题解 树论
首先暴力 DP 是容易的，考虑优化
一开始的思路是考虑质因子集合，后来发现取值范围是个区间
但是把转移写出来：
\[f_{i,j}\gets f_{i,j}*\sum\limits_{k=l_v}^{r_v}[\gcd(j, k)=1]f_{v, k}
\]
发现可以莫反
然后就没有了
复杂度 \(O(nV\log V)\)
- 互质一类条件先想想能不能套个莫比乌斯反演
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 50010
#define pb push_back
#define ll long long
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n;
bool npri[N];
int pri[N], low[N], lowp[N], mu[N], l[N], r[N], pcnt;
int head[N], deg[N], ecnt, lim;
const ll mod=1e9+7;
struct edge{int to, next;}e[N<<1];
inline void add(int s, int t) {e[++ecnt]={t, head[s]}; head[s]=ecnt;}
inline void md(ll& a, ll b) {a+=b; a=a>=mod?a-mod:a;}
inline ll qpow(ll a, ll b) {ll ans=1; for (; b; a=a*a%mod,b>>=1) if (b&1) ans=ans*a%mod; return ans;}
namespace force{
ll f[55][N], g[55][N], rec[55][N][55], rk[55][55], pre[55], suf[55], cnt[55];
bool check(int a, int b) {
for (int t=a; t>1; t/=lowp[t])
if (b%low[t]==0) return 0;
return 1;
}
void dfs1(int u, int fa) {
for (int i=l[u]; i<=r[u]; ++i) f[u][i]=1;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
dfs1(v, u);
rk[u][v]=++cnt[u];
for (int j=l[u]; j<=r[u]; ++j) {
ll sum=0;
for (int k=l[v]; k<=r[v]; ++k) if (check(j, k)) md(sum, f[v][k]);
// cout<<"sum: "<<u<<' '<<v<<' '<<j<<endl;
f[u][j]=f[u][j]*sum%mod;
rec[u][j][cnt[u]]=sum;
}
}
}
void dfs2(int u, int fa) {
if (!fa) {for (int j=l[u]; j<=r[u]; ++j) g[u][j]=1;}
for (int j=l[u]; j<=r[u]; ++j) {
pre[0]=suf[cnt[u]+1]=1;
for (int i=1; i<=cnt[u]; ++i) pre[i]=pre[i-1]*rec[u][j][i]%mod;
for (int i=cnt[u]; i; --i) suf[i]=suf[i+1]*rec[u][j][i]%mod;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
for (int k=l[v]; k<=r[v]; ++k) if (check(j, k))
g[v][k]=(g[v][k]+g[u][j]*pre[rk[u][v]-1]%mod*suf[rk[u][v]+1])%mod;
}
}
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
dfs2(v, u);
}
}
void solve() {
//cout<<double(sizeof(rec)+sizeof(f)*2)/1000/1000<<endl;
dfs1(1, 0); dfs2(1, 0);
#if 0
cout<<"---f---"<<endl; for (int i=1; i<=n; ++i) {for (int j=1; j<=lim; ++j) cout<<f[i][j]<<' '; cout<<endl;}
cout<<"---g---"<<endl; for (int i=1; i<=n; ++i) {for (int j=1; j<=lim; ++j) cout<<g[i][j]<<' '; cout<<endl;}
#endif
for (int i=1; i<=n; ++i) {
ll ans=0;
for (int j=l[i]; j<=r[i]; ++j)
ans=(ans+j*f[i][j]%mod*g[i][j])%mod;
printf("%lld\n", (ans%mod+mod)%mod);
}
}
}
namespace task{
vector<int> div[N];
ll h[N][55], tem[N];
ll f[55][N], g[55][N], rec[55][N];
bool check(int a, int b) {
for (int t=a; t>1; t/=lowp[t])
if (b%low[t]==0) return 0;
return 1;
}
void dfs1(int u, int fa) {
vector<int> son;
for (int i=l[u]; i<=r[u]; ++i) f[u][i]=1;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
son.pb(v);
dfs1(v, u);
}
// cout<<"calc: "<<u<<endl;
for (int i=1; i<=r[u]; ++i) for (auto j:son) h[i][j]=0;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
for (int d=1; d<=r[u]; ++d) {
int tl=(l[v]-1)/d+1, tr=r[v]/d;
for (int k=tl*d; k<=tr*d; k+=d)
h[d][v]=(h[d][v]+f[v][k])%mod;
}
}
for (int i=1; i<=r[u]; ++i) for (auto j:son) h[i][j]*=mu[i];
// cout<<"h: "; for (int i=1; i<=r[u]; ++i) cout<<h[i][2]<<' '; cout<<endl;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
for (int j=l[u]; j<=r[u]; ++j) {
ll sum=0;
for (auto d:div[j]) sum=(sum+h[d][v])%mod;
f[u][j]=f[u][j]*sum%mod;
}
}
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
for (int j=l[u]; j<=r[u]; ++j) {
ll sum=0;
for (auto d:div[j]) sum=(sum+h[d][v])%mod;
rec[v][j]=f[u][j]*qpow(sum, mod-2)%mod;
}
}
}
void dfs2(int u, int fa) {
// cout<<"dfs2: "<<u<<endl;
// cout<<"rec: "; for (int i=1; i<=lim; ++i) cout<<rec[u][i]<<' '; cout<<endl;
if (!fa) {for (int j=l[u]; j<=r[u]; ++j) g[u][j]=1;}
else {
for (int i=1; i<=r[u]; ++i) tem[i]=0;
for (int d=1; d<=r[u]; ++d) {
int tl=(l[fa]-1)/d+1, tr=r[fa]/d;
for (int k=tl*d; k<=tr*d; k+=d)
tem[d]=(tem[d]+g[fa][k]*rec[u][k])%mod;
tem[d]*=mu[d];
}
for (int j=l[u]; j<=r[u]; ++j) {
for (auto d:div[j])
g[u][j]=(g[u][j]+tem[d])%mod;
}
}
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
dfs2(v, u);
}
}
void solve() {
//cout<<double(sizeof(rec)+sizeof(f)*2)/1000/1000<<endl;
for (int i=1; i<=lim; ++i) for (int j=1; i*j<=lim; ++j) div[i*j].pb(i);
// cout<<"div: "<<endl;
// for (int i=1; i<=lim; ++i) {cout<<i<<": "; for (auto it:div[i]) cout<<it<<' '; cout<<endl;}
dfs1(1, 0); dfs2(1, 0);
#if 0
cout<<"---f---"<<endl; for (int i=1; i<=n; ++i) {for (int j=1; j<=lim; ++j) cout<<f[i][j]<<' '; cout<<endl;}
cout<<"---g---"<<endl; for (int i=1; i<=n; ++i) {for (int j=1; j<=lim; ++j) cout<<g[i][j]<<' '; cout<<endl;}
#endif
for (int i=1; i<=n; ++i) {
ll ans=0;
for (int j=l[i]; j<=r[i]; ++j)
ans=(ans+j*f[i][j]%mod*g[i][j])%mod;
printf("%lld\n", (ans%mod+mod)%mod);
}
}
}
signed main()
{
freopen("tree.in", "r", stdin);
freopen("tree.out", "w", stdout);
n=read();
memset(head, -1, sizeof(head));
for (int i=1; i<=n; ++i) l[i]=read();
for (int i=1; i<=n; ++i) lim=max(lim, r[i]=read());
for (int i=1,u,v; i<n; ++i) {
u=read(); v=read();
add(u, v); add(v, u);
}
mu[1]=1;
for (int i=2; i<=lim; ++i) {
if (!npri[i]) pri[++pcnt]=low[i]=lowp[i]=i, mu[i]=-1;
for (int j=1,x; j<=pcnt&&i*pri[j]<=lim; ++j) {
npri[x=i*pri[j]]=1;
if (!(i%pri[j])) {
low[x]=pri[j];
lowp[x]=lowp[i]*pri[j];
}
else low[x]=lowp[x]=pri[j], mu[x]=-mu[i];
}
}
// force::solve();
task::solve();
return 0;
}
