hdu 5919--Sequence II（主席树--求区间不同数个数+区间第k大）

题目链接

Problem Description

Mr. Frog has an integer sequence of length n, which can be denoted as a1,a2,⋯,an There are m queries.

In the i-th query, you are given two integers li and ri. Consider the subsequence ali,ali+1,ali+2,⋯,ari.

We can denote the positions(the positions according to the original sequence) where an integer appears first in this subsequence as p(i)1,p(i)2,⋯,p(i)ki (in ascending order, i.e.,p(i)1<p(i)2<⋯<p(i)ki).

Note that ki is the number of different integers in this subsequence. You should output p(i)⌈ki2⌉for the i-th query.

 

Input

In the first line of input, there is an integer T (T≤2) denoting the number of test cases.

Each test case starts with two integers n (n≤2×105) and m (m≤2×105). There are n integers in the next line, which indicate the integers in the sequence(i.e., a1,a2,⋯,an,0≤ai≤2×105).

There are two integers li and ri in the following m lines.

However, Mr. Frog thought that this problem was too young too simple so he became angry. He modified each query to l‘i,r‘i(1≤l‘i≤n,1≤r‘i≤n). As a result, the problem became more exciting.

We can denote the answers as ans1,ans2,⋯,ansm. Note that for each test case ans0=0.

You can get the correct input li,ri from what you read (we denote them as l‘i,r‘i)by the following formula:

li=min{(l‘i+ansi−1) mod n+1,(r‘i+ansi−1) mod n+1}

ri=max{(l‘i+ansi−1) mod n+1,(r‘i+ansi−1) mod n+1}

 

Output

You should output one single line for each test case.

For each test case, output one line “Case #x: p1,p2,⋯,pm”, where x is the case number (starting from 1) and p1,p2,⋯,pm is the answer.

 

Sample Input

2

5 2

3 3 1 5 4

2 2

4 4

5 2

2 5 2 1 2

2 3

2 4

 

Sample Output

Case #1: 3 3

Case #2: 3 1

 

Hint

 

 

 

题意：一个有n个数的序列，现在有m次询问，每次给一个区间（l，r），设区间中有k个不同的数，它们在区间中第一次出现的位置为p1,p2 ,……,pk 并且将它们排序p1<p2<……<pk,现在求p[（k+1）/2]的值？

 

思路：主席树记录当前数出现的位置,即每次在当前数出现的位置（下标）+1，对于序列数a[1]~a[n]建立主席树时，如果当前这个数之前出现过，那么在当前这个版本线段树上对前一次出现的位置-1，在当前位置+1，这样就可以求出区间不同数的个数。现在要求出区间第k大，那么可以从a[n]~a[1]建立线段树，然后直接求k大就行。

 

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
typedef long long LL;
const int N=2e5+;
int a[N],ans[N];
int t[N],tot;
map<int,int>mp;
struct Node
{
    int l,r;
    int num;
}tr[*N];
void init()
{
    tot=;
    mp.clear();
}
int build(int l,int r)
{
    int ii=tot++;
    tr[ii].num=;
    if(l<r)
    {
        int mid=(l+r)>>;
        tr[ii].l=build(l,mid);
        tr[ii].r=build(mid+,r);
    }
    return ii;
}
int update(int now,int l,int r,int x,int y)
{
    int ii=tot++;
    tr[ii].num=tr[now].num+y;
    tr[ii].l=tr[now].l;
    tr[ii].r=tr[now].r;
    if(l<r)
    {
        int mid=(l+r)>>;
        if(x<=mid) tr[ii].l=update(tr[now].l,l,mid,x,y);
        else       tr[ii].r=update(tr[now].r,mid+,r,x,y);
    }
    return ii;
}
int query(int now,int l,int r,int L,int R)
{
    if(L<=l&&r<=R) return tr[now].num;
    int mid=(l+r)>>;
    int sum=;
    if(mid>=L) sum+=query(tr[now].l,l  ,mid,L,R);
    if(mid<R)  sum+=query(tr[now].r,mid+,r,L,R);
    return sum;
}
int finds(int now,int l,int r,int k)
{
    if(l==r) return l;
    int mid=(l+r)>>;
    if(tr[tr[now].l].num>=k) return finds(tr[now].l,l,mid,k);
    else return finds(tr[now].r,mid+,r,k-tr[tr[now].l].num);
}
int main()
{
    int T,Case=; cin>>T;
    while(T--)
    {
        init();
        int n,m; scanf("%d%d",&n,&m);
        for(int i=;i<=n;i++) scanf("%d",&a[i]);
        t[n+]=build(,n);
        for(int i=n;i>=;i--)
        {
            if(mp.count(a[i]))
            {
                int tmp=update(t[i+],,n,mp[a[i]],-);
                t[i]=update(tmp,,n,i,);
            }
            else t[i]=update(t[i+],,n,i,);
            mp[a[i]]=i;
        }
        ans[]=;
        for(int i=;i<=m;i++)
        {
            int x,y; scanf("%d%d",&x,&y);
            int l=min((x+ans[i-])%n+,(y+ans[i-])%n+);
            int r=max((x+ans[i-])%n+,(y+ans[i-])%n+);
            int k=(query(t[l],,n,l,r)+)/;
            ans[i]=finds(t[l],,n,k);
        }
        printf("Case #%d:",Case++);
        for(int i=;i<=m;i++) printf(" %d",ans[i]);
        puts("");
    }
    return ;
}
/**
10 4
1 1 1 1 1 1 1 1 1 1
3 6
6 8
7 10
2 5
*/