• 周二. 11月 29th, 2022

5G编程聚合网

5G时代下一个聚合的编程学习网

热门标签

21牛客多校第五场

admin

11月 28, 2021

A

圆方树 咕

B

显然开盒子的顺序应按照(w)升序,而(hint)若选择使用应该在一开始就使用

在使用(hint)的情况下,因为知道(01)的总数,每种情况应当在最后一段连续的(0/1)前终止

(100cdots 0)此类情况,在(0)处截止,其花费为(sum_i)即到(0)(w)的前缀和

而此类情况的概率为后(n-i)一样且第(i)个不同,(p=frac{2}{2^{n-i+1}}=frac{1}{2^{n-i}})

最后比较使用(hint)的答案与暴力全开哪个较小

#include<bits/stdc++.h>
#define inf 2139062143
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define MAXN 100100
#define MOD 998244353
#define Fill(a,x) memset(a,x,sizeof(a))
#define rep(i,s,t) for(int i=(s),i##end=(t);i<=i##end;++i)
#define dwn(i,s,t) for(int i=(s),i##end=(t);i>=i##end;--i)
#define ren for(int i=fst[x];i;i=nxt[i])
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-(b)+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n;db x,w[MAXN],ans;
int main()
{
	n=read();scanf("%lf",&x);
	rep(i,1,n) scanf("%lf",&w[i]);
	sort(w+1,w+n+1);rep(i,1,n) w[i]+=w[i-1];
	db pw=0.5;
	dwn(i,n-1,1) ans+=w[i]*pw,pw*=0.5;
	ans=min(ans+x,w[n]);
	printf("%.7lf
",ans);
}

C

由于调和级数,需要(O(1))查询对(i=t)的情况终止点在哪里

考虑先预处理每个位置的(W,L)前缀和以及每个(W,L)位置,这样可以找到从位置(x)开始后先到(i)球的位置

若此时(W,L)的差(>1)则显然胜负已分,否则需要处理一个从每个点开始何时破局,设为(nxt_x)

对位置(x)来说,若接下来是两球不同,则仍然是平局,(nxt_x=nxt_{x+2});否则在(x+1)时刻该局结束

预处理完毕后模拟

#include<bits/stdc++.h>
#define inf 2139062143
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define MAXN 1001001
#define MOD 998244353
#define Fill(a,x) memset(a,x,sizeof(a))
#define rep(i,s,t) for(int i=(s),i##end=(t);i<=i##end;++i)
#define dwn(i,s,t) for(int i=(s),i##end=(t);i>=i##end;--i)
#define ren for(int i=fst[x];i;i=nxt[i])
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-(b)+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,cnt[MAXN][2],pos[MAXN][2],nxt[MAXN],ans;
char s[MAXN];
int main()
{
	n=read();scanf("%s",s+1);
	rep(i,1,n)
	{
		rep(j,0,1) cnt[i][j]=cnt[i-1][j];
		if(s[i]=='L') pos[++cnt[i][0]][0]=i;
		else pos[++cnt[i][1]][1]=i;
	}
	nxt[n]=nxt[n+1]=n+1;
	dwn(i,n-1,1) nxt[i]= s[i]==s[i+1]?i+1:nxt[i+2];
	int p1,p2,p,c1,c2,pw=1,res;
	rep(t,1,n)
	{
		res=0;rep(i,0,n)
		{
			p1=p2=n+1;
			if(cnt[i][0]+t<=cnt[n][0]) p1=pos[cnt[i][0]+t][0];
			if(cnt[i][1]+t<=cnt[n][1]) p2=pos[cnt[i][1]+t][1];
			p=min(p1,p2);if(p==n+1) break;
			c1=cnt[p][0]-cnt[i][0],c2=cnt[p][1]-cnt[i][1];
			if(abs(c1-c2)==1) p=nxt[p];
			if(p==n+1) break;
			c1=cnt[p][0]-cnt[i][0],c2=cnt[p][1]-cnt[i][1];
			res+=(c1<c2);i=p-1;
		}
		ans=pls(ans,mul(res,pw)),pw=mul(pw,n+1);
	}
	printf("%d
",ans);
}

D

可以在每次找到一对相同(a_i=b_j)时统计答案,设(f[i][j])表示前面范围内以(i,j)结尾的公共子序列个数,(g[i][j])表示后面范围内符合条件的点对个数

(f[i][j])简单(dp)即可,(g[i][j])的计算过程中,每次找到一对符合答案的(a_i<b_j)需要(+sumlimits_{i=0}^{min(x,y)} inom{x}{i}inom{y}{i})这样的东西

不妨令(xle y)(sumlimits_{i=0}^{x} inom{x}{i}inom{y}{i}=sumlimits_{i=0}^{x} inom{x}{x-i}inom{y}{i}=sumlimits_{i=0}^{x} inom{x+y}{x})

预处理阶乘后直接(dp)

#include<bits/stdc++.h>
#define inf 2139062143
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define MAXN 100100
#define MOD 1000000007
#define Fill(a,x) memset(a,x,sizeof(a))
#define rep(i,s,t) for(int i=(s),i##end=(t);i<=i##end;++i)
#define dwn(i,s,t) for(int i=(s),i##end=(t);i>=i##end;--i)
#define ren for(int i=fst[x];i;i=nxt[i])
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-(b)+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,dp[5050][5050],f[5050][5050];
int fac[MAXN],ifac[MAXN],ans;
char a[5050],b[5050];
int qp(int x,int t,int res=1)
{
	for(;t;t>>=1,x=mul(x,x)) if(t&1) res=mul(res,x);
	return res;
}
#define inv(x) qp(x,MOD-2)
int C(int n,int m)
{
	return mul(fac[n],mul(ifac[n-m],ifac[m]));
}
int main()
{
	scanf("%s%s",a+1,b+1);n=strlen(a+1),m=strlen(b+1);
	ifac[0]=fac[0]=1;rep(i,1,1e4+10) fac[i]=mul(fac[i-1],i),ifac[i]=inv(fac[i]);
	dwn(i,n,1) dwn(j,m,1)
	{
		f[i][j]=mns(pls(f[i][j+1],f[i+1][j]),f[i+1][j+1]);
		if(a[i]<b[j]) f[i][j]=pls(f[i][j],C(m-j+n-i,m-j));
	}
	ans=f[1][1];
	rep(i,1,n) rep(j,1,m) 
	{
		if(a[i]==b[j])
		{
			  dp[i][j]=pls(dp[i-1][j-1],1);
			ans=pls(ans,mul(dp[i][j],f[i+1][j+1]));
		}
		dp[i][j]=pls(mns(pls(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1]),dp[i][j]);
	}
	printf("%d
",ans);
}

E

(dp)

F

对于在凸包内的点,固定(x),(y)后均为凸函数(

三分套三分然后暴力(check)

#include<bits/stdc++.h>
#define inf 2139062143
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define MAXN 100100
#define MOD 998244353
#define Fill(a,x) memset(a,x,sizeof(a))
#define rep(i,s,t) for(int i=(s),i##end=(t);i<=i##end;++i)
#define dwn(i,s,t) for(int i=(s),i##end=(t);i>=i##end;--i)
#define ren for(int i=fst[x];i;i=nxt[i])
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-(b)+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const ld eps=1e-9,pi=acos(-1);
int n;ld xx[110],yy[110],xl=10000,xr=-10000,yl=10000,yr=-10000;
struct vec{ld x,y;}v[110];
ld operator ^ (vec &a,vec &b){return a.x*b.x+a.y*b.y;}
ld operator & (vec &a,vec &b){return a.x*b.y-a.y*b.x;}
inline ld getLen(vec &a){return sqrt(a.x*a.x+a.y*a.y);}
inline ld getAngle(vec &a,vec &b)
{
	ld res=a^b;res/=getLen(a)*getLen(b);
	return acos(res);
}
inline int ok(ld x,ld y)
{
	rep(i,1,n) if((v[i]&v[i+1])<0-eps) return 0;
	return 1;
}
inline ld cheq(ld x,ld y)
{
	ld res=4;
	rep(i,1,n) v[i]={xx[i]-x,yy[i]-y};v[n+1]=v[1];
	if(!ok(x,y)) return -inf;
	rep(i,1,n) res=min(res,getAngle(v[i],v[i+1]));
	return res;
}
inline ld calc(ld x)
{
	ld l=yl,r=yr,ml,mr,al,ar;
	while(r-l>eps)
	{
		ml=(l*2+r)/3,mr=(l+r*2)/3;
		al=cheq(x,ml),ar=cheq(x,mr);
		if(al>ar) r=mr;else l=ml;
	}
	return cheq(x,(l+r)/2);
}
int main()
{
	n=read();
	rep(i,1,n)
		xx[i]=read(),yy[i]=read(),
		xl=min(xl,xx[i]),xr=max(xr,xx[i]),
		yl=min(yl,yy[i]),yr=max(yr,yy[i]);
	ld l=xl,r=xr,ml,mr,al,ar;
	while(r-l>eps)
	{
		ml=(l*2+r)/3,mr=(l+r*2)/3;
		al=calc(ml),ar=calc(mr);
		if(al>ar) r=mr;else l=ml;
	}
	printf("%.10Lf
",calc((l+r)/2)*180./pi);
}

G

(F_S)表示该位为(1)的质因数取满其余随意组合出的数和(a)的最小差,(G_S)表示和(b)的差,答案即为所有(F_{i}+G_{U-i})(min)

而搜索的时候可以搜索出: 该位为(1)的质因数取满其余不取满组合出的数和(a)的最小差,记为(f_S)

(F=f)(F_S)可以更新所有(F_{S-(1<<j)})(O(n2^n))计算出所有(F)(G)同理

最后统计答案

#include<bits/stdc++.h>
#define inf 2139062143
#define ll __int128
#define db double
#define ld long double
#define ull unsigned long long
#define MAXN 100100
#define MOD 998244353
#define Fill(a,x) memset(a,x,sizeof(a))
#define rep(i,s,t) for(int i=(s),i##end=(t);i<=i##end;++i)
#define dwn(i,s,t) for(int i=(s),i##end=(t);i>=i##end;--i)
#define ren for(int i=fst[x];i;i=nxt[i])
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-(b)+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void write(ll x)
{
	int st[40],tp=0;
	if(!x) {putchar('0');return ;}
	while(x) st[++tp]=x%10,x/=10;
	while(tp) putchar(st[tp--]+'0');
}
int n;ll w[20],tot[20],a,b;
ll f[1<<18],g[1<<18],ans;
void dfs(int x,int st,ll now)
{
	if(x==n+1) 
	{
		if(now>=a) f[st]=min(f[st],now-a);
		if(now>=b) g[st]=min(g[st],now-b);
		return ;
	}
	dfs(x+1,st,now);
	rep(i,1,tot[x]-1) {now*=w[x];dfs(x+1,st,now);}
	now*=w[x],st|=(1<<x-1);dfs(x+1,st,now);
}
int main()
{
	n=read();rep(i,1,n) w[i]=read(),tot[i]=read();
	a=read(),b=read(),ans=1;
	int mxs=(1<<n)-1;rep(i,1,33) ans=ans*10;
	rep(i,0,mxs) f[i]=g[i]=ans;dfs(1,0,1);
	dwn(i,mxs,0) rep(j,0,n-1) if(!((i>>j)&1))
		f[i]=min(f[i|(1<<j)],f[i]),g[i]=min(g[i|(1<<j)],g[i]);
	rep(i,0,mxs) ans=min(ans,f[i]+g[mxs^i]);
	write(ans);
}

H

考虑如下形式的矩阵,在任何情况下均满足条件:

[egin{pmatrix}
0&1&0&1&0&cdots\
0&1&0&1&0&cdots\
1&0&1&0&1&cdots\
1&0&1&0&1&cdots\
vdots&vdots&vdots&vdots&vdots&ddots
end{pmatrix}
]

#include<bits/stdc++.h>
#define inf 2139062143
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define MAXN 100100
#define MOD 998244353
#define Fill(a,x) memset(a,x,sizeof(a))
#define rep(i,s,t) for(int i=(s),i##end=(t);i<=i##end;++i)
#define dwn(i,s,t) for(int i=(s),i##end=(t);i>=i##end;--i)
#define ren for(int i=fst[x];i;i=nxt[i])
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-(b)+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m; 
int main()
{
	n=read(),m=read();
	rep(i,0,n-1) {rep(j,0,m-1) printf("%d",(i/2+j)&1);puts("");}
}

I

还挺裸的回滚莫队,用链表维护一下每个区间的左右端点,每次更新答案总复杂度(O(nsqrt{n}+k))

回滚莫队时可以把一个更改操作的指针和值一起放到(vector)里,写起来非常方便

#include<bits/stdc++.h>
#define inf 2139062143
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define MAXN 100100
#define MOD 998244353
#define Fill(a,x) memset(a,x,sizeof(a))
#define rep(i,s,t) for(int i=(s),i##end=(t);i<=i##end;++i)
#define dwn(i,s,t) for(int i=(s),i##end=(t);i>=i##end;--i)
#define ren for(int i=fst[x];i;i=nxt[i])
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-(b)+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void inc(int &x,int y){x=pls(x,y);}
int n,Q,sz,g[MAXN],bl[MAXN],ans[MAXN],cnt[MAXN],res,l[MAXN],r[MAXN];
vector<pair<int*,int> > opt;
struct Ask{int id,l,r,k;}q[MAXN];
bool operator < (Ask &a,Ask &b){return bl[a.l]==bl[b.l]?a.r<b.r:a.l<b.l;}
inline void mdf(int *x,int w){opt.pb({x,*x});*x=w;}
inline void add(int x)
{
	if(cnt[g[x]]) return ;x=g[x];mdf(&cnt[x],1);
	//cout<<"add: "<<x<<endl;
	int pre=l[x-1],suf=r[x+1];
	if(pre==-1) pre=x;if(suf==-1) suf=x;
	mdf(&r[pre],suf);mdf(&l[suf],pre);
	if(suf-pre+1>res) mdf(&res,suf-pre+1);
}
void del(int t)
{
	while(opt.size()>t)
		{*opt.back().fi=opt.back().se;opt.pop_back();}
}
int main()
{
	n=read(),Q=read(),sz=sqrt(n+5);Fill(l,0xff);Fill(r,0xff);
	rep(i,1,n) g[i]=read(),bl[i]=i/sz;
	rep(i,1,Q) q[i].l=read(),q[i].r=read(),q[i].k=read()-1,q[i].id=i;
	sort(q+1,q+Q+1);int ed,pos,tag,pw;
	rep(i,1,Q)
	{
		if(i==1||bl[q[i].l]!=bl[q[i-1].l])
			{ed=bl[q[i].l]*sz+sz,pos=ed+1;del(0);}
		rep(j,pos,q[i].r) add(j);if(q[i].r>=pos) pos=q[i].r+1;tag=opt.size();
		dwn(j,min(ed,q[i].r),q[i].l) add(j);
		pw=1;inc(ans[q[i].id],res);
		rep(j,1,q[i].k)
		{
			pw=mul(pw,n+1);add(q[i].l-j);add(q[i].r+j);
			inc(ans[q[i].id],mul(res,pw));
		}
		del(tag);
	}
	rep(i,1,Q) printf("%d
",ans[i]);
}

J

(KM)裸题

#include<bits/stdc++.h>
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define MAXN 350
#define MOD 998244353
#define Fill(a,x) memset(a,x,sizeof(a))
#define rep(i,s,t) for(int i=(s),i##end=(t);i<=i##end;++i)
#define dwn(i,s,t) for(int i=(s),i##end=(t);i>=i##end;--i)
#define ren for(int i=fst[x];i;i=nxt[i])
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-(b)+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const ll inf=213906214300000000LL;
namespace KM
{
	int n,vx[MAXN],vy[MAXN],linky[MAXN],pre[MAXN];
	ll slack[MAXN],dbx[MAXN],dby[MAXN],dis[MAXN][MAXN],ans;
	void bfs(int k)
	{
	    int py=0,px,yy=0;ll tmp;
	    linky[0]=k;Fill(slack,0x3f);Fill(pre,0);
	    do
	    {
	        px=linky[py],tmp=inf,vy[py]=1;
			rep(i,1,n) if(!vy[i])
	        {
				if(dbx[px]+dby[i]-dis[px][i]<slack[i])
	            	slack[i]=dbx[px]+dby[i]-dis[px][i],pre[i]=py;
	            if(slack[i]<tmp) tmp=slack[i],yy=i;
	        }
	        rep(i,0,n)
	            if(vy[i])
	                dbx[linky[i]]-=tmp,dby[i]+=tmp;
	            else slack[i]-=tmp;
	        py=yy;
	    }while(linky[py]);
	    while(py) linky[py]=linky[pre[py]],py=pre[py];
	}
	ll solve()
	{
	    Fill(linky,0);rep(i,1,n) {Fill(vy,0);bfs(i);}
	    ans=0;rep(i,1,n) ans+=dis[linky[i]][i];return ans;
	}
};
int n,x[MAXN],y[MAXN],z[MAXN],v[MAXN];
ll ans;
int main()
{
	n=read();KM::n=n;
	rep(i,1,n) x[i]=read(),y[i]=read(),z[i]=read(),v[i]=read();
	rep(i,1,n)
	{
		ans+=x[i]*x[i]+y[i]*y[i]+z[i]*z[i],KM::dbx[i]=-inf;
		rep(j,1,n)
			KM::dis[i][j]=-2*(i-1)*v[j]*z[j]-1LL*(i-1)*(i-1)*v[j]*v[j],
			KM::dbx[i]=max(KM::dbx[i],KM::dis[i][j]);
	}
	ans-=KM::solve();
	printf("%lld
",ans);
}

K

考虑对于每个左端点(l)找到一个(r)使得([l,r])符合条件,则([l,r+1],dots [l,n])这些区间均符合条件

容易发现当(l)右移时,(r)也一定右移,不可能左移;因此可以使用滑动窗口

对于区间内的最大值与最小值,可以用单调队列维护

#include<bits/stdc++.h>
#define inf 2139062143
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define MAXN 100100
#define MOD 998244353
#define Fill(a,x) memset(a,x,sizeof(a))
#define rep(i,s,t) for(int i=(s),i##end=(t);i<=i##end;++i)
#define dwn(i,s,t) for(int i=(s),i##end=(t);i>=i##end;--i)
#define ren for(int i=fst[x];i;i=nxt[i])
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-(b)+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,mn[MAXN],mx[MAXN],g[MAXN],h1,t1,h2,t2;
ll ans;
int main()
{
	n=read(),m=read();rep(i,1,n) g[i]=read();
	rep(t,1,m)
	{
		k=read();h1=h2=1,t1=t2=0;ans=0;int l=1;
		rep(i,1,n)
		{
			while(h1<=t1&&g[mx[t1]]<=g[i]) t1--;mx[++t1]=i;
			while(h2<=t2&&g[mn[t2]]>=g[i]) t2--;mn[++t2]=i;
			while(h1<=t1&&h2<=t2&&g[mx[h1]]-g[mn[h2]]>k)
			{
				ans+=n-i+1,l++;
				if(mx[h1]<l) h1++;
				if(mn[h2]<l) h2++;
			}
		}
		printf("%lld
",ans);
	}
}

发表回复

您的电子邮箱地址不会被公开。 必填项已用*标注