19暑假主席树D（区间不重复数字个数模板

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
询问一个区间中数字第一次出现位置的中位数
主席树可以求出来一个区间中出现数字个数，所以先把数字个数求出来，再求由左边界一直到n范围中的第（k+1）/2小即可

#include<bits/stdc++.h>
#define N 200005
using namespace std;
typedef long long ll;
int T,n,m,cnt,pos,a[N],b[N],rt[N],f[N],ans[N];
struct Tree
{
    int ls,rs,sum;
}tr[N*40];
void Insert(int pre,int &node,int x,int l,int r)
{
    node = ++cnt;
    tr[node] = tr[pre];
    tr[node].sum += x;
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(pos <= mid) Insert(tr[pre].ls,tr[node].ls,x,l,mid);
    else Insert(tr[pre].rs,tr[node].rs,x,mid+1,r);
}
int Ask(int node,int x,int l,int r)
{
    if(l == r) return tr[node].sum;
    int mid = (l+r)>>1;
    if(x <= mid) return Ask(tr[node].ls,x,l,mid);
    else return tr[tr[node].ls].sum + Ask(tr[node].rs,x,mid+1,r);
}
int Query(int node,int k,int l,int r)
{
	if(l == r) return l;
	int mid = (l+r)>>1;
	if(tr[tr[node].ls].sum >= k) return Query(tr[node].ls,k,l,mid);
	else return Query(tr[node].rs,k-tr[tr[node].ls].sum,mid+1,r);
}
int main()
{
    scanf("%d",&T);
    for(int t = 1;t <= T;t++)
    {
        cnt = 0;
        memset(f,0,sizeof(f));
        scanf("%d%d",&n,&m);
        for(int i = 1;i <= n;i++) scanf("%d",&a[i]);
        tr[n+1].ls=tr[n+1].rs=tr[n+1].sum=0;
	    rt[n+1]=0;
        for(int i = n;i;i--)
        {
            pos = f[a[i]];
            if(pos)
            {
                Insert(rt[i+1],rt[i],-1,1,n);
                pos = i;
                Insert(rt[i],rt[i],1,1,n);
            }
            else
            {
                pos = i;
                Insert(rt[i+1],rt[i],1,1,n);
            }
            f[a[i]] = i;
        }
        int pre = 0;
        for(int i = 1;i <= m;i++)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            l = (l + pre) % n + 1,r = (r + pre) % n + 1;
            if(l > r) swap(l,r);
            int k = Ask(rt[l],r,1,n);
            pre = Query(rt[l],(k+1)/2,1,n);
            ans[i]=pre;
        }
        printf("Case #%d:",t);
			for(int i=1;i<=m;i++)
			printf(" %d",ans[i]);
		    printf("\n");
    }
}