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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF809E Surprise me!

LuoGu: CF809E Surprise me!

CF: E. Surprise me!

权值为 $[1, n]$ ,将所有点权值换成编号。考虑将 $\varphi(ij)$ 拆开。即 $\varphi(i j) = \dfrac{\varphi(i)\varphi(j)\gcd(i, j)} {\varphi(\gcd(i, j))}$ 。

上莫反:

$$
\begin{aligned}
&\sum_{i = 1} ^ n\sum_{j = 1} ^ n \frac{\varphi(i) \varphi(j) \gcd(i, j)} {\varphi(\gcd(i, j))} \text{dist}(i, j)\\
=& \sum_{d = 1} ^ n \frac{d}{\varphi(d)} \sum_{i = 1} ^ n \sum_{j = 1} ^ n [\gcd(i, j) = d] \varphi(i) \varphi(j) \text{dist}(i, j)\\
=& \sum_{d = 1} ^ n \frac d{\varphi(d)} \sum_{k = 1} ^ {\frac nd} \sum_{kd | i} \sum_{kd | j} \mu(k) \varphi(i) \varphi(j) \text{dist}(i, j)\\
=& \sum_{T = 1} ^ n \sum_{d | T} \frac{d\mu(\frac Td)}{\varphi(d)} \sum_{T | i} \sum_{T | j} \varphi(i) \varphi(j) \text{dist}(i, j)
\end{aligned}
$$

可以用 $O(n \ln n)$ 的时间预处理出 $\frac{d\mu(\frac Td)}{\varphi(d)}$ 。

对于每个 $T$ ,注意到可用的点只有 $\frac n T$ 个,考虑建出虚树,对于每个点 $i$ ,考虑求出 $\varphi(j) \text{dist}(i, j)$ ,可以用换根DP。

查看代码
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#include <cstdio>
#include <vector>
#include <algorithm>
#define pb push_back
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
flag |= c == '-';
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar('0' + x % 10);
}
typedef long long LL;
const int N = 2e5 + 10, M = 20, mod = 1e9 + 7;
bool vis[N];
int top, stk[N];
int n, d[N], A[N], dfn[N], w[N], r[N], s[N];
int cnt, p[N], mu[N], phi[N], f[N];
vector <int> h, e[N];
struct Edge { int v, w; };
vector <Edge> g[N];
int binpow (int b, int k = mod - 2)
{
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % mod)
if (k & 1) res = (LL)res * b % mod;
return res;
}
namespace LCA
{
int stmp, lg[N << 1], st[N << 1][M];
int dmin (int a, int b) { return d[a] < d[b] ? a : b; }
void dfs (int u = 1)
{
st[dfn[u] = ++stmp][0] = u;
for (int v : e[u]) if (!dfn[v])
{
d[v] = d[u] + 1;
dfs(v);
st[++stmp][0] = u;
}
}
int lca (int a, int b)
{
if ((a = dfn[a]) > (b = dfn[b])) swap(a, b);
int k = lg[b - a + 1];
return dmin(st[a][k], st[b - (1 << k) + 1][k]);
}
void init ()
{
dfs();
for (int i = 2; i <= stmp; ++i)
lg[i] = lg[i >> 1] + 1;
for (int k = 1; 1 << k <= stmp; ++k)
for (int i = 1; i + (1 << k) - 1 <= stmp; ++i)
st[i][k] = dmin(st[i][k - 1], st[i + (1 << k - 1)][k - 1]);
}
}
void init ()
{
vis[1] = true;
mu[1] = 1;
phi[1] = 1;
for (int i = 2; i <= n; ++i)
{
if (!vis[i])
{
p[++cnt] = i;
mu[i] = -1;
phi[i] = i - 1;
}
for (int j = 1; j <= cnt && i * p[j] <= n; ++j)
{
vis[i * p[j]] = true;
if (i % p[j] == 0)
{
mu[i * p[j]] = 0;
phi[i * p[j]] = phi[i] * p[j];
break;
}
mu[i * p[j]] = mu[i] * mu[p[j]];
phi[i * p[j]] = phi[i] * phi[p[j]];
}
}
for (int i = 1; i <= n; ++i)
for (int j = 1; i * j <= n; ++j)
(f[i * j] += (LL)i * mu[j] * binpow(phi[i]) % mod) %= mod;
}
void dfs1 (int u = 1)
{
r[u] = w[u], s[u] = 0;
for (Edge i : g[u])
{
dfs1(i.v);
(r[u] += r[i.v]) %= mod;
(s[u] += ((LL)r[i.v] * i.w + s[i.v]) % mod) %= mod;
}
}
void dfs2 (int u = 1)
{
for (Edge i : g[u])
{
s[i.v] = (s[u] + (r[1] - 2ll * r[i.v]) * i.w) % mod;
dfs2(i.v);
}
}
int calc ()
{
sort(h.begin(), h.end(), [&](int a, int b) { return dfn[a] < dfn[b]; });
g[stk[top = 1] = 1].clear();
w[1] = h[0] == 1 ? phi[1] : 0;
auto add = [&](int a, int b) { g[a].pb((Edge){b, d[b] - d[a]}); };
for (int i : h) if (i ^ 1)
{
int t = LCA::lca(i, stk[top]);
if (t ^ stk[top])
{
for (; top > 1 && dfn[t] < dfn[stk[top - 1]]; --top)
add(stk[top - 1], stk[top]);
if (dfn[t] > dfn[stk[top - 1]])
{
g[t].clear(); w[t] = 0;
add(t, stk[top]);
stk[top] = t;
}
else add(stk[top - 1], stk[top]), --top;
}
g[stk[++top] = i].clear(); w[i] = phi[i];
}
for (; top > 1; --top)
add(stk[top - 1], stk[top]);
dfs1(), dfs2();
int res = 0;
for (int i : h)
(res += (LL)phi[i] * s[i] % mod) %= mod;
return res;
}
int main ()
{
read(n);
init();
for (int i = 1; i <= n; ++i) read(A[i]);
for (int i = 1, a, b; i < n; ++i)
{
read(a, b), a = A[a], b = A[b];
e[a].pb(b), e[b].pb(a);
}
LCA::init();
int res = 0;
for (int d = 1; d <= n; ++d)
{
h.clear();
for (int i = d; i <= n; i += d) h.pb(i);
(res += (LL)calc() * f[d] % mod) %= mod;
}
write(((LL)res * binpow((LL)n * (n - 1) % mod) % mod + mod) % mod);
return 0;
}