字典树

重点
1. 区间修改用懒标记
2. 在每次分裂之前，都需要进行pushdown
3. pushdown操作，每次将本层都add向下传递，下一层的add和sum都增加，但是最开始从上向下传递终止时，本层的sum时已经累加过的，因为sum是属性需要提前维护

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010;
typedef long long ll;
int a[N];
int n, m;
struct node {
    int l, r;
    ll sum, add;
} tr[N * 4];

void pushup(int u) {
    tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}

void pushdown(int u) {
    auto &l = tr[u << 1], &r = tr[u << 1 | 1];
    l.add += tr[u].add, l.sum += tr[u].add * (l.r - l.l + 1);
    r.add += tr[u].add, r.sum += tr[u].add * (r.r - r.l + 1);
    tr[u].add = 0;
}

void build(int u, int l, int r) {
    if (l == r) tr[u] = {l, r, a[l], 0};
    else {
        tr[u] = {l, r};
        int mid = l + r >> 1;
        build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }
}

void modify(int u, int l, int r, int d) {
    if (tr[u].l >= l && tr[u].r <= r) tr[u].add += d, tr[u].sum += (ll)(tr[u].r - tr[u].l + 1) * d;
    else {
        pushdown(u);  //这里不可以省略，每次分裂之前都要向下传递
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) modify(u << 1, l, r, d);
        if (r > mid) modify(u << 1 | 1, l, r, d);
        pushup(u);
    }
}

ll query(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) return tr[u].sum;
    else {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        ll sum = 0;
        if (l <= mid) sum += query(u << 1, l, r);
        if (r > mid) sum += query(u << 1 | 1, l, r);
        //pushup(u);
        return sum;
    }
}

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);

    build(1, 1, n);

    char s[2];
    int l, r, d;
    while (m--) {
        scanf("%s%d%d", s, &l, &r);
        if (*s == 'C') {
            scanf("%d", &d);
            modify(1, l, r, d);
        } else {
            printf("%lld\n", query(1, l, r));   
        }
    }

    return 0;
}