题干：

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

解题报告：

 

AC代码：（树状数组，mp存位置，其实也可以用一个pre数组去存位置，(需要先都初始化成-1)，用时分别是170ms和150ms）

#include<bits/stdc++.h>

using namespace std;
map<int ,int> mp;
struct Node {
	int l,r;
	int id;
} node[200000 + 5];
int n,m,c[30000 + 5],ans[200000 + 5];
int a[30000 + 5];
int lowbit(int x) {
	return x&-x;
}
void update(int x,int val) {
	while(x<=n) {
		c[x] += val;
		x+=lowbit(x);
	}
}
int query(int x) {
	int res = 0;
	while(x>0) {
		res += c[x];
		x-=lowbit(x);
	}
	return res;
}
bool cmp(const Node & a,const Node & b) {
    return a.r < b.r;
}
int main()
{
	while(~scanf("%d",&n) ) {
		memset(c,0,sizeof c);
		mp.clear();
		for(int i = 1; i<=n; i++) {
			scanf("%d",&a[i]);
		}
		scanf("%d",&m);
		for(int i = 1; i<=m; i++) {
			scanf("%d%d",&node[i].l,&node[i].r);
			node[i].id = i;
		}
		int cur = 1;
		sort(node+1,node+m+1,cmp);
		for(int i = 1; i<=m; i++) {
			while(cur <= node[i].r) {
				if(mp[a[cur] ] == 0) update(cur,1);
				else {
					update(mp[a[cur] ] ,-1);
					update(cur,1);
				}
				mp[a[cur] ] = cur;
				cur++;
			}
			ans[node[i].id] = query(node[i].r) - query(node[i].l - 1);
		}
		for(int i = 1; i<=m; i++) {
            printf("%d\n",ans[i]);
        }
		
	}
	return 0 ;
}

AC代码2：

#include<bits/stdc++.h>

using namespace std;
const int MAX = 30000 + 5;
struct TREE {
	int l,r;
	int val;
} tree[MAX * 40];
int n,tot;
int pre[1000000 + 5],root[MAX],a[MAX];
int build(int l,int r) {
	int ne = ++tot;
	tree[ne].val = 0;
	tree[ne].l=tree[ne].r = 0;
	if(l == r) return ne;
	int m = (l+r)/2;
	tree[ne].l = build(l,m);
	tree[ne].r = build(m+1,r);
	return ne;
}
int update(int pos,int c,int val,int l,int r) {
	int ne = ++tot;
	tree[ne] = tree[c];
	tree[ne].val += val;
	if(l == r) return ne;
	int m = (l+r)/2;
	if(m >= pos) tree[ne].l = update(pos,tree[c].l,val,l,m);
	else tree[ne].r = update(pos,tree[c].r,val,m+1,r);
	return ne; 
	
}
int query(int pos,int c,int l,int r) {
	if(l == r) return tree[c].val;
	int m = (l+r)/2;
	if(m>=pos) return tree[tree[c].r].val + query(pos,tree[c].l,l,m);
	else return query(pos,tree[c].r,m+1,r);
}
int main()
{
	cin>>n;
	memset(pre,-1,sizeof pre);
	for(int i = 1; i<=n; i++) scanf("%d",a+i);
	root[0] = build(1,n);
	//构造主席树 
	for(int i = 1; i<=n; i++) {
		if(pre[a[i]] == -1) {
			root[i] = update(i,root[i-1],1,1,n);
		}
		else {
			int tmp = update(pre[a[i]],root[i-1],-1,1,n);
			root[i] = update(i,tmp,1,1,n);	
		}
		pre[a[i]]=i;
	}
	int q,x,y;
	cin>>q;
	while(q--) {
		scanf("%d%d",&x,&y);
		printf("%d\n",query(x,root[y],1,n));
	}
	return 0 ;
}

莫队算法：（还未做、。、）