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 

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = 2e6 + ;
 int n, m, L, R, sz, a[maxn];
 int sum[maxn], ans, ANS[maxn];
 struct node {
     int l, r, id;
     node() {}
     node(int l, int r, int id): l(l), r(r), id(id) {}
     bool operator <(const node & a)const {
         if (l / sz == a.l / sz) return r < a.r;
         return l < a.l;
     }
 } qu[maxn];
 void add(int x) {
     if (sum[a[x]] == ) ans++;
     sum[a[x]]++;
 }
 void del(int x) {
     sum[a[x]]--;
     if (sum[a[x]] == ) ans--;
 }
 int main() {
     scanf("%d", &n);
     for (int i =  ; i <= n ; i++)
         scanf("%d", &a[i]);
     scanf("%d", &m);
     for (int i =  ; i <= m ; i++) {
         scanf("%d%d", &qu[i].l, &qu[i].r);
         qu[i].id = i;
     }
     sz = (int)sqrt(n);
     sort(qu + , qu + m + );
     L = , R = ;
     for (int i =  ; i <= m ; i++) {
         while(L > qu[i].l) add(--L);
         while(R < qu[i].r) add(++R);
         while(L < qu[i].l) del(L++);
         while(R > qu[i].r) del(R--);
         ANS[qu[i].id] = ans;
     }
     for (int i =  ; i <= m ; i++)
         printf("%d\n", ANS[i]);
     return ;
 }