2021.3.21 死に逝く君、WAに芽吹く憎悪

T3

题意:给定长度为n排列a，每次可以交换相邻2数，代价为1。可以随时停止操作，代价为\(\sum_{i=1}^n[a_i!=i]\)。求最小的代价，输出操作位置

我们不难证明，调整法有着极高的正确率，且期望操作次数极低

因为我们只有当每次交换会影响終态代价时才会换

而出现错误仅当最优策略为一次冒泡(或局部冒泡)时，所以当\(n\rightarrow \infty\)趋近为0

#include<bits/stdc++.h>
using namespace std;
# define ll long long
# define read read1<ll>()
# define Type template<typename T>
Type T read1(){
	T t=0;char k;bool vis=0;
	do (k=getchar())=='-'&&(vis=1);while('0'>k||k>'9');
	while('0'<=k&&k<='9')t=(t<<3)+(t<<1)+(k^'0'),k=getchar();
	return vis?-t:t;
}
# define fre(k) freopen(k".in","r",stdin);freopen(k".out","w",stdout)
int s,a[200005],sta[200005];
int main(){//fre("pm");
	s=read;
	for(int i=1;i<=s;++i)
		a[i]=read;int x=0;
	while(1){
		for(int i=1;i<s;++i)
			if(a[i]==i+1||a[i+1]==i){
				sta[++*sta]=i;
				swap(a[i],a[i+1]);
			}if(x==*sta)break;x=*sta;
		for(int i=s;--i;)
			if(a[i]==i+1||a[i+1]==i){
				sta[++*sta]=i;
				swap(a[i],a[i+1]);
			}if(x==*sta)break;x=*sta;
	}
	printf("%d\n",*sta);
	for(int i=1;i<=*sta;++i)printf("%d\n",sta[i]);
	return 0;
}

T2

题意：\(\prod_{x_1,x_2...x_n\in [1,m]}lcm(x)^{\gcd(x)}\)

\(n\leq 10^8,m\leq 200000\)

\[\prod_{i=1}^m\prod_{a_i\leq m}lcm(a)^{gcd(a)[gcd(a)=i]}\\ =\prod_{i=1}^m\prod_{a_i\leq \frac{m}{i}}(i*lcm(a))^{i[gcd(a)=1]}\\ =\prod_{i=1}^m\prod_{a_i\leq \frac{m}{i}}(i*lcm(a))^{i\sum_{d|gcd(a)}\mu(d)}\\ =\prod_{i=1}^m\prod_{d=1}^{\frac{m}{i}}\prod_{a_i\leq \frac{m}{id}}(i*d*lcm(a))^{i\mu(d)}\\ =\prod_{T=1}^m\prod_{a_i\leq \frac{m}{T}}(T*lcm(a))^{\sum_{d|T}{T\over d}\mu(d)} \\ \]

\[=\prod_{T=1}^m\prod_{a_i\leq \frac{m}{T}}(T*lcm(a))^{\varphi(T)} \\ =\prod_{T=1}^m(T^{\lfloor\frac{m}{T}\rfloor^n}\prod_{a_i\leq \frac{m}{T}}lcm(a))^{\varphi(T)} \\ \]

#include<bits/stdc++.h>
using namespace std;
# define ll long long
# define read read1<ll>()
# define Type template<typename T>
Type T read1(){
	T t=0;char k;bool vis=0;
	do (k=getchar())=='-'&&(vis=1);while('0'>k||k>'9');
	while('0'<=k&&k<='9')t=(t<<3)+(t<<1)+(k^'0'),k=getchar();
	return vis?-t:t;
}
# define fre(k) freopen(k".in","r",stdin);freopen(k".out","w",stdout)
int s;
ll qkpow(ll n,int m,int mod){
	ll t=1;
	for(;m;m>>=1,n=n*n%mod)
		(m&1)&&(t=t*n%mod);
	return t;
}int pri[20005],phi[200005];
bool vis[200005];
void prep(const int N=200000){phi[1]=1;
	for(int i=2;i<=N;++i){
		if(!vis[i])pri[++pri[0]]=i,phi[i]=i-1;
		for(int j=1;j<=pri[0]&&pri[j]*i<=N;vis[i*pri[j]]=1,++j)
			if(!(i%pri[j]))phi[i*pri[j]]=phi[i]*pri[j];
			else phi[i*pri[j]]=phi[i]*(pri[j]-1);
	}
}int h[205],n;
# define mod 998244353
ll solve(int m){ll q=1;
	for(int i=1;pri[i]<=m;++i){
		int tot=0,v=1;
		while(1ll*pri[i]*v<=m)
			++tot,v*=pri[i];
		int sum=h[tot]=m/v,t=0;
		for(int j=tot;v/=pri[i],j--;sum+=h[j])
			h[j]=m/v-sum;sum=qkpow(h[0],n,mod-1);
		for(int j=1;j<=tot;++j){
			h[j]+=h[j-1];
			int tv=qkpow(h[j],n,mod-1)-sum+(mod-1);tv+1>=mod&&(tv-=mod-1);
			t=(t+1ll*j*tv)%(mod-1);
			sum=(sum+tv)%(mod-1);
		}q=q*qkpow(pri[i],t,mod)%mod;
	}
	return q;
}
int main(){
	fre("lg");
	prep();
	n=read;s=read;
	ll ans=1;
	for(int T=1;T<=s;++T){
		ll t=qkpow(qkpow(T,qkpow(s/T,n,mod-1),mod)*solve(s/T)%mod,phi[T],mod);
		ans=ans*t%mod;
	}cout<<ans;
	return 0;
}

T1

题意：给出序列n个数的值域区间，求此序列的期望逆序对数

题不是很难，考虑计算第\(j\)个数对第\(i\)个数产生的期望贡献

分类讨论

  1. \(l_i\leq l_j \leq r_i < r_j\)

\[(r_j-r_i)\rightarrow(r_j-l_j)\And(l_j-l_i)(r_j-l_j)\\ {{(2r_j-r_i-l_j)(r_i-l_j)\over 2}+l_j(r_j-l_j)-l_i(r_j-l_j)\over (r_i-l_i)(r_j-l_j)}\\ ={{(2r_j-l_j)(r_i-l_j)-r_i(r_i-l_j)\over 2}+l_j(r_j-l_j)-l_i(r_j-l_j)\over (r_i-l_i)(r_j-l_j)}\\ ={{2r_ir_j-(2r_j-l_j)l_j-r_i^2\over 2}+l_j(r_j-l_j)-l_i(r_j-l_j)\over (r_i-l_i)(r_j-l_j)}\\ \]

  1. \(l_j<l_i\leq r_j\leq r_i\)

\[0\rightarrow (r_j-l_i)\\ {(r_j-l_i)^2\over 2(r_j-l_j)(r_i-l_i)}\\ ={r_j^2+l_i^2-2l_ir_j\over 2(r_j-l_j)(r_i-l_i)}\\ ={r_j^2+(l_i^2-2l_ir_j)\over 2(r_j-l_j)(r_i-l_i)} \]

  1. \(l_i\leq l_j<r_j\leq r_i\)

\[0\rightarrow (r_j-l_j)|(l_j-l_i)(r_j-l_j)\\ {{(r_j-l_j)^2\over 2}+l_j(r_j-l_j)-l_i(r_j-l_j)\over (r_j-l_j)(r_i-l_i)} \]

  1. \(l_j<l_i<r_i<r_j\)

\[(r_j-r_i)\rightarrow (r_j-l_i)\\ {(2r_j-l_i-r_i)(r_i-l_i)\over 2(r_j-l_j)(r_i-l_i)}\\ ={2r_j(r_i-l_i)-(l_i+r_i)(r_i-l_i)\over 2(r_j-l_j)(r_i-l_i)} \]

  1. \(r_i< l_j\)

    直接查

下面这份代码因实现的问题可能会被卡时空，但是它是对的

#include<bits/stdc++.h>
using namespace std;
# define ll long long
# define read read1<ll>()
# define Type template<typename T>
Type T read1(){
	T t=0;char k;bool vis=0;
	do (k=getchar())=='-'&&(vis=1);while('0'>k||k>'9');
	while('0'<=k&&k<='9')t=(t<<3)+(t<<1)+(k^'0'),k=getchar();
	return vis?-t:t;
}
# define fre(k) freopen(k".in","r",stdin);freopen(k".out","w",stdout)
# define mod 998244353ll
# define inv2 499122177ll
struct A{
	int x,y;
	A(int _x=0,int _y=0):x(_x),y(_y){}
	A& operator +=(const A &b){return (x+=b.x)>=mod&&(x-=mod),(y+=b.y)>=mod&&(y-=mod),*this;}
	A& operator -=(const A &b){return (x-=b.x)<0&&(x+=mod),(y-=b.y)<0&&(y+=mod),*this;}
	A operator +(A b)const{return b+=*this;}
	A operator -(A b)const{return A(x,y)-=b;}
};
struct B{
	int x,y,z;
	B(int _x=0,int _y=0,int _z=0):x(_x),y(_y),z(_z){}
	B& operator +=(const B &b){return (x+=b.x)>=mod&&(x-=mod),(y+=b.y)>=mod&&(y-=mod),(z+=b.z)>=mod&&(z-=mod),*this;}
	B& operator -=(const B &b){return (x-=b.x)<0&&(x+=mod),(y-=b.y)<0&&(y+=mod),(z-=b.z)<0&&(z+=mod),*this;}
	B operator +(B b)const{return b+=*this;}
	B operator -(B b)const{return B(x,y,z)-=b;}
};
Type struct node{
	node *l,*r;
	T v;
	node():v(){l=r=NULL;}
};
node<B>*a[200005],*b[200005];
node<A>*d[200005],*c[200005],*g[200005];
int s;
Type void add(node<T>*& x,int w,T v,int l,int r){
	if(!x)x=new node<T>;
	x->v+=v;
	if(l==r)return;
	int mid=l+r>>1;
	if(w<=mid)add(x->l,w,v,l,mid);
	else add(x->r,w,v,mid+1,r);
}
Type T query(node<T>* x,int l,int r,int tl,int tr){
	if(r<l)return T();
	if(!x)return T();
	if(l==tl&&r==tr)return x->v;
	int mid=tl+tr>>1;
	if(r<=mid)return query(x->l,l,r,tl,mid);
	if(mid<l)return query(x->r,l,r,mid+1,tr);
	return query(x->l,l,mid,tl,mid)+query(x->r,mid+1,r,mid+1,tr);
}
int fu[200005];
Type void add(node<T> **a,int x,int y,T t){while(x<=*fu)add(a[x],y,t,1,*fu),x+=x&-x;}
Type T query(node<T> **a,int x,int l,int r){
	T t;
	while(x)t+=query(a[x],l,r,1,*fu),x&=x-1;
	return t;
}Type T query(node<T> **a,int xl,int xr,int l,int r){return query(a,xr,l,r)-query(a,xl-1,l,r);}
ll qkpow(ll n,ll m){
	ll t=1;
	for(;m;m>>=1){
		if(m&1)t=t*n%mod;
		n=n*n%mod;
	}
	return t;
}
pair<int,int>q[100005];
int main(){fre("rng");
	s=read;ll t=0,ans=0;B k(0,0,0);A v(0,0);
	for(int i=1;i<=s;++i)fu[++fu[0]]=read,fu[++fu[0]]=read,q[i]=make_pair(fu[fu[0]-1],fu[fu[0]]);
	sort(fu+1,fu+*fu+1);
	*fu=unique(fu+1,fu+*fu+1)-fu-1;
	for(int i=1;i<=s;++i){
		int l=q[i].first,r=q[i].second,xl=lower_bound(fu+1,fu+*fu+1,l)-fu,xr=lower_bound(fu+1,fu+*fu+1,r)-fu;
		k=query(a,xl,xr,xr+1,*fu);
		t=(t+k.x+1ll*r*k.y-1ll*l*k.z-inv2*r%mod*r%mod*query(d,xl,xr,xr+1,*fu).y)%mod;
		k=query(b,xl,xr,1,xl-1);
		t=(t+k.x+inv2*l%mod*l%mod*k.z-1ll*l*k.y)%mod;
		v=query(c,xl,xr,xl,xr);
		t=(t+v.x-1ll*l*v.y+mod)%mod;
		v=query(d,1,xl-1,xr+1,*fu);
		t=(t+1ll*(r-l)*v.x-inv2*(l+r)%mod*(r-l)%mod*v.y)%mod;
		ll tv=qkpow(r-l,mod-2);
		add(a,xl,xr,B((1ll*l*(r-l)+(mod-2*r+l)*l%mod*inv2)%mod*tv%mod,r*tv%mod,(r-l)*tv%mod));
		add(b,xr,xl,B(1ll*r*r%mod*tv%mod*inv2%mod,r*tv%mod,tv));
		add(c,xl,xr,A((inv2*(r-l)%mod*(r-l)+1ll*l*(r-l))%mod*tv%mod,(r-l)*tv%mod));
		add(d,xl,xr,A(r*tv%mod,tv));
		ans=(ans+t*tv)%mod;
		v=query(g,xr+1,*fu,xr+1,*fu);
		ans=(ans+v.x)%mod;
		add(g,xl,xr,A(1));t=0;
	}printf("%lld\n",(ans+mod)%mod);
	return 0;
}

虽然可以用差分改成\(n\log_2n\),但我这份打的是\(n\log_2^2n\)的树套树 ☕

考场上头铁了，所以就盯着这道题打。debug不出来，想打下道题，但是一想到已经打出来了，就不由得不爽了起来，所以我就收获了0分的好成绩😄

posted @ 2021-03-22 21:20  ファイナル  阅读(92)  评论(0)    收藏  举报