P4168-蒲公英-分块

蒲公英
暴力分块思想。分块的思想与莫队相同。它能将时间和空间复杂度均摊XD
belong表示所属区块，num维护区间颜色出现次数，maxx维护区间max值。查询时只需要比较两端的区块即可。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<utility>
#include<cstring>
#include<cctype>
using namespace std;
inline int read(){
	int x=0,w=0;char c=getchar();
	while(!isdigit(c))w|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<3)+(x<<1)+(c^48),c=getchar();
	return w?-x:x;
}
typedef pair<int,int> pii;

namespace star
{
	const int maxn=40010,block=1500,maxm=50;
	int n,m,a[maxn],cnt,cn=1;
	int belong[maxn],beg[maxm],end[maxm],num[maxm][maxm][maxn],maxx[maxm][maxm];
	int ed[maxn],time=0;
	int color[maxn],id[maxn];
	int con[maxn];
	pii p[maxn];
	inline void solve(){
		int l=0,r=0,ans=0;
		while(m--){
			time++;
			l=(read()+ans-1)%n+1,r=(read()+ans-1)%n+1;
			if(l>r)swap(l,r);
			ans=0;
			int mx=0,mxid=0;
			if(belong[l]==belong[r]){
				for(int i=l;i<=r;i++)
				{
					if(ed[id[i]]!=time)con[id[i]]=0,ed[id[i]]=time;
					con[id[i]]++;
					if(con[id[i]]>mx or (con[id[i]]==mx and id[i]<mxid))mxid=id[i],mx=con[mxid];
				}
				ans=color[mxid];
				printf("%d\n",ans);
			}else{
				int L=belong[l]+1,R=belong[r]-1;
				mxid=maxx[L][R];mx=num[L][R][mxid];
				for(int i=l;i<beg[L];i++)
				{
					if(ed[id[i]]!=time)con[id[i]]=0,ed[id[i]]=time;
					con[id[i]]++;
					if(con[id[i]]+num[L][R][id[i]]>mx or (con[id[i]]+num[L][R][id[i]]==mx and id[i]<mxid))mxid=id[i],mx=con[mxid]+num[L][R][id[i]];
				}
				for(int i=end[R]+1;i<=r;i++)
				{
					if(ed[id[i]]!=time)con[id[i]]=0,ed[id[i]]=time;
					con[id[i]]++;
					if(con[id[i]]+num[L][R][id[i]]>mx or (con[id[i]]+num[L][R][id[i]]==mx and id[i]<mxid))mxid=id[i],mx=con[mxid]+num[L][R][id[i]];
				}
				ans=color[mxid];
				printf("%d\n",ans);
			}
		}
	}
	
	inline void makeblock(){
		beg[1]=1;
		cn=1;
		for(int i=1;i<=n;i++){
			if(!(i%block)){
				end[cn]=i-1;
				cn++;
				beg[cn]=i;
			}
			belong[i]=cn;
		}
		if(n%block)cn++;
		end[cn]=n;
		for (int i = 1; i <= cn; i++)
			for (int j = i; j <= cn; j++)
			{
				int makk = 0;
				for (int k = beg[i]; k <= end[j]; k++)
					num[i][j][id[k]]++;
				for (int k = 1; k <= cnt; k++){
					if(makk<num[i][j][k]) {
						makk=num[i][j][k];
						maxx[i][j]=k;
					}
				}
			}
	}
	
	inline void cried()
	{
		n=read(),m=read();
		for(int i=1;i<=n;i++)
			p[i]=make_pair(read(),i);
		sort(p+1,p+n+1);
		for(int i=1;i<=n;i++){
			if(p[i].first!=p[i-1].first or i==1)cnt++;
			color[cnt]=p[i].first;
			id[p[i].second]=cnt;
		}
		makeblock();
		solve();
	}
}
int main()
{
	star::cried();
	return 0;
}