D-query

Description
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

题目大意
在一个长度为n的数组中给定m次查询,每一次查询输入l和r，表示数组区间【l,r】。要求输出每一次查询时在要求的区间里不同数的种类。

代码

#include
#include
#include
#include
#include
typedef long long ll;
using namespace std;
int n,m;
struct node
{
    int l,r,id;
};
node a[200005];
int sz,sum[30005],ans,Ans[200005],num[1000005];
bool cmp(node a,node b)
{
    if(a.l/sz==b.l/sz)
    return a.rreturn a.lint main()
{
    int i,j;
    while(cin>>n)
    {
        memset(num,0,sizeof(num));
        sz=sqrt(n*1.0);
        for(i=1;i<=n;i++)
         scanf("%d",&sum[i]);  

        cin>>m; 
        for(i=1;i<=m;i++)
        {
            cin>>a[i].l>>a[i].r;
            a[i].id=i;
        }

        sort(a+1,a+1+m,cmp);
        int l=1,r=0;
        ans=0;
        for(i=1;i<=m;i++)
        {
            int id=a[i].id;
            if(a[i].l==a[i].r)
            {
                Ans[id]=1;
                continue;
            }
            if(rfor(j=r+1;j<=a[i].r;j++)
                {
                    if(num[sum[j]]==0)
                    {
                        ans++;
                    }
                    num[sum[j]]++;
                }
            }
            else
            {
                for(j=r;j>a[i].r;j--)
                {
                    num[sum[j]]--;
                    if(num[sum[j]]==0)
                    ans--;
                }
            }
            r=a[i].r;

            if(lfor(j=l;jif(num[sum[j]]==0)
                    ans--;
                }
            }
            else
            {
                for(j=l-1;j>=a[i].l;j--)
                {
                    if(num[sum[j]]==0)
                    ans++;
                    num[sum[j]]++;
                }
            }
            l=a[i].l;
            Ans[id]=ans;
        }
        for(i=1;i<=m;i++)
        printf("%d\n",Ans[i]);  
    }
    return 0;
}