0分求调

#include<bits/stdc++.h>
#define lc x<<1
#define rc x<<1|1
#define ll long long
using namespace std;
struct node {
    ll val, l, r, lazy;
} tr[100005 << 2];
ll n, m, k, a[100005], op, x, y;
void pushUp(int x) {
    tr[x].val = tr[lc].val + tr[rc].val;
}
void pushDown(int x) {
    if (tr[x].lazy == 0) return;
    tr[lc].val += tr[x].lazy*(tr[x].r - tr[x].l + 1);
    tr[rc].val += tr[x].lazy*(tr[x].r - tr[x].l + 1);
    tr[lc].lazy += tr[x].lazy;
    tr[rc].lazy += tr[x].lazy;
    tr[x].lazy = 0;
}
void build(int x, int l, int r) {
    tr[x].l = l, tr[x].r = r;
    if (l == r) {
        tr[x].val = a[l];
        return;
    }
    int m = (l + r) >> 1;
    build(lc, l, m);
    build(rc, m + 1, r);
    pushUp(x);
}
ll quary(int x, int l, int r) {
    if (tr[x].l >= l && tr[x].r <= r)
        return tr[x].val;
    pushDown(x);
    ll sum = 0;
    int m = (tr[x].l + tr[x].r) >> 1;
    if (m >= l) sum += quary(lc, l, r);
    if (m < r) sum += quary(rc, l, r);
    return sum;
}
void update(int x, int l, int r, int k) {
    if (l <= tr[x].l && tr[x].r <= r) {
        tr[x].val += k*(tr[x].r - tr[x].l + 1);
        tr[x].lazy += k;
        return;
    }
    pushDown(x);
    int m = (tr[x].l + tr[x].r) >> 1;
    if (l <= m) update(lc, l, r, k);
    if (m < r) update(rc, l, r, k);
    pushUp(x);
}
int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    build(1, 1, n);
    for (int i = 1; i <= m; i++) {
        cin >> op;
        if (op == 1) {
            cin >> x >> y >> k;
            update(1, x, y, k);
        } else {
            cin >> x >> y;
            cout << quary(1, x, y) << endl;
        }
    }
    return 0;
}