可以先离散化，然后开线段树维护。我们对于栈的查找操作可以理解成找到插入的时间点，然后再找到对应的元素。
对于插入和删除操作，可以认为是在指定时间以后+1或者-1。对于查询操作，我们首先可以确定对应时间栈内元素个数，然后用线段树查询在当前时间之前最后一个小于这个个数的时间点t，t+1就是插入的时间。
整体复杂度O(n log(n))

#include <bits/stdc++.h>
using namespace std;
#define ls (rt << 1)
#define rs (rt << 1 | 1)
const int maxn = 2e5 + 10;
struct Node
{
    int lb, rb;
}Tree[maxn << 2];
struct Query 
{
    int op, t, v;
}ask[maxn];
int n, b[maxn], val[maxn];
void pushup(Node & x, const Node & l, const Node & r)
{
    if(l.rb <= r.lb)
    {
        x.lb = l.lb + r.lb - l.rb;
        x.rb = r.rb;
    }
    else
    {
        x.lb = l.lb;
        x.rb = l.rb + r.rb - r.lb;
    }
}
void update(int rt, int l, int r, int x, bool typ)
{
    if(l == r)
    {
        if(typ)
            Tree[rt].lb = 1;
        else
            Tree[rt].rb = 1;
        return;
    }
    int mid = l + r >> 1;
    if(x <= mid) update(ls, l, mid, x, typ);
    else update(rs, mid + 1, r, x, typ);
    pushup(Tree[rt], Tree[ls], Tree[rs]);
}
Node query(int rt, int l, int r, int a, int b)
{
    if(a <= l && r <= b) return Tree[rt];
    int mid = l + r >> 1;
    if(b <= mid) return query(ls, l, mid, a, b);
    if(a > mid) return query(rs, mid + 1, r, a, b);
    Node ret;
    pushup(ret, query(ls, l, mid, a, b), query(rs, mid + 1, r, a, b));
    return ret;
}
int solve(int rt, int l, int r, int a, int b, int x)
{
    if(l == r) return val[l];
    int mid = l + r >> 1;
    if(b <= mid) return solve(ls, l, mid, a, b, x);
    if(a > mid) return solve(rs, mid + 1, r, a, b, x);
    Node tmp = query(rs, mid + 1, r, mid + 1, b);
    if(tmp.rb >= x) return solve(rs, mid + 1, r, a, b, x);
    else return solve(ls, l, mid, a, b, x + tmp.lb - tmp.rb);
}
int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
    {
        scanf("%d%d", &ask[i].op, &ask[i].t);
        if(ask[i].op == 0) scanf("%d", &ask[i].v);
        b[i] = ask[i].t;
    }
    sort(b + 1, b + n + 1);
    for(int i = 1; i <= n; ++i) 
    {
        ask[i].t = lower_bound(b + 1, b + n + 1, ask[i].t) - b;
        if(ask[i].op == 0)
        {
            val[ask[i].t] = ask[i].v;
            update(1, 1, n, ask[i].t, 0);
        }
        else if(ask[i].op == 1)
            update(1, 1, n, ask[i].t, 1);
        else
            printf("%d\n", solve(1, 1, n, 1, ask[i].t, 1));
    }
    return 0;
}