https://ac.nowcoder.com/acm/contest/24154/E

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 100010;

int n, m;
LL w[N];
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 &root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
    if (root.add)
    {
        left.add += root.add, left.sum += (LL)(left.r - left.l + 1) * root.add;
        right.add += root.add, right.sum += (LL)(right.r - right.l + 1) * root.add;
        root.add = 0;
    }
}

void build(int u, int l, int r)
{
    if (l == r) tr[u] = {l, r, w[r], 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].sum +=  (LL)((tr[u].r - tr[u].l + 1) * d);
        tr[u].add += d;
    }
    else    // 一定要分裂
    {
        pushdown(u);
        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;

    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);
    return sum;
}


int main()
{
    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; i ++ ) scanf("%d", &w[i]);

    build(1, 1, n);

    int op;
    int l, r, d;

    while (m -- )
    {
        cin >> op >> l >> r;
        if (op == 1)
        {
            cin >> d;
            modify(1, l, r, -d);
        }
        else printf("%lld\n", query(1, l, r));
    }

    return 0;
}

1.题目后台数据不严谨，理论上血量可以为最大值，所以LL定义，但是成功ac的代码里可以不用LL
2.在做这道题时按照yxc的 一个简单的整数问题2 https://www.acwing.com/problem/content/244/
来写，在modify强制转换tr[u].sum += (LL)(tr[u].r - tr[u].l + 1) *d;这一步中，反而过不去最后一个案例，就在于强制转换出了问题 ,所以出现int long long 运算，之后我会优先把全部先定义成long long
相似题型 https://www.acwing.com/problem/content/1266/