SPOJ DQUERY

DQUERY - D-query

 

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 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct Query
{
    int l,r,id;
    bool operator < (const Query& rhs)const
    {
        return r<rhs.r;
    }
};
const int N=3e4+5;
int n,q,sum[N],last[1000005],ans[200005],a[N];
Query Q[200005];
void update(int pos,int ope)
{
    if(ope)
        for(int i=pos;i<=n;i+=i&(-i))
            sum[i]+=1;
    else
        for(int i=pos;i<=n;i+=i&(-i))
            sum[i]-=1;
}
int query(int pos)
{
    int res=0;
    for(int i=pos;i>0;i-=i&(-i))
        res+=sum[i];
    return res;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    scanf("%d",&q);
    for(int i=0;i<q;i++)
        scanf("%d%d",&Q[i].l,&Q[i].r),Q[i].id=i;
    sort(Q,Q+q);
    for(int i=1,k=0;i<=n;i++)
    {
        if(last[a[i]])update(last[a[i]],0);
        update(i,1);
        last[a[i]]=i;
        while(Q[k].r==i)ans[Q[k].id]=query(i)-query(Q[k].l-1),k++;
    }
    for(int i=0;i<q;i++)printf("%d
",ans[i]);
    return 0;
}