spoj 3267

离线+树状数组   给出N (1 <= N <= 30000) 个正整数 (<= 10 ^ 6) 和Q (1 <= Q <= 200000) 个询问，每次查询区间 [i, j] 内有多少个不同的数。

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<stdio.h>
#include<string.h>
#include<algorithm>
#define lowbit(x) ((x)&(-x))
using namespace std;
typedef struct query
{
    int x,y;
    int id;
    inline bool operator < (query t) const  
    {
        return y < t.y;
    }
}query;
query Q[200001]; int a[30001],pre[1000001]={0};
int ans[200001]; int cc[30010]={0};

void update(int a,int d)
{
    while(a<=30000)
    {
        cc[a]+=d;
        a+=lowbit(a);
    }
}

int getsum(int a)
{
    int sum=0;
    while(a>0)
    {
        sum+=cc[a];
        a-=lowbit(a);
    }
    return sum;
}

int main()
{
    int n,i,j,q,c,d,cur;
    scanf("%d",&n);
    memset(pre,0,sizeof(pre));
    memset(cc,0,sizeof(cc));
    for(i=1;i<=n;i++)  scanf("%d",a+i);
    scanf("%d",&q);
    for(i=1;i<=q;i++)
    {
        scanf("%d%d",&c,&d);
        Q[i].x=c; Q[i].y=d; Q[i].id=i;
    }
    sort(Q+1,Q+q+1);//按照询问的右端点排序
    for(i=1,cur=1;i<=q;i++)  // 在每个询问的右端点前面插入比它序号小的数
    {
        for(;cur<=Q[i].y;cur++)
        {
            update(pre[a[cur]]+1,1);  //在[pre[a[cur]]+1,cur]这一段区间加上1
            update(cur+1,-1);
            pre[a[cur]]=cur;//记录上一次出现a[cur]值的下标号
        }
        ans[Q[i].id]=getsum(Q[i].x);
    }
    for(j=1;j<=q;j++)
        printf("%d\n",ans[j]);
    return 0;
}