（原创）D-query SPOJ - DQUERY（莫队）统计不同数的数量

A - D-query

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 
解题思路：这道题就是给你n个数，q次查询，查询l到r区间有多少个不同的数字；此题用莫队算法；
代码如下：

 #include<iostream>
 #include<stdio.h>
 #include<algorithm>
 #include<cmath>
 using namespace std;

 const int maxn = ;
 int n ;
 int m ;
 int a[maxn];
 int ans[maxn];
 int vis[];
 int block[maxn];
 int blocksize ;
 int count1 = ;
 struct query{
     int l ;
     int r ;
     int id ;
 }q[maxn];
 bool cmp(query a ,query b)
 {
     if(block[a.l]==block[b.l])
         return a.r<b.r;
     return block[a.l]<block[b.l];
 }
 void add(int num)
 {
     if(vis[a[num]]==)
         count1++;
     vis[a[num]]++;
 }
 void remove(int num){
     if(vis[a[num]]==)
         count1--;
     vis[a[num]]--;
 }
 void solve()
 {
     int r = ;
     int l = ;
     for(int i =  ; i < m;i++)
     {
         while(q[i].r > r)
         {
             r++;
             add(r);
         }
         while(q[i].r<r)
         {
             remove(r);
             r--;
         }
         while(q[i].l>l)
         {
             remove(l);
             l++;
         }
         while(q[i].l<l)
         {
             l--;
             add(l);
         }
         ans[q[i].id] = count1;
     }
 }
 int main()
 {
     scanf("%d",&n);
     blocksize = sqrt(n);
     for(int i =  ; i <= n ; i++)
     {
         scanf("%d",&a[i]);
         block[i] = (i-)/blocksize + ;
     }
     scanf("%d",&m);
     for(int i = ; i < m;i++)
     {
         scanf("%d%d",&q[i].l,&q[i].r);
         q[i].id = i;
     }
     sort(q,q+m,cmp);
     solve();
     for(int i =  ; i < m ;i++ )
     {
         printf("%d\n",ans[i]);
     }
     return ;
 }