Comet OJ Contest #2

　　A：暴力，显然每两次至少翻一倍。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
ll n,s;
signed main()
{
	cin>>n;
	s=1;
	for (int i=2;;i++)
	{
		if (i==2) s++;
		else s+=s/2;
		if (s>n) {cout<<i;return 0;}
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　B：列出柿子发现是二次函数。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int l,r,L,R;
double ans,a,b,u;
double calc(double x){return (-x*x+b*x-l*(L+R)*0.5)/(r-l);}
signed main()
{
	l=read(),r=read(),L=read(),R=read();
	a=-1,b=(L+R)*0.5+l,u=-b/(2*a);
	double ans=0;
	if (l<=u&&u<=r) ans=calc(u);
	else if (u>r) ans=calc(r);
	printf("%.4f",max(0.0,ans));
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　C：考虑大小为i的点集有多大的概率是独立集，则要求其内部C(i,2)条边均被破坏，概率显然为(x/y)^C(i,2)，累加各大小点集贡献即可。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define P 998244353
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,x,y,ans,fac[N],inv[N];
int ksm(int a,ll k)
{
	int s=1;
	for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
	return s;
}
int Inv(int a){return ksm(a,P-2);}
int calc(int k)
{
	return 1ll*ksm(x,1ll*k*(k-1)/2)*Inv(ksm(y,1ll*k*(k-1)/2))%P;
}
int C(int n,int m){return 1ll*fac[n]*inv[m]%P*inv[n-m]%P;}
signed main()
{
	cin>>n>>x>>y;
	fac[0]=1;for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%P;
	inv[0]=inv[1]=1;for (int i=2;i<=n;i++) inv[i]=P-1ll*(P/i)*inv[P%i]%P;
	for (int i=2;i<=n;i++) inv[i]=1ll*inv[i]*inv[i-1]%P;
	for (int i=0;i<=n;i++) ans=(ans+1ll*C(n,i)*calc(i))%P;
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D：对于一个连通块的所有直径，其中点一定相同。考虑枚举中点，这个中点可以是某个点也可以是某条边，以其为树根。然后若要统计直径为i的点集数量，只要保证所选点深度均<=i/2，并且跨过树根有至少两个深度为i/2的点。随便怎么算。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define P 998244353
#define N 2010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,p[N],deep[N],cnt[N],tot[N],qwq[N],ctrb[N],pw[N],ans[N],t;
struct data{int to,nxt;
}edge[N<<1];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void dfs(int k,int from)
{
	cnt[deep[k]=deep[from]+1]++;
	for (int i=p[k];i;i=edge[i].nxt)
	if (edge[i].to!=from) dfs(edge[i].to,k);
}
void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int calc(int k,int *cnt){return 1ll*pw[cnt[k-1]]*(pw[cnt[k]-cnt[k-1]]-1)%P;}
signed main()
{
	n=read();
	for (int i=1;i<n;i++)
	{
		int x=read(),y=read();
		addedge(x,y),addedge(y,x);
	}
	pw[0]=1;for (int i=1;i<=n;i++) pw[i]=2ll*pw[i-1]%P;
	for (int i=1;i<=n;i++)
	{
		deep[i]=0;
		memset(tot,0,sizeof(tot));
		memset(ctrb,0,sizeof(ctrb));
		memset(qwq,0,sizeof(qwq));
		for (int j=p[i];j;j=edge[j].nxt)
		{
			memset(cnt,0,sizeof(cnt));
			dfs(edge[j].to,i);
			for (int k=1;k<=n/2;k++) cnt[k]+=cnt[k-1];
			for (int k=1;k<=n/2;k++) ctrb[k]=1ll*ctrb[k]*pw[cnt[k]]%P;
			for (int k=1;k<=n/2;k++) inc(ctrb[k],1ll*qwq[k]*calc(k,cnt)%P);
			for (int k=1;k<=n/2;k++) qwq[k]=(1ll*qwq[k]*pw[cnt[k-1]]+1ll*pw[tot[k-1]]*calc(k,cnt))%P;
			for (int k=1;k<=n/2;k++) tot[k]+=cnt[k];
		}
		for (int j=1;j<=n/2;j++) inc(ans[j*2],2ll*ctrb[j]%P);
	}
	for (int i=1;i<=n;i++)
		for (int j=p[i];j;j=edge[j].nxt)
		if (j&1)
		{
			int x=i,y=edge[j].to;
			deep[y]=-1;memset(cnt,0,sizeof(cnt));
			dfs(x,y);
			for (int k=1;k<=n/2;k++) cnt[k]+=cnt[k-1];
			for (int k=0;k<=n/2;k++) tot[k]=cnt[k];
			deep[x]=-1;memset(cnt,0,sizeof(cnt));
			dfs(y,x);
			for (int k=1;k<=n/2;k++) cnt[k]+=cnt[k-1];
			for (int k=0;k<=n/2;k++) inc(ans[k*2+1],1ll*calc(k,cnt)*calc(k,tot)%P);
		}
	for (int i=1;i<n;i++) cout<<ans[i]<<endl;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　E：先考虑树，显然有设f[i]为i最终醒来的概率，则初始f[i]=p_i，依次合并子树有f[i]=f[i]+(1-f[i])*f[son]*s_son。dp完后即转化为若干个环的问题。环上容易想到类似的做法，求出f[i]为i被自身或子树内点唤醒的概率后，有f[i]=f[i]+(1-f[i])*g[from[i]]*s_from[i]，即为答案。注意这里的g[i]指to[i]不被自身或子树内点唤醒的前提下i醒来的概率，而不能直接用f[i]代替做高斯消元。不考虑复杂度的话，这个g[i]可以通过枚举是谁唤醒他的求出。观察一下式子容易发现加加减减乘乘除除就可以递推出所有g[i]了。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define P 998244353
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,a[N],b[N],p[N],to[N],from[N],f[N],g[N],ans[N],degree[N],q[N],t,cnt;
bool flag[N];
struct data{int to,nxt;
}edge[N];
int ksm(int a,int k)
{
	assert(k>=0);
	int s=1;
	for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
	return s;
}
int inv(int a){return ksm(a,P-2);}
void dfs(int k)
{
	f[k]=a[k];
	for (int i=p[k];i;i=edge[i].nxt)
	if (!flag[edge[i].to])
	{
		dfs(edge[i].to);
		f[k]=(f[k]+1ll*(P+1-f[k])*f[edge[i].to]%P*b[edge[i].to])%P;
	}
	ans[k]=f[k];
}
void topsort()
{
	int head=0,tail=0;
	for (int i=1;i<=n;i++) degree[to[i]]++;
	for (int i=1;i<=n;i++) if (!degree[i]) q[++tail]=i;
	while (head<tail)
	{
		int x=q[++head];
		for (int i=p[x];i;i=edge[i].nxt)
		{
			degree[edge[i].to]--;
			if (!degree[edge[i].to]) q[++tail]=edge[i].to;
		}
	}
	for (int i=1;i<=n;i++) if (degree[i]) flag[i]=1;
}
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
signed main()
{
	n=read();
	for (int i=1;i<=n;i++)
	{
		int x=read(),y=read();
		a[i]=1ll*x*inv(y)%P;
	}
	for (int i=1;i<=n;i++)
	{
		to[i]=read();
		addedge(i,to[i]);
	}
	for (int i=1;i<=n;i++)
	{
		int x=read(),y=read();
		b[i]=1ll*x*inv(y)%P;
	}
	topsort();
	memset(p,0,sizeof(p));t=0;
	for (int i=1;i<=n;i++) addedge(to[i],i);
	for (int i=1;i<=n;i++) if (flag[i]) from[to[i]]=i;
	for (int i=1;i<=n;i++) if (flag[i]) dfs(i);
	for (int i=1;i<=n;i++)
	if (flag[i])
	{
		int x=f[i],s=f[i];
		for (int j=from[i];j!=to[i];j=from[j])
		{
			x=1ll*x*b[j]%P*(P+1-f[to[j]])%P*f[j]%P*inv(f[to[j]])%P;
			s=(s+x)%P;
		}
		g[i]=s;
		for (int j=to[i];j!=i;j=to[j])
		{
			s=(s-x+P)%P;
			s=1ll*s*b[from[j]]%P*(P+1-f[j])%P;
			x=1ll*x*b[from[j]]%P*(P+1-f[j])%P;
			x=1ll*x*inv(b[to[j]])%P*inv(P+1-f[to[to[j]]])%P;
			x=1ll*x*inv(f[to[j]])%P;
			x=1ll*x*(f[to[to[j]]])%P;
			s=(s+f[j])%P;
			g[j]=s;
		}
		for (int j=i;flag[j];j=to[j])
		{
			flag[j]=0;
			ans[j]=(f[j]+1ll*(P+1-f[j])*g[from[j]]%P*b[from[j]])%P;
		}
	}
	for (int i=1;i<=n;i++) printf("%d ",ans[i]);
	return 0;
	//NOTICE LONG LONG!!!!!
}