個人感覺可能是最不需要腦子寫的方法

不過也不太好調

就是用一個普通的線段樹維護這個序列，但是對於線段樹的每一個區間，再開一個動態開點的權值線段樹，裏面存儲這個區間所有元素值

單點修改只會涉及到log棵權值線段樹的單點修改（不用打lazy太棒了 log^2

查詢區間內x的排名相當於查詢區間內<x的數的個數+1，我們把區間分成log個外層線段樹上的區間，然後在每個權值線段樹上統計即可，復雜度log^2

查詢排名為x的數比較麻煩，我們直接二分，復雜度log^3

查詢前驅後繼：由於線段樹維護的區間，總區間是把這log個區間相加，所以我們再每個權值線段樹查詢下前驅後繼再合並就行，前驅取max，後繼取min

至於怎麽查詢，可以在線段樹上二分

代碼寫的特別亂...

#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]內數的個數
    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);
}

//---- 外面線段樹

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的最大一撮數中最小的一個
        {
            // 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;
}
 

luogu3380 樹套樹之線段樹套線段樹