个人感觉可能是最不需要脑子写的方法

不过也不太好调

就是用一个普通的线段树维护这个序列，但是对于线段树的每一个区间，再开一个动态开点的权值线段树，里面存储这个区间所有元素值

单点修改只会涉及到log棵权值线段树的单点修改（不用打lazy太棒了 log^2

查询区间内x的排名相当于查询区间内<x的数的个数+1，我们把区间分成log个外层线段树上的区间，然后在每个权值线段树上统计即可，复杂度log^2

查询排名为x的数比较麻烦，我们直接二分，复杂度log^3

查询前驱后继：由于线段树维护的区间，总区间是把这log个区间相加，所以我们再每个权值线段树查询下前驱后继再合并就行，前驱取max，后继取min

至于怎么查询，可以在线段树上二分

代码写的特别乱...

#include <cstdio>
#include <functional>
using namespace std;



int l[17000010], r[17000010], tree[17000010], tot;
int rt[200010], init[50010], fuck = 100000000;
int s[10000010], top;

int n, m;

void chenge(int &x, int cl, int cr, int pos, int val)
{
    if (x == 0)
    {
        if (top > 0) x = s[top--];
        else x = ++tot;
    }
    if (tot % 100000 == 0) fprintf(stderr, "(%d, %d)\n", tot, top);
    if (cl == cr) {tree[x] += val; if (tree[x] == 0) s[++top] = x, x = 0; return; }
    int mid = (cl + cr) / 2;
    if (pos > mid) chenge(r[x], mid + 1, cr, pos, val);
    else chenge(l[x], cl, mid, pos, val);
    tree[x] = tree[l[x]] + tree[r[x]];
    if (l[x] == 0 && r[x] == 0) { s[++top] = x, x = 0; }
}

int query(int x, int cl, int cr, int pos)
{
    if (x == 0 || cl == cr) return 0;
    int mid = (cl + cr) / 2;
    if (pos > mid)
        return tree[l[x]] + query(r[x], mid + 1, cr, pos);
    else return query(l[x], cl, mid, pos);
}

int qrange(int x, int cl, int cr, int L, int R)
{
    if (x == 0) return 0;
    if (R < cl || cr < L) return 0;
    if (L <= cl && cr <= R) return tree[x];
    int mid = (cl + cr) / 2;
    return qrange(l[x], cl, mid, L, R) + qrange(r[x], mid + 1, cr, L, R);
}

int getnumber(int x, int cl, int cr, int rank)
{
    if (cl == cr) { return cl; }
    int mid = (cl + cr) / 2;
    if (rank <= tree[l[x]]) return getnumber(l[x], cl, mid, rank);
    else return getnumber(r[x], mid + 1, cr, rank - tree[l[x]]);
}

int getnumber2(int x, int cl, int cr, int rank)
{
    if (cl == cr) { return cl; }
    int mid = (cl + cr) / 2;
    if (rank <= tree[r[x]]) return getnumber2(r[x], mid + 1, cr, rank);
    else return getnumber2(l[x], cl, mid, rank - tree[r[x]]);
}

int getprev(int rt, int pos)
{
    int tot = qrange(rt, 0, fuck, 0, pos - 1); // [1, pos - 1]内数的个数
    if (tot == 0) return -2147483647;
    return getnumber(rt, 0, fuck, tot);
}

int getnext(int rt, int pos)
{
    int tot = qrange(rt, 0, fuck, pos + 1, fuck);
    if (tot == 0) return 2147483647;
    return getnumber2(rt, 0, fuck, tot);
}

//---- 外面线段树

void build(int x, int l, int r)
{
    for (int i = l; i <= r; i++)
        chenge(rt[x], 0, fuck, init[i], 1);
    if (l == r) return;
    int mid = (l + r) / 2;
    build(x * 2, l, mid);
    build(x * 2 + 1, mid + 1, r);
}

int qrank(int x, int cl, int cr, int L, int R, int k)
{
    if (R < cl || cr < L) return 0;
    if (L <= cl && cr <= R) return query(rt[x], 0, fuck, k);
    int mid = (cl + cr) / 2;
    return qrank(x * 2, cl, mid, L, R, k) + qrank(x * 2 + 1, mid + 1, cr, L, R, k);
}

void change(int x, int cl, int cr, int pos, int val)
{
    chenge(rt[x], 0, fuck, init[pos], -1);
    chenge(rt[x], 0, fuck, val, 1);
    if (cl == cr) return;
    int mid = (cl + cr) / 2;
    if (pos > mid) change(x * 2 + 1, mid + 1, cr, pos, val);
    else change(x * 2, cl, mid, pos, val);
}

int qprev(int x, int cl, int cr, int L, int R, int k)
{
    if (R < cl || cr < L) return -2147483647;
    if (L <= cl && cr <= R)
    {
        int res = getprev(rt[x], k);
        return res;
    }
    int mid = (cl + cr) / 2;
    return max(qprev(x * 2, cl, mid, L, R, k), qprev(x * 2 + 1, mid + 1, cr, L, R, k));
}

int qnext(int x, int cl, int cr, int L, int R, int k)
{
    if (R < cl || cr < L) return 2147483647;
    if (L <= cl && cr <= R) return getnext(rt[x], k);
    int mid = (cl + cr) / 2;
    return min(qnext(x * 2, cl, mid, L, R, k), qnext(x * 2 + 1, mid + 1, cr, L, R, k));
}

int main()
{
    // printf("%f\n", (3 * sizeof(l) + sizeof(s) + sizeof(rt)) / 1000000.);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d", &init[i]);
    build(1, 1, n);
    for (int opd, l, r, k, i = 1; i <= m; i++)
    {
        scanf("%d%d%d", &opd, &l, &r);
        if (opd != 3) scanf("%d", &k);
        if (opd == 1)
        {
            printf("%d\n", qrank(1, 1, n, l, r, k) + 1);
        }
        if (opd == 2) // query值<k的最大一撮数中最小的一个
        {
            // for (int i = 0; i <= 10; i++) printf("query(%d) = %d\n", i, qrank(1, 1, n, l, r, i));
            int cl = 0, cr = 100000000;
            while (cl < cr)
            {
                int mid = (cl + cr + 1) / 2;
                if (qrank(1, 1, n, l, r, mid) < k) cl = mid;
                else cr = mid - 1;
            }
            // int ans = qrank(1, 1, n, l, r, cl);
            // cl = 0, cr = 100000000;
            // while (cl < cr)
            // {
                // int mid = (cl + cr) / 2;
                // if (qrank(1, 1, n, l, r, mid) >= ans) cr = mid;
                // else cl = mid + 1;
            // }
            printf("%d\n", cl);
        }
        if (opd == 3)
        {
            change(1, 1, n, l, r);
            init[l] = r;
        }
        if (opd == 4)
        {
            printf("%d\n", qprev(1, 1, n, l, r, k));
        }
        if (opd == 5)
        {
            printf("%d\n", qnext(1, 1, n, l, r, k));
        }
    }
    return 0;
}