【P1972 [SDOI2009]HH的项链】莫队1000ms卡过

P1972
鉴于这种问题应该离线或者树状数组处理线段树处理 根据的原则是如果一个项链中要元素
那么我们一定是取最右边的那个
但是写了发莫队水过去了
当复习莫队了

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
//biubiubiuAK!zlsAK!NitrogensAK!zzqAK!
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 500005;
int cnt[1000005],Ans = 0,block;
int arr[MAX_N],ans[MAX_N];
int read(){
    int ret=0;char ch=getchar();bool f=1;
    for(;!isdigit(ch);ch=getchar()) f^=!(ch^'-');
    for(;isdigit(ch);ch=getchar()) ret=(ret<<1)+(ret<<3)+ch-48;
    return f?ret:-ret;
}
inline void write(int qw)
{
    if(qw<0){putchar('-'); qw = -qw;}
    if(qw>9)write(qw/10);
    putchar(qw%10+'0');
}
struct node
{
    int l,r,id;
}Q[MAX_N];
inline bool cmp(node a,node b)
{
    return(a.l/block)^(b.l/block)?a.l<b.l:(((a.l/block)&1)?a.r<b.r:a.r>b.r);
}
inline void add(int x)
{
    if(!cnt[arr[x]]) Ans++;
    cnt[arr[x]]++;
}
inline void del(int x)
{
    cnt[arr[x]]--;
    if(!cnt[arr[x]]) Ans--;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,m;
    n = read();for(register int i = 1;i<=n;++i) arr[i] = read();
    m = read();
    block = n/sqrt(m*2/3);
    for(register int i = 1;i<=m;++i) Q[i].l = read(),Q[i].r=  read(),Q[i].id = i;
    sort(Q+1,Q+1+m,cmp);
    int ql = 0,qr = 0;
    Ans = 0;
    for(register int i = 1;i<=m;++i)
    {
        while(qr<Q[i].r) add(++qr);
        while(ql>Q[i].l) add(--ql);
        while(qr>Q[i].r) del(qr--);
        while(ql<Q[i].l) del(ql++);
        ans[Q[i].id] = Ans;
    }
    for(register int i = 1;i<=m;++i)
        write(ans[i]),printf("\n");
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}