树状数组

$$∑_{i=1}^x∑_{j=1}^ib_j = ∑_{i=1}^x(x−i+1)×b_i = (x+1)∑_{i=1}^xb_i−∑_{i=1}^x i×b_i $$

因此只需维护两个树状数组即可

一个是差分数组d[i]的树状数组tr1[i]，一个是i*d[i]的树状数组tr2[i]。

#include <bits/stdc++.h>
using namespace std;
#define lowbit(x) (x & (-x))
typedef long long ll;
const int N = 1e6 + 10, INF = 0x3f3f3f3f;
int n, m, a[N];
ll tr1[N], tr2[N];
void add(ll tr[], int x, ll y) {
    for (int i = x; i <= n; i += lowbit(i))
        tr[i] += y;
}
ll query(ll tr[], ll x) {
    ll res = 0;
    for (int i = x; i; i -= lowbit(i))
        res += tr[i];
    return res;
}
ll pre(int x) {
    return query(tr1, x) * (x + 1) - query(tr2, x);
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        ll b = a[i] - a[i - 1];
        add(tr1, i, b), add(tr2, i, i * b);
    }
    char op[10];
    int l, r, d;
    while (m--) {
        scanf("%s%d%d", op, &l, &r);
        if (*op == 'Q') {
            printf("%lld\n", pre(r) - pre(l - 1));
        } else {
            scanf("%d", &d);
            add(tr1, l, d), add(tr2, l, l * d);
            add(tr1, r + 1, -d), add(tr2, r + 1, -(r + 1) * d);
        }
    }
    return 0;
}