] = ++dfc] = u;
ieu[leu[u] = ++euc] = u;
for (auto v : G[u]) if (v != fz)
DFS(v, u), ieu[++euc] = u;
reu[u] = euc;
}
inline int ckdep(csi &i, csi &j) {
return dep[i] < dep[j] ? i : j;
}
int lg[MN * 2], ST[19][MN * 2];
void _ST() {
lg[0] = -1;
for (int i = 1; i <= euc; ++i) {
lg[i] = lg[i >> 1] + 1;
ST[0][i] = ieu[i];
}
for (int j = 0; j < lg[euc]; ++j)
for (int i = 2 << j; i <= euc; ++i)
ST[j + 1][i] = ckdep(ST[j][i - (1 << j)], ST[j][i]);
}
inline int rmq(csi &l, csi &r) {
csi &g = lg[r - l + 1];
return ckdep(ST[g][l + (1 << g) - 1], ST[g][r]);
}
inline int lca(csi &u, csi &v) {
if (reu[u] < leu[v]) return rmq(reu[u], leu[v]);
if (reu[v] < leu[u]) return rmq(reu[v], leu[u]);
return ckdep(u, v);
}
std::vector<int> V[MN];
int stk[MN], tp, st[MN], t;
std::vector<int> T[MN];
int val[MN];
inline void AddP(csi &x, csi &y) {
if (y) T[x].push_back(y), val[y] = dep[y] - dep[x];
}
int it[MN], sphi[MN], Sum;
inline void DP(csi &u) {
sphi[u] = it[u] ? phi[A[u]] : 0;
for (auto v : T[u]) {
DP(v);
sphi[u] += sphi[v];
if (sphi[u] >= Mod) sphi[u] -= Mod;
}
Sum = (Sum - (LL)sphi[u] * sphi[u] % Mod * val[u]) % Mod;
}
void vtree() {
for (int id = 1; id <= N; ++id) {
int u = idf[id], k = A[u];
for (auto i : F[k]) V[i].push_back(u);
}
for (int i = 1; i <= N; ++i) {
if (!f[i]) continue;
int sz = V[i].size(), x;
int S1 = 0, S2 = 0;
for (int j = 0; j < sz; ++j) {
int u = V[i][j];
it[u] = 1;
st[++t] = u;
S1 += phi[A[u]];
if (S1 >= Mod) S1 -= Mod;
S2 = (S2 + (LL)phi[A[u]] * dep[u]) % Mod;
if (tp) {
csi &l = lca(stk[tp], u), &dl = dep[l];
for (x = 0; tp && dep[stk[tp]] > dl; x = stk[tp--])
AddP(stk[tp], x);
if (stk[tp] != l) stk[++tp] = l, st[++t] = l;
AddP(stk[tp], x);
stk[++tp] = u;
}
else stk[++tp] = u;
}
for (x = 0; tp; x = stk[tp--]) AddP(stk[tp], x);
val[x] = dep[x];
Sum = (LL)S1 * S2 % Mod;
DP(x);
Ans = (Ans + (LL)f[i] * (Sum + Mod)) % Mod;
while (t) it[st[t]] = 0, T[st[t--]].clear();
}
}
int main() {
scanf("%d", &N);
for (int i = 1; i <= N; ++i) scanf("%d", &A[i]);
Init(N);
for (int i = 1, x, y; i < N; ++i)
scanf("%d%d", &x, &y),
G[x].push_back(y),
G[y].push_back(x);
DFS(1, 0);
_ST();
vtree();
printf("%lld
", (LL)Ans * qPow((LL)N * (N - 1) / 2 % Mod, Mod - 2) % Mod);
return 0;
}