D-query(莫队)

Given a sequence of n numbers a1, a2, ..., an 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 ai, ai+1, ..., aj.

Input

  • Line 1: n (1 ≤ n ≤ 30000).
  • Line 2: n numbers a1, a2, ..., an (1 ≤ ai ≤ 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 ai, ai+1, ..., aj in a single line.

     

Example

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

Output
3
2
3 

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <iomanip>
 6 #include <set>
 7 #include <map>
 8 #include <vector>
 9 #include <queue>
10 #include <cmath>
11 #define N 30005
12 #define ll long long
13 using namespace std;
14 
15 int n,m,k,L,R;
16 int arr[N],cnt[1000005];
17 int res;
18 int sum[200005];
19 
20 struct Node{
21     int l,r,b,ans;
22     bool operator<(const Node&X)const{
23         if(ans!=X.ans) return ans<X.ans;
24         return r<X.r;
25     }
26 }A[200005];
27 
28 void add(int ee){
29     if(cnt[arr[ee]]==0) res++;
30     cnt[arr[ee]]++;
31 }
32 
33 void del(int ee){
34     cnt[arr[ee]]--;
35     if(cnt[arr[ee]]==0) res--;
36 }
37 
38 
39 int main(){
40     scanf("%d",&n);
41     for(int i=1;i<=n;i++) scanf("%d",&arr[i]);
42     scanf("%d",&m);
43     int big=(int)sqrt(n);
44     for(int i=1;i<=m;i++){
45         scanf("%d%d",&A[i].l,&A[i].r);
46         A[i].b=i;A[i].ans=(A[i].l-1)/big+1;
47     }
48     sort(A+1,A+1+m);
49     L=1,R=0;
50     for(int i=1;i<=m;i++){
51         while(L<A[i].l) del(L++);
52         while(L>A[i].l) add(--L);
53         while(R<A[i].r) add(++R);
54         while(R>A[i].r) del(R--);
55         sum[A[i].b]=res;
56     }
57     for(int i=1;i<=m;i++) printf("%d\n",sum[i]);
58     return 0;
59 }
View Code

 

posted @ 2019-08-16 16:03  厂长在线养猪  Views(119)  Comments(0Edit  收藏  举报