2019.9.22
summary
- Back to the first half of the states do question is difficult to know the melon still want positive solutions
- The sub did not get to take
- Daily melon want
- Error computing space
== Ouyang said that all cases go through what state of mind numb like fried
football match
https://i.loli.net/2019/09/22/JKP8ufGywEjphBe.jpg
https://i.loli.net/2019/09/22/jvQPI7EoaY12sxO.jpg
Skull pain
\(ans=\sum C_n^i((pa*(1-qb))^i*(pb*qb)^{n*P-2}\)
Kaka also on the pretreatment of a space == so the inverse factorial fac
#include<bits/stdc++.h>
#define ll long long
#define rg register
const int N=1e7+5,M=100+5,P=1e9+7,INF=1e9+7,inf=0x3f3f3f3f;
int n,p,q,pa,pb,qa,qb,facn,sum=0,ans=0,ifac[N];
template <class t>void rd(t &x){
x=0;int w=0;char ch=0;
while(!isdigit(ch)) w|=ch=='-',ch=getchar();
while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
x=w?-x:x;
}
int qpow(int a,int b){
int ret=1;
while(b){
if(b&1) ret=(ll)ret*a%P;
a=(ll)a*a%P,b>>=1;
}
return ret;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("T1.txt","r",stdin);
//freopen("attack.out","w",stdout);
#endif
rd(n),rd(pa),rd(pb),rd(qa),rd(qb);
if(!pa||qa==qb) return puts("0"),0;
p=(ll)pa*qpow(pb,P-2)%P,q=(ll)qa*qpow(qb,P-2)%P;
facn=1;
for(rg int i=2;i<=n;++i) facn=(ll)facn*i%P;
ifac[n]=qpow(facn,P-2);
for(rg int i=n;i;--i) ifac[i-1]=(ll)ifac[i]*i%P;//i!的逆
int nwp=qpow((P+1-p)%P,n),nwq=qpow((P+1-q)%P,n);
int inv1=(ll)p*qpow((P+1-p)%P,P-2)%P;
int inv2=(ll)q*qpow((P+1-q)%P,P-2)%P;
for(int i=0;i<=(n>>1);++i)
ifac[n-i]=(ll)ifac[i]*ifac[n-i]%P,ifac[i]=ifac[n-i];
for(int i=1;i<=n;++i){
sum=(sum+(ll)nwq*ifac[i-1])%P,nwp=(ll)nwp*inv1%P;
ans=(ans+(ll)nwp*sum%P*ifac[i])%P,nwq=(ll)nwq*inv2%P;
}
ans=(ll)ans*facn%P;
if(p==1) ans=sum;
ans=(ll)ans*facn%P;
printf("%d\n",ans);
return 0;
}civilization
In fact, want to come out on a very simple examination room to find the midpoint of two points to jump, but did not think the tree and then cross the tree line and to maintain its
In other violence hit on the test results did not expect to jump on violence and then jump to the mid-point to cut the lead to duplication during which there are a lot of low-level errors plus root \ (QAQ \)
After seeing the solution to a problem that jumps to the midpoint of the operation to know how to do it all of a sudden hit out
#include<bits/stdc++.h>
#define ll long long
#define ls (o<<1)
#define rs (o<<1|1)
using namespace std;
const int N=500000+5,M=100+5,P=7,INF=1e9+7,inf=0x3f3f3f3f;
int n,m,ans;
template <class t>void rd(t &x){
x=0;int w=0;char ch=0;
while(!isdigit(ch)) w|=ch=='-',ch=getchar();
while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
x=w?-x:x;
}
int head[N],tot=0;
struct edge{int v,nxt;}e[N<<1];
void add(int u,int v){e[++tot]=(edge){v,head[u]},head[u]=tot;}
int idx=0,dfn[N],id[N],son[N],f[N],dep[N],sz[N],top[N];
void dfs1(int u,int fa){
dep[u]=dep[fa]+1,f[u]=fa,sz[u]=1;
for(int i=head[u],v,mxs=-1;i;i=e[i].nxt)
if((v=e[i].v)!=fa){
dfs1(v,u),sz[u]+=sz[v];
if(mxs<sz[v]) mxs=sz[v],son[u]=v;
}
}
void dfs2(int u,int topf){
top[u]=topf,dfn[u]=++idx,id[idx]=u;
if(!son[u]) return;
dfs2(son[u],topf);
for(int i=head[u],v;i;i=e[i].nxt)
if((v=e[i].v)!=f[u]&&v!=son[u]) dfs2(v,v);
}
struct node{int sum,tg;}t[N<<2];
void pup(int o){t[o].sum=t[ls].sum+t[rs].sum;}
void updnode(int o,int l,int r,int k){t[o].sum=(r-l+1)*k,t[o].tg=k;}
void pudw(int o,int l,int r){
if(t[o].tg==-1) return;
int mid=l+r>>1;
updnode(ls,l,mid,t[o].tg),updnode(rs,mid+1,r,t[o].tg);
t[o].tg=-1;
}
void mdf(int o,int l,int r,int x,int y,int k){
if(l>y||r<x) return;
if(x<=l&&r<=y){updnode(o,l,r,k);return;}
pudw(o,l,r);
int mid=l+r>>1;
mdf(ls,l,mid,x,y,k),mdf(rs,mid+1,r,x,y,k);
pup(o);
}
int LCA(int x,int y){
while(top[x]!=top[y]){
if(dep[top[x]]<dep[top[y]]) swap(x,y);
x=f[top[x]];
}
return dep[x]<dep[y]?x:y;
}
int jump(int x,int k){
while(dep[x]-dep[top[x]]+1<=k){
k-=(dep[x]-dep[top[x]]+1);
x=f[top[x]];
}
return id[dfn[x]-k];
}
int main(){
freopen("T2.txt","r",stdin);
// freopen("civilization.out","w",stdout);
rd(n),rd(m);
for(int i=1,u,v;i<n;++i) rd(u),rd(v),add(u,v),add(v,u);
dfs1(1,0),dfs2(1,1);
for(int i=1;i<=(n<<2);++i) t[i].tg=-1;
while(m--){
int tt,rt;rd(tt),rd(rt);
mdf(1,1,n,1,n,1);
for(int i=2,x,lca,L,R,mid;i<=tt;++i){
rd(x);lca=LCA(rt,x);
L=dep[rt]-dep[lca]-1,R=dep[x]-dep[lca]-1,mid=L+R+1>>1;
if(mid<=R){
mid=jump(x,mid);
mdf(1,1,idx,dfn[mid],dfn[mid]+sz[mid]-1,0);
}
else{
mid=jump(rt,L+R+1-mid);
mdf(1,1,idx,1,dfn[mid]-1,0),
mdf(1,1,idx,dfn[mid]+sz[mid],idx,0);
}
}
printf("%d\n",t[1].sum);
}
return 0;
}