牛客14527 随机树
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题解
在DFS序序列上维护每种素数的幂次之和
代码
#include
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
ll c, f(1);
for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
for(;isdigit(c);c=getchar())x=x*10+c-0x30;
return f*x;
}
struct SegmentTree
{
int sum[maxn<<2], L[maxn<<2], R[maxn<<2];
void pushup(ll o)
{
sum[o]=sum[o<<1]+sum[o<<1|1];
}
void build(ll o, ll l, ll r)
{
ll mid(l+r>>1);
L[o]=l, R[o]=r;
if(l==r)return;
build(o<<1,l,mid);
build(o<<1|1,mid+1,r);
pushup(o);
}
ll Sum(ll o, ll l, ll r)
{
ll mid(L[o]+R[o]>>1), ans(0);
if(l<=L[o] and r>=R[o])return sum[o];
if(l<=mid)ans+=Sum(o<<1,l,r);
if(r>mid)ans+=Sum(o<<1|1,l,r);
return ans;
}
void add(ll o, ll pos, ll v)
{
ll mid(L[o]+R[o]>>1);
if(L[o]==R[o]){sum[o]+=v;return;}
if(pos<=mid)add(o<<1,pos,v);
else add(o<<1|1,pos,v);
pushup(o);
}
}segtree[6];
struct Graph
{
int etot, head[maxn], to[maxe], next[maxe], w[maxe];
void clear(int N)
{
for(int i=1;i<=N;i++)head[i]=0;
etot=0;
}
void adde(int a, int b, int c=0){to[++etot]=b;w[etot]=c;next[etot]=head[a];head[a]=etot;}
#define forp(_,__) for(auto p=__.head[_];p;p=__.next[p])
}G;
struct Easy_Tree
{
int depth[maxn], dist[maxn], tid[maxn], rtid[maxn], tim, size[maxn], rev[maxn];
void dfs(int pos, int pre, Graph& G)
{
tid[pos]=++tim;
rev[tid[pos]]=pos;
size[pos]=1;
forp(pos,G)if(G.to[p]!=pre)
{
depth[G.to[p]]=depth[pos]+1;
dist[G.to[p]]=dist[pos]+G.w[p];
dfs(G.to[p],pos,G);
size[pos]+=size[G.to[p]];
}
rtid[pos]=tim;
}
void run(Graph& G, int root)
{
tim=0;
depth[root]=1;
dfs(1,0,G);
}
}et;
struct EasyMath
{
ll prime[maxn], phi[maxn], mu[maxn];
bool mark[maxn];
ll fastpow(ll a, ll b, ll c)
{
ll t(a%c), ans(1ll);
for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
return ans;
}
void exgcd(ll a, ll b, ll &x, ll &y)
{
if(!b){x=1,y=0;return;}
ll xx, yy;
exgcd(b,a%b,xx,yy);
x=yy, y=xx-a/b*yy;
}
ll inv(ll x, ll p) //p是素数
{return fastpow(x%p,p-2,p);}
ll inv2(ll a, ll p)
{
ll x, y;
exgcd(a,p,x,y);
return (x+p)%p;
}
void shai(ll N)
{
ll i, j;
for(i=2;i<=N;i++)mark[i]=false;
*prime=0;
phi[1]=mu[1]=1;
for(i=2;i<=N;i++)
{
if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
for(j=1;j<=*prime and i*prime[j]<=N;j++)
{
mark[i*prime[j]]=true;
if(i%prime[j]==0)
{
phi[i*prime[j]]=phi[i]*prime[j];
break;
}
mu[i*prime[j]]=-mu[i];
phi[i*prime[j]]=phi[i]*(prime[j]-1);
}
}
}
ll CRT(vector<ll> a, vector<ll> m) //要求模数两两互质
{
ll M=1, ans=0, n=a.size(), i;
for(i=0;i<n;i++)M*=m[i];
for(i=0;i<n;i++)(ans+=a[i]*(M/m[i])%M*inv2(M/m[i],m[i]))%=M;
return ans;
}
}em;
ll getcnt(ll n, ll p)
{
ll ret=0;
while(n%p==0)n/=p, ret++;
return ret;
}
#define mod 1000000007ll
int main()
{
ll n=read(), i, j;
vector<ll> pr={2,3,5,7,11,13};
rep(i,1,n-1)
{
ll u=read()+1, v=read()+1;
G.adde(u,v);
}
et.run(G,1);
rep(j,0,5)segtree[j].build(1,1,n);
rep(i,1,n)
{
ll T=read();
rep(j,0,5)
{
ll cnt = getcnt(T,pr[j]);
segtree[j].add(1,et.tid[i],cnt);
}
}
ll q=read();
char s[10];
while(q--)
{
scanf("%s",s);
if(s[0]=='S')
{
ll u=read()+1, x=read();
rep(j,0,5)
{
ll cnt = getcnt(x,pr[j]);
segtree[j].add(1,et.tid[u],cnt);
}
}
else
{
ll u=read()+1;
ll ans=1, ans2=1;
rep(j,0,5)
{
ll P = segtree[j].Sum(1,et.tid[u],et.rtid[u]);
// printf("%lld^%lld\n",pr[j],P);
ans *= em.fastpow(pr[j],P,mod);
ans %= mod;
ans2 *= P+1;
ans2 %= mod;
}
printf("%lld %lld\n",ans,ans2);
}
}
return 0;
}