我太菜了，只能用线段树把这个题目给水过了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long

using namespace std;

const int N = 1e5 + 10;

char op;
int n, m;
int x, y, k;
int a[N];
struct Tree {
    int l, r;
    int sum, lazytag;
}tree[N * 4];

void pushdown(int cnt) {
    if (tree[cnt].lazytag) {
        tree[cnt << 1].lazytag += tree[cnt].lazytag;
        tree[cnt << 1 | 1].lazytag += tree[cnt].lazytag;
        tree[cnt << 1].sum += tree[cnt].lazytag * (tree[cnt << 1].r - tree[cnt << 1].l + 1);
        tree[cnt << 1 | 1].sum += tree[cnt].lazytag * (tree[cnt << 1 | 1].r - tree[cnt << 1 | 1].l + 1);
        tree[cnt].lazytag = 0;
    }
}

void build(int l, int r, int cnt) {
    tree[cnt].l = l, tree[cnt].r = r, tree[cnt].lazytag = 0;
    if (l == r) {
        tree[cnt].sum = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, cnt << 1), build(mid + 1, r, cnt << 1 | 1);
    tree[cnt].sum = tree[cnt << 1].sum + tree[cnt << 1 | 1].sum;
    return;
}

void modify(int l, int r, int k, int cnt) {
    if (tree[cnt].l >= l && tree[cnt].r <= r) {
        tree[cnt].sum += k * (tree[cnt].r - tree[cnt].l + 1);
        tree[cnt].lazytag += k;
        return;
    }
    pushdown(cnt);
    int mid = (tree[cnt].l + tree[cnt].r) >> 1;
    if (l <= mid) modify(l, r, k, cnt << 1);
    if (r > mid) modify(l, r, k, cnt << 1 | 1);
    tree[cnt].sum = tree[cnt << 1].sum + tree[cnt << 1 | 1].sum;
    return;
}

int query(int l, int r, int cnt) {
    if (tree[cnt].l >= l && tree[cnt].r <= r) return tree[cnt].sum;
    pushdown(cnt);
    int mid = (tree[cnt].l + tree[cnt].r) >> 1;
    int ans = 0;
    if (l <= mid) ans += query(l, r, cnt << 1);
    if (r > mid) ans += query(l, r, cnt << 1 | 1);
    return ans; 
}

signed main(void) {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    cin >> n >> m;

    for (int i = 1; i <= n; i ++ ) cin >> a[i];

    build(1, n, 1);

    while (m -- ) {
        cin >> op;
        if (op == 'Q') {
            cin >> x >> y;
            cout << query(x, y, 1) << '\n';
        }
        else {
            cin >> x >> y >> k;
            modify(x, y, k, 1);
        }
    }

    return 0;
}