D-query SPOJ - DQUERY（莫队）统计不同数的数量

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 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 2e6 + 10;
 5 int n, m, L, R, sz, a[maxn];
 6 int sum[maxn], ans, ANS[maxn];
 7 struct node {
 8     int l, r, id;
 9     node() {}
10     node(int l, int r, int id): l(l), r(r), id(id) {}
11     bool operator <(const node & a)const {
12         if (l / sz == a.l / sz) return r < a.r;
13         return l < a.l;
14     }
15 } qu[maxn];
16 void add(int x) {
17     if (sum[a[x]] == 0) ans++;
18     sum[a[x]]++;
19 }
20 void del(int x) {
21     sum[a[x]]--;
22     if (sum[a[x]] == 0) ans--;
23 }
24 int main() {
25     scanf("%d", &n);
26     for (int i = 1 ; i <= n ; i++)
27         scanf("%d", &a[i]);
28     scanf("%d", &m);
29     for (int i = 1 ; i <= m ; i++) {
30         scanf("%d%d", &qu[i].l, &qu[i].r);
31         qu[i].id = i;
32     }
33     sz = (int)sqrt(n);
34     sort(qu + 1, qu + m + 1);
35     L = 1, R = 0;
36     for (int i = 1 ; i <= m ; i++) {
37         while(L > qu[i].l) add(--L);
38         while(R < qu[i].r) add(++R);
39         while(L < qu[i].l) del(L++);
40         while(R > qu[i].r) del(R--);
41         ANS[qu[i].id] = ans;
42     }
43     for (int i = 1 ; i <= m ; i++)
44         printf("%d\n", ANS[i]);
45     return 0;
46 }