#541. 「LibreOJ NOIP Round #1」七曜圣贤

题目描述

https://loj.ac/problem/541

题解

一直在往数据随机的方向想

把<=A中的删除操作看成元素的话，那么就是求一个队列里的最小值，单调队列维护

>A的部分再维护一个未选的最小值，选了之后就和上面一样了，最后再取min

code

#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define mod 998244353
#define ll long long
#define file
using namespace std;

int d[2100001][2],d2[2100001],Q,A,B,C,D,n,i,j,k,l,h,t,ans,x,T,t1,t2;
bool bz1[2100001],bz2[2100001],Bz;
unsigned int seed;
char st[21],ch;
ll Ans;

unsigned int Random()
{
	seed=seed^(seed<<13);
	seed=seed^(seed>>17);
	seed=seed^(seed<<5);
	return seed;
}

int main()
{
	freopen("sage.in","r",stdin);
	#ifdef file
	freopen("sage.out","w",stdout);
	#endif
	
	scanf("%d",&Q);
	for (;Q;--Q)
	{
		scanf("%d%u%d%d%d%d",&n,&seed,&A,&B,&C,&D),T=A+1;
		memset(bz1,0,max(A,B)+2),memset(bz2,0,max(A,B)+2);
		if (A>=0)
		memset(bz1,1,A+1),memset(bz2,1,A+1);
		ans=A+1;
		
		Ans=t=0,h=1;t1=1,t2=0;
		fo(i,1,n)
		{
			if (!(Random()%C))
			x=-1;
			else
			x=Random()%B;
			
			Bz=0;
			if (!D)
			{
				if (x>=0)
				{
					if (!bz1[x])
					{
						bz1[x]=bz2[x]=1;
						while (bz1[T]) ++T;
						Bz=1;
					}
					else
					if (bz2[x])
					{
						bz2[x]=0;
						while (h<=t && d[t][0]>x) --t;
						++t,d[t][0]=x,d[t][1]=++t2,d2[t2]=x;
						Bz=1;
					}
					else
					if (h<=t)
					{
						if (d[h][1]==t1) ++h;
						bz2[d2[t1++]]=1;
						Bz=1;
					}
				}
				else
				if (h<=t)
				{
					if (d[h][1]==t1) ++h;
					bz2[d2[t1++]]=1;
					Bz=1;
				}
				if (h>t) ans=T; else ans=min(T,d[h][0]);
			}
			else
			if (x>=0)
			{
				if (!bz2[x])
				{
					bz2[x]=1;
					while (bz2[ans]) ++ans;
					Bz=1;
				}
			}
			
			if (Bz)
			Ans^=1ll*ans*(1ll*i*i+i*7)%mod;
//			cout<<ans<<endl;
		}
		printf("%lld
",Ans);
	}
	
	fclose(stdin);
	fclose(stdout);
	return 0;
}