bzoj4241: 历史研究

看到dalao们都捉了此题

第一眼：这tm不就是区间众数嘛

然后脑子有坑去学回滚莫队

没学会，大力搞分块我还能在线呢~~~~

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x;
}
int a[110000],lslen,ls[110000];
int block,st[110000];
int s[410][110000],c[110000],tim,ti[110000];
LL mx[410][410];

int main()
{
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    int n=read(),Q=read();
    lslen=0;
    for(int i=1;i<=n;i++)
        a[i]=read(), ls[++lslen]=a[i];
        
    sort(ls+1,ls+lslen+1);
    lslen=unique(ls+1,ls+lslen+1)-ls-1;
    for(int i=1;i<=n;i++)
        a[i]=lower_bound(ls+1,ls+lslen+1,a[i])-ls;
        
    block=int(sqrt(double(n+1)))+1;
    for(int i=1;i<=n;i++)st[i]=(i-1)/block+1;
    
    memset(s,0,sizeof(s));
    for(int i=1;i<=block;i++)
    {
        LL ans=0,id=i;
        for(int j=(i-1)*block+1;j<=n;j++)
        {
            s[i][a[j]]++;
            if( (LL(s[i][a[j]]))*(LL(ls[a[j]]))>ans )
                ans=(LL(s[i][a[j]]))*(LL(ls[a[j]]));
                
            if(j%block==0)mx[i][id]=ans,id++;
        }
    }
    
    int l,r;
    tim=0;memset(ti,0,sizeof(ti));
    for(int i=1;i<=Q;i++)
    {
        l=read();r=read();
        if(st[l]==st[r])
        {
            LL ans=0;tim++;
            for(int j=l;j<=r;j++)
            {
                if(ti[a[j]]!=tim)
                    ti[a[j]]=tim, c[a[j]]=0;
                c[a[j]]++;
                if( (LL(c[a[j]]))*(LL(ls[a[j]]))>ans )
                    ans=(LL(c[a[j]]))*(LL(ls[a[j]]));
            }
            printf("%lld\n",ans);
        }
        else
        {
            LL ans=mx[st[l]+1][st[r]-1];
            tim++;
            for(int j=l;j<=st[l]*block;j++)
            {
                if(ti[a[j]]!=tim)
                    ti[a[j]]=tim, c[a[j]]=0;
                c[a[j]]++;
                if( (LL(c[a[j]]+s[st[l]+1][a[j]]-s[st[r]][a[j]]))*(LL(ls[a[j]]))>ans )
                    ans=(LL(c[a[j]]+s[st[l]+1][a[j]]-s[st[r]][a[j]]))*(LL(ls[a[j]]));
            }
            for(int j=(st[r]-1)*block+1;j<=r;j++)
            {
                if(ti[a[j]]!=tim)
                    ti[a[j]]=tim, c[a[j]]=0;
                c[a[j]]++;
                if( (LL(c[a[j]]+s[st[l]+1][a[j]]-s[st[r]][a[j]]))*(LL(ls[a[j]]))>ans )
                    ans=(LL(c[a[j]]+s[st[l]+1][a[j]]-s[st[r]][a[j]]))*(LL(ls[a[j]]));
            }
            printf("%lld\n",ans);
        }
    }
    
    return 0;
}