題解：解法一：莫隊

解法二：按區間左端點排序，讓區間內最左邊的貝殼對答案產生貢獻，樹狀數組維護，轉移對答案產生貢獻的貝殼位置

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;
const int maxn=1000009;

int n,m;
int a[maxn];
int nex[maxn];
int ans[maxn];

int tong[maxn];

inline int lowbit(int x){
	return x&(-x);
}
int c[maxn];
int add(int x,int val){
	while(x<=1000001){
		c[x]+=val;
		x+=lowbit(x);
	}
}
int query(int x){
	int ret=0;
	while(x){
		ret+=c[x];
		x-=lowbit(x);
	}
	return ret;
}

struct Section{
	int l,r,id;
}sec[maxn];
int cmp(const Section &t1,const Section &t2){
	if(t1.l==t2.l)return t1.r<t2.r;
	else return t1.l<t2.l;
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d",&a[i]);
		if(a[i]==0)a[i]=1000001;
	}
	for(int i=n;i>=1;--i){
		if(!tong[a[i]]){
			nex[i]=n+1;
		}else{
			nex[i]=tong[a[i]];
		}
		tong[a[i]]=i;
	}
	scanf("%d",&m);
	for(int i=1;i<=m;++i){
		scanf("%d%d",&sec[i].l,&sec[i].r);
		sec[i].id=i;
	}
	sort(sec+1,sec+1+m,cmp);
	
	memset(tong,0,sizeof(tong));
	for(int i=1;i<=n;++i){
		if(!tong[a[i]])add(i,1);
		tong[a[i]]=1;
	}
	
	int head=1;
	for(int i=1;i<=m;++i){
		while(head<sec[i].l){
			add(head,-1);
			add(nex[head],1);
			++head;
		}
		ans[sec[i].id]=query(sec[i].r)-query(sec[i].l-1);
	}
	for(int i=1;i<=m;++i)printf("%d\n",ans[i]);
	return 0;
}

BZOJ：1878: [SDOI2009]HH的項鏈