D-query SPOJ

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 + 10;
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]] == 0) ans++;
    sum[a[x]]++;
}
void del(int x) {
    sum[a[x]]--;
    if (sum[a[x]] == 0) ans--;
}
int main() {
    scanf("%d", &n);
    for (int i = 1 ; i <= n ; i++)
        scanf("%d", &a[i]);
    scanf("%d", &m);
    for (int i = 1 ; i <= m ; i++) {
        scanf("%d%d", &qu[i].l, &qu[i].r);
        qu[i].id = i;
    }
    sz = (int)sqrt(n);
    sort(qu + 1, qu + m + 1);
    L = 1, R = 0;
    for (int i = 1 ; 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 = 1 ; i <= m ; i++)
        printf("%d
", ANS[i]);
    return 0;
}