SPOJ

D-query

Time Limit: 227MS

Memory Limit: 1572864KB

64bit IO Format: %lld & %llu

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Description

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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 

Hint

Added by:	Duc
Date:	2008-10-26
Time limit:	0.227s
Source limit:	50000B
Memory limit:	1536MB
Cluster:	Cube (Intel Pentium G860 3GHz)
Languages:	All except: ERL JS NODEJS PERL 6 VB.net
Resource:	© VNOI

给出若干个数，给出q次询问，每次询问有一个区间，问在这个区间内有多少个不同的数，输出其数量

利用主席树套模板即可

//主席树
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
/*
 * 给出一个序列，查询区间内有多少个不相同的数
 */
const int MAXN = 30010;
const int M = MAXN * 100;
int n,q,tot;
int a[MAXN];
int T[M],lson[M],rson[M],c[M];
int build(int l,int r)
{
    int root = tot++;
    c[root] = 0;
    if(l != r)
    {
        int mid = (l+r)>>1;
        lson[root] = build(l,mid);
        rson[root] = build(mid+1,r);
    }
    return root;
}
int update(int root,int pos,int val)
{
    int newroot = tot++, tmp = newroot;
    c[newroot] = c[root] + val;
    int l = 1, r = n;
    while(l < r)
    {
        int mid = (l+r)>>1;
        if(pos <= mid)
        {
            lson[newroot] = tot++; rson[newroot] = rson[root];
            newroot = lson[newroot]; root = lson[root];
            r = mid;
        }
        else
        {
            rson[newroot] = tot++; lson[newroot] = lson[root];
            newroot = rson[newroot]; root = rson[root];
            l = mid+1;
        }
        c[newroot] = c[root] + val;
    }
    return tmp;
}
int query(int root,int pos)
{
    int ret = 0;
    int l = 1, r = n;
    while(pos < r)
    {
        int mid = (l+r)>>1;
        if(pos <= mid)
        {
            r = mid;
            root = lson[root];
        }
        else
        {
            ret += c[lson[root]];
            root = rson[root];
            l = mid+1;
        }
    }
    return ret + c[root];
}

int main(){
    while(scanf("%d",&n) == 1)
    {
        tot = 0;
        for(int i = 1;i <= n;i++)
            scanf("%d",&a[i]);
        T[n+1] = build(1,n);
        map<int,int>mp;
        for(int i = n;i>= 1;i--)
        {
            if(mp.find(a[i]) == mp.end())
            {
                T[i] = update(T[i+1],i,1);
            }
            else
            {
                int tmp = update(T[i+1],mp[a[i]],-1);
                T[i] = update(tmp,i,1);
            }
            mp[a[i]] = i;
        }
        scanf("%d",&q);
        while(q--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            printf("%d
",query(T[l],r));
        }
    }
    return 0;
}