ittle w and Discretization

主席樹維護Mex。

每個右端點 r 維護出一棵 在[1, r ] 區間中 其他所有的 值離這個 r 最近的的位置是多少。

然後詢問區間[L,R]的時候，從rt[R] 出發，然後如果左兒子的中所有出線位置的最小值 >= L, 則說明他們所有的點都出線在這個區間內了，然後往右走。

否則就說明左邊有點沒有出線過，需要往左走。

程式碼：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define
 
 LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int
 
,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N =3e5+ 100;
struct Node{
    int ls, rs, mn, num;
    Node(){ls = rs = mn = num = 0;}
}tr[N<<5];
int rt[N];
int tot;
int build(int l, int
 
 r){
    int x = ++tot;
    if(l == r) return x;
    int m = l+r >> 1;
    tr[x].ls = build(l,m);
    tr[x].rs = build(m+1,r);
    return x;
}
int Update(int L, int p, int lst, int l, int r){
    int x = ++tot;
    tr[x] = tr[lst];
    if(l == r){
        tr[x].mn = p;
        ++tr[x].num;
        return x;
    }
    int m = l+r >> 1;
    if(L <= m) tr[x].ls = Update(L, p, tr[lst].ls, l, m);
    else tr[x].rs = Update(L, p, tr[lst].rs, m+1, r);
    tr[x].mn = min(tr[tr[x].ls].mn, tr[tr[x].rs].mn);
    tr[x].num = tr[tr[x].ls].num + tr[tr[x].rs].num;
    return x;
}
int Query(int L, int x, int x2){
//    cout << L << ' ' << tr[x].mn << ' ' << tr[x].num - tr[x2].num << endl;
    if(tr[x].mn >= L) return tr[x].num - tr[x2].num;
    if(!tr[x].ls) return 0;
    int ret = 0;
    if(tr[tr[x].ls].mn >= L) ret = tr[tr[x].ls].num-tr[tr[x2].ls].num + Query(L,tr[x].rs,tr[x2].rs);
    else ret = Query(L,tr[x].ls,tr[x2].ls);
    return ret;
}
void Q(int p, int l, int r){
    cout << l << " " << r <<  " " << tr[p].mn << ' ' << tr[p].num << endl;
    if(l == r) return ;
    int m = l+r >> 1;
    Q(tr[p].ls, l, m);
    Q(tr[p].rs, m+1, r);
}
int a[N];
int main(){
    int n, m;
    scanf("%d", &n);
    rt[0] = build(1,n);
    for(int i = 1; i <= n; ++i){
        scanf("%d", &a[i]);
        if(a[i] <= n) rt[i] = Update(a[i], i, rt[i-1], 1, n);
        else rt[i] = rt[i-1];
    }
//    Q(rt[4],1,n);
    scanf("%d", &m);
    int L, R;
    for(int i = 1; i <= m; ++i){
        scanf("%d%d", &L, &R);
        printf("%d\n", (R-L+1)-Query(L, rt[R], rt[L-1]));
    }
    return 0;
}