D-query（莫队）

English

Vietnamese

Given a sequence of n numbers a₁, a₂, ..., a_n and a number of d-queries. A d-query is a pair (i, j) (1 ≤ i ≤ j ≤ n). For each d-query (i, j), you have to return the number of distinct elements in the subsequence a_i, a_i+1, ..., a_j.

Input

• Line 1: n (1 ≤ n ≤ 30000).
• Line 2: n numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁶).
• Line 3: q (1 ≤ q ≤ 200000), the number of d-queries.
• In the next q lines, each line contains 2 numbers i, j representing a d-query (1 ≤ i ≤ j ≤ n).

Output

• For each d-query (i, j), print the number of distinct elements in the subsequence a_i, a_i+1, ..., a_j in a single line.

Example

Input
5
1 1 2 1 3
3
1 5
2 4
3 5

Output
3
2
3 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <cmath>
#define N 30005
#define ll long long
using namespace std;

int n,m,k,L,R;
int arr[N],cnt[1000005];
int res;
int sum[200005];

struct Node{
    int l,r,b,ans;
    bool operator<(const Node&X)const{
        if(ans!=X.ans) return ans<X.ans;
        return r<X.r;
    }
}A[200005];

void add(int ee){
    if(cnt[arr[ee]]==0) res++;
    cnt[arr[ee]]++;
}

void del(int ee){
    cnt[arr[ee]]--;
    if(cnt[arr[ee]]==0) res--;
}


int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&arr[i]);
    scanf("%d",&m);
    int big=(int)sqrt(n);
    for(int i=1;i<=m;i++){
        scanf("%d%d",&A[i].l,&A[i].r);
        A[i].b=i;A[i].ans=(A[i].l-1)/big+1;
    }
    sort(A+1,A+1+m);
    L=1,R=0;
    for(int i=1;i<=m;i++){
        while(L<A[i].l) del(L++);
        while(L>A[i].l) add(--L);
        while(R<A[i].r) add(++R);
        while(R>A[i].r) del(R--);
        sum[A[i].b]=res;
    }
    for(int i=1;i<=m;i++) printf("%d
",sum[i]);
    return 0;
}