【luoguSP3267】--莫队，不同数字个数

题意翻译

给出一个长度为n 的数列，a1​​ a2​​ ,...an​ ，有q 个询问，每个询问给出数对(i,j)，需要你给出ai​​

ai+1​​ ，...，aj​​ 这一段中有多少不同的数字

题目描述

English VietnameseGiven a sequence of n numbers a 1  a 2 ..., 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 _{i} i​ , a i+1 _{i+1} i+1​ , ..., a j _{j} j​ .

输入格式

• Line 1: n (1 ≤ n ≤ 30000).
• Line 2: n numbers a 1, a 2, ..., a n _ n​ (1 ≤ a i ≤ 10 ^ 6 ).
• 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).

输出格式

• For each d-query (i, j), print the number of distinct elements in the subsequence a 1, a i+1 ​ , ..., a j in a single line.

输入输出样例

输入 #1

5
1 1 2 1 3
3
1 5
2 4
3 5

输出 #1

3
2
3
思路：    裸的莫队.维护数出现的次数即可
代码：

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<cmath>
#define N 1000000
struct node {
	int l,r,pos,id;
	bool operator < (const node &a)const
	{
		if(pos==a.pos)return r<a.r;
		return pos<a.pos;
	}
}e[N];
int n,m,ta[N],a[N],ans,Ans[N];
int l=1,r=0;
using namespace std;
void add(int x)
{
	ta[a[x]]++;
	if(ta[a[x]]==1)ans++;
}
void del(int x)
{
	ta[a[x]]--;
	if(ta[a[x]]==0)ans--;
}
int main()
{
	scanf("%d",&n);
	int len=sqrt(n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	scanf("%d",&m);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&e[i].l,&e[i].r);
		e[i].pos=(e[i].l-1)/len+1;
		e[i].id=i;
	}
	sort(e+1,e+m+1);
	for(int i=1;i<=m;i++)
	{
		while(l<e[i].l)del(l++);
		while(r>e[i].r)del(r--);
		while(l>e[i].l)add(--l);
		while(r<e[i].r)add(++r);
		Ans[e[i].id]=ans;
	}
	for(int i=1;i<=m;i++)printf("%d
",Ans[i]);
	return 0;
}