大意: 给定树, 点$i$的点权为$a_i$, 求$sumlimits_{a_i perp a_j}dis(i,j)$
中等难度可以枚举每条边的贡献, 维护子树内每个数出现次数$a$, 转化为求$sumlimits_{i=1}^{500}sumlimits_{j=1}^{500}([gcd(i,j)=1]a_i(tot_i-a_i))$, 反演一下可以$O(500log500)$计算.
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
const int N = 1e4+10, M = 510;
int n, a[N], c[M];
vector<int> g[N];
int f[N][M], mu[M];
ll ans;
void dfs(int x, int fa) {
for (int y:g[x]) if (y!=fa) {
dfs(y,x);
REP(i,1,500) {
f[x][i]+=f[y][i];
ll a = 0, b = 0;
for (int j=i; j<=500; j+=i) a+=f[y][j],b+=(c[j]-f[y][j]);
ans += mu[i]*a*b;
}
}
++f[x][a[x]];
}
int main() {
mu[1]=1;
REP(i,1,500) for (int j=2*i;j<=500;j+=i) mu[j]-=mu[i];
scanf("%d", &n);
REP(i,1,n) scanf("%d",a+i),++c[a[i]];
REP(i,2,n) {
int u, v;
scanf("%d%d", &u, &v);
g[u].pb(v),g[v].pb(u);
}
dfs(1,0);
printf("%lld
", ans);
}
困难难度就对每个因子建虚树, 最后容斥统计答案.
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
const int N = 1e5+10;
int n, cnt, sum, vis[N], a[N], mu[N], mi[N], L[N], R[N];
int c[N], sz[N], dep[N], top[N], fa[N], son[N], s[N];
vector<int> g[N], f[N], v;
ll ans;
void dfs(int x, int d, int f) {
L[x]=++*L,dep[x]=d,fa[x]=f;
for (int y:g[x]) if (y!=f) {
dfs(y,d+1,x),sz[x]+=sz[y];
if (sz[y]>sz[son[x]]) son[x]=y;
}
R[x]=*L;
}
void dfs(int x, int tf) {
top[x]=tf;
if (son[x]) dfs(son[x],tf);
for (int y:g[x]) if (!top[y]) dfs(y,y);
}
int lca(int x, int y) {
while (top[x]!=top[y]) {
if (dep[top[x]]<dep[top[y]]) swap(x,y);
x = fa[top[x]];
}
return dep[x]<dep[y]?x:y;
}
void fac(int x, int num, int id) {
if (x==v.size()) f[num].pb(id);
else fac(x+1,num,id),fac(x+1,num*v[x],id);
}
bool cmp(int x, int y) {return L[x]<L[y];}
void DP(int x, int z) {
for (int y:g[x]) {
DP(y,z);
ans+=z*(ll)c[y]*(sum-c[y])*(dep[y]-dep[x]);
c[x]+=c[y];
}
c[x]+=vis[x];
}
void solve(vector<int> a, int z) {
for (int x:a) vis[x]=1;
sum = a.size();
sort(a.begin(),a.end(),cmp);
int sz = a.size();
REP(i,1,sz-1) a.pb(lca(a[i],a[i-1]));
sort(a.begin(),a.end(),cmp);
a.erase(unique(a.begin(),a.end()),a.end());
s[cnt=1]=a[0],sz=a.size();
REP(i,1,sz-1) {
while (cnt>=1) {
if (L[s[cnt]]<=L[a[i]]&&L[a[i]]<=R[s[cnt]]) {
g[s[cnt]].pb(a[i]);
break;
}
--cnt;
}
s[++cnt]=a[i];
}
DP(s[1],z);
for (int x:a) g[x].clear(),vis[x]=c[x]=0;
}
int main() {
mu[1] = mi[1] = 1;
REP(i,2,N-1) mi[i]=i,mu[i]=-1;
REP(i,2,N-1) for(int j=2*i;j<N;j+=i) mu[j]-=mu[i],mi[j]=min(mi[j],i);
scanf("%d", &n);
REP(i,1,n) {
int t, x;
scanf("%d", &t);
v.clear();
while (t!=1) {
v.pb(x=mi[t]);
while (t%x==0) t/=x;
}
fac(0,1,i);
}
REP(i,2,n) {
int u, v;
scanf("%d%d", &u, &v);
g[u].pb(v),g[v].pb(u);
}
dfs(1,0,0),dfs(1,1);
REP(i,1,n) g[i].clear();
REP(i,1,N-1) if (mu[i]&&f[i].size()) solve(f[i],mu[i]);
printf("%lld
", ans);
}