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+5;
int a[N],ans[N];
int t[N],tot;
map<int,int>mp;
struct Node
{
    int l,r;
    int num;
}tr[40*N];
void init()
{
    tot=0;
    mp.clear();
}
int build(int l,int r)
{
    int ii=tot++;
    tr[ii].num=0;
    if(l<r)
    {
        int mid=(l+r)>>1;
        tr[ii].l=build(l,mid);
        tr[ii].r=build(mid+1,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)>>1;
        if(x<=mid) tr[ii].l=update(tr[now].l,l,mid,x,y);
        else       tr[ii].r=update(tr[now].r,mid+1,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)>>1;
    int sum=0;
    if(mid>=