【省選模擬】20/06/19

$A$

• 容易發現，答案即爲 $\frac{n!}{\prod size_i}$，考慮從根開始往下走，一定存在一個子樹是滿兒叉樹，將其係數預處理出來就可以 $O(\log n)$ 詢問，階乘需要打一個表

#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;
cs int Mod = 1e9 + 7;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int dec(int a, int b){ return a - b < 0 ? a - b + Mod : a - b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
void Add(int &a, int b){ a = add(a,b); }
void Dec(int &a, int b){ a = dec(a,b); }
void Mul(int &a, int b){ a = mul(a,b); }
int ksm(int a, int b){ int as=1; for(;b;b>>=1,Mul(a,a)) if(b&1) Mul(as,a); return as; }
int fc[1050] = {};
int ans[40] = {};
cs int S = 1e6;
int fac(int n){ 
	int t = n / S, ans = fc[t];
	for(int i = t * S + 1; i <= n; i++) Mul(ans,i);
	return ans; 
} int n; 
int main(){
	freopen("heap.in","r",stdin);
	freopen("heap.out","w",stdout);
	scanf("%d",&n);
	int d = 0, x = n; 
	while(x) ++d, x>>=1; --d;
	int Ans = fac(n);
	if(!(n & (n+1))) return mul(Ans,ans[d]),0;
	x = n; while(x > 1){
		Mul(Ans, ksm(x,Mod-2));
		int t = x - (1 << d);
		if(t >= (1<<(d-1))) Mul(Ans, d?ans[d-1]:1), x -= (1<<d);
		else Mul(Ans, d>1?ans[d-2]:1), x -= (1<<(d-1)); --d;
	} cout << Ans;
	return 0;
}

$B$

• 考慮一次詢問，若知道 $S_{l-1}$，就可以知道 $S_r$，$S$ 爲異或前綴和，那麼我們將 $l-1$ 與 $r$ 連邊，最終就是一棵最小生成樹，在 $trie$ 樹上貪心，考慮高位的兩個子樹一定只有一對連邊，啓發式合併在另一個裏面查就可以了

#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;
namespace IO{
	cs int Rlen=1<<22|1;
	inline char gc(){
		static char buf[Rlen],*p1,*p2;
		(p1==p2)&&(p2=(p1=buf)+fread(buf,1,Rlen,stdin));
		return p1==p2?EOF:*p1++;
	} int gi(){
		int x=0; char c=gc();
		while(!isdigit(c))c=gc();
		while(isdigit(c)) x=(((x<<2)+x)<<1)+(c^48),c=gc();
		return x;
	} 
} using namespace IO;
typedef long long ll;
cs int N = 1e5 + 50;
cs int INF = 2147483647;
int n, a[N], rt[N];
cs int M = N << 5;
int ch[M][2], nd, vl[M]; bool ed[M]; 
void ins(int x){
	int u=0; for(int i=30,c;~i;i--){
		c=x>>i&1; if(!ch[u][c])
		ch[u][c]=++nd; u=ch[u][c];
	} vl[u]=x; ed[u]=true;
}
int sz[M], ban, now; ll Ans;
int qry(int x){
	int u=0; for(int i=30,c;~i;i--){
		c=x>>i&1;if(!ch[u][c])c^=1;
		if(ban==u)c^=1;u=ch[u][c];
	} return vl[u];
}
void _cope(int u){
	if(ch[u][0])_cope(ch[u][0]);
	if(ch[u][1])_cope(ch[u][1]);
	if(ed[u])now=min(now,vl[u]^qry(vl[u]));
}
void cope(int u){
	if(ch[u][0])cope(ch[u][0]);
	if(ch[u][1])cope(ch[u][1]);
	sz[u]=sz[ch[u][0]]+sz[ch[u][1]]+1;
	if(ch[u][0]&&ch[u][1]){
		ban = u; now = INF;
		if(sz[ch[u][0]]<sz[ch[u][1]])_cope(ch[u][0]);
		else _cope(ch[u][1]); Ans+=(ll)now;
	} 
}
int main(){
	freopen("secret.in","r",stdin);
	freopen("secret.out","w",stdout);
	n=gi();
	for(int i=1; i<=n; i++) 
	a[i]=gi()^a[i-1];
	for(int i=0; i<=n; i++) ins(a[i]);
	cope(0); cout<<Ans; return 0;
}

$C$

• 枚舉 $B$ 的根，記 $dp_{i,j}$ 爲到 $i$ 拼出 $j$ 的子樹的方案數，子樹同構要除一個階乘的係數

#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;
namespace IO{
	cs int Rlen=1<<22|1;
	inline char gc(){
		static char buf[Rlen],*p1,*p2;
		(p1==p2)&&(p2=(p1=buf)+fread(buf,1,Rlen,stdin));
		return p1==p2?EOF:*p1++;
	} int gi(){
		int x=0; char c=gc();
		while(!isdigit(c))c=gc();
		while(isdigit(c)) x=(((x<<2)+x)<<1)+(c^48),c=gc();
		return x;
	} 
} using namespace IO;
typedef long long ll;
typedef unsigned long long ull;
cs int Mod = 1e9 + 7;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int dec(int a, int b){ return a - b < 0 ? a - b + Mod : a - b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
void Add(int &a, int b){ a = add(a,b); }
void Dec(int &a, int b){ a = dec(a,b); }
void Mul(int &a, int b){ a = mul(a,b); }
int ksm(int a, int b){ int as=1; for(;b;b>>=1,Mul(a,a)) if(b&1) Mul(as,a); return as; }
cs int N = 1 << 12 | 5, M = 20;
int n, m; vector<int> A[N], B[M], C[M];
ull h[M], pw[M<<1]; int sz[M];
bool cmp(int a, int b){ return h[a] < h[b]; }
int dp[N][M], coe[M], ifc[M];
void pre_dfs(int u, int f){
	h[u] = 0; sz[u] = 1; C[u].clear();
	for(int e=0,v;e<(int)B[u].size();e++)
	if((v=B[u][e])!=f) pre_dfs(v,u),C[u].pb(v);
	sort(C[u].begin(),C[u].end(),cmp);
	for(int e=0,v; e<(int)C[u].size(); e++){
		v=C[u][e];  sz[u] += sz[v];
		h[u] = h[u] * pw[sz[v]<<1] + h[v];
	} h[u] = h[u] * pw[sz[u]+sz[u]-1] + 1;
	coe[u] = 1;
	for(int e=0,v,t=0; e<(int)C[u].size(); e=t){
		v=C[u][e]; while(t<(int)C[u].size()
		&&h[v]==h[C[u][t]]) ++t;
		Mul(coe[u],ifc[t-e]); 
	} 
}
void dfs(int u, int fa){
	for(int e=0,v;e<(int)A[u].size();e++)
	if((v=A[u][e])!=fa) dfs(v,u);
	for(int i=1; i<=m; i++){
		static int ic[M]; int ct = 0;
		for(int j=1; j<=m; j++) ic[j]=-1;
		for(int e=0;e<(int)C[i].size();e++)
		ic[C[i][e]] = ct++;
		if(!ct){ dp[u][i] = 1; continue; }
		static int f[2][N]; int S = 1<<ct, now = 0, nxt = 1; 
		for(int j=0; j<S; j++) f[0][j] = 0; f[0][0] = 1;
		for(int e=0,v;e<(int)A[u].size();e++)
		if((v=A[u][e])!=fa){
			memset(f[nxt],0,sizeof(f[nxt]));
			for(int i=1,t; i<=m; i++)
			if(~(t=ic[i])){
				for(int j=1<<t;j<S;j=(j+1)|(1<<t))
				Add(f[nxt][j],mul(f[now][j^(1<<t)],dp[v][i]));
			} swap(now, nxt);
			for(int o=0; o<S; o++) Add(f[now][o],f[nxt][o]);
		} dp[u][i] = mul(coe[i],f[now][S-1]);
	} 
}
int main(){
	freopen("tree.in","r",stdin);
	freopen("tree.out","w",stdout);
	n=gi();
	for(int i=1,u,v;i<n;i++)
	u=gi(),v=gi(),A[u].pb(v),A[v].pb(u);
	m=gi();
	for(int i=1,u,v;i<m;i++)
	u=gi(),v=gi(),B[u].pb(v),B[v].pb(u);
	pw[0]=1; ifc[0]=1;
	for(int i=1; i<=m+m; i++)pw[i]=pw[i-1]*223;
	for(int i=1; i<=m; i++)ifc[i]=mul(ifc[i-1],ksm(i,Mod-2));
	int ans=0;
	map<ull,int> mp;
	for(int r=1;r<=m;r++){
		pre_dfs(r,0); 
		if(mp.count(h[r])) continue;
		mp[h[r]]=true;
		memset(dp,0,sizeof(dp));
		dfs(1,0);
		for(int i=1; i<=n; i++)
		Add(ans, dp[i][r]);
	} cout << ans; 
}