本题就是一个简单的线段树模版，我们只需要维护一下区间最大值，区间最小值和区间和即可。

std食用警告：为了偷懒，std中的query函数使用了相当多的三目运算符，使代码十分丑陋。

#include <iostream>
using namespace std;
typedef long long ll;
const int N = 100010;
const ll inf = 0x3f3f3f3f3f3f3f3f;

ll n, m;
ll a[N];

struct segment_tree {
    ll l, r;
    ll sum, maxn, minn;
    ll laz;
    segment_tree *ls, *rs;
    
    void push_up() {
        sum = ls->sum + rs->sum;
        maxn = max(ls->maxn, rs->maxn);
        minn = min(ls->minn, rs->minn);
    }
    
    void push_down() {
        if (laz) {
            ls->laz += laz;
            rs->laz += laz;
            ls->sum += (ls->r - ls->l + 1) * laz;
            rs->sum += (rs->r - rs->l + 1) * laz;
            ls->maxn += laz;
            rs->maxn += laz;
            ls->minn += laz;
            rs->minn += laz;
        }
    }
    
    void update(ll L, ll R, ll k) {
        if (L <= l && R >= r) {
            laz += k;
            sum += (r - l + 1) * k;
            maxn += k;
            minn += k;
            return; 
        }
        push_down();
        if (L <= ls->r) ls->update(L, R, k);
        if (R >= rs->l) rs->update(L, R, k);
        push_up();
    }
    
    ll query(ll L, ll R, ll op) {
        if (L <= l && R >= r) return (op != 4 ? (op == 1 ? maxn : minn) : sum);
        push_down();
        ll res = (op == 1 || op == 4 ? 0 : inf);
        if (L <= ls->r) (op == 4 ? res += ls->query(L, R, op) : res = (op == 1 ? max(res, ls->query(L, R, op)) : min(res, ls->query(L, R, op))));
        if (R >= rs->l) (op == 4 ? res += rs->query(L, R, op) : res = (op == 1 ? max(res, rs->query(L, R, op)) : min(res, rs->query(L, R, op))));
        return res;
    }
} tree[N << 1], *pos = tree;

segment_tree *build(ll l, ll r) {
    auto tmp = pos++;
    tmp->l = l, tmp->r = r;
    tmp->minn = inf;
    if (l != r) {
        ll mid = (l + r) >> 1;
        tmp->ls = build(l, mid);
        tmp->rs = build(mid + 1, r);
        tmp->push_up();
    } else tmp->sum = tmp->maxn = tmp->minn = a[l];
    return tmp;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    auto rt = build(1, n);
    while (m--) {
        ll op, l, r;
        cin >> op >> l >> r;
        if (op == 1 || op == 2) cout << rt->query(l, r, op) << '\n';
        else if (op == 3) {
            ll k = rt->query(l, r, 1) - rt->query(l, r, 2);
            rt->update(l, r, k);
        } else cout << rt->query(l, r, op) / (r - l + 1) << '\n';
    }
    return 0;
}