A Simple Problem with Integers POJ - 3468 线段树区间修改+区间查询

//add，懒标记，给以当前节点为根的子树中的每一个点加上add(不包含根节点)
//
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = ;
int n, m;
int w[N];
struct Node
{
    int l, r;
    //总和
    //如果只考虑当前节点及子节点上的标记，当前区间和是多少  ，没考虑所有祖先节点上的标记
    LL sum;
    //懒标记
    //给当前区间的所有儿子加上add
    LL add;
}tr[N * ];
//用子节点的信息来计算父节点的信息
void pushup(int u)
{
    tr[u].sum = tr[u << ].sum + tr[u <<  | ].sum;
}
void pushdown(int u)
{
    Node &root = tr[u], &left = tr[u << ], &right = tr[u <<  | ];
    //如果当前根节点有标记，往下传，还要清空
    if (root.add)
    {
        left.add += root.add;
        left.sum += (LL)(left.r - left.l + ) * root.add;
        right.add += root.add;
        right.sum += (LL)(right.r - right.l + ) * root.add;
        root.add = ;
    }
}
void build(int u, int l, int r)
{
    if (l == r)
        tr[u] = {l, r, w[r], };
    else
    {
        tr[u] = {l, r};
        int mid = l + r >> ;
        build(u << , l, mid);
        build(u <<  | , mid + , r);
        pushup(u);
    }
}
void modify(int u, int l, int r, int d)
{
    if (tr[u].l >= l && tr[u].r <= r)
    {
        //总和
        tr[u].sum += (LL)(tr[u].r - tr[u].l + ) * d;
        //懒标记
        tr[u].add += d;
    }
    // 区间太大，一定要分裂
    else
    {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> ;
        if (l <= mid)
            modify(u << , l, r, d);
        if (r > mid)
            modify(u <<  | , l, r, d);
        //当前区间和发生变化，需要向上传
        pushup(u);
    }
}
LL query(int u, int l, int r)
{
    if (tr[u].l >= l && tr[u].r <= r)
        return tr[u].sum;
    //查询子区间
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> ;
    LL sum = ;
    if (l <= mid)
        sum = query(u << , l, r);
    if (r > mid)
        sum += query(u <<  | , l, r);
    return sum;
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = ; i <= n; i ++ )
        scanf("%d", &w[i]);
    build(, , n);
    char op[];
    int l, r, d;
    while (m -- )
    {
        scanf("%s%d%d", op, &l, &r);
        if (*op == 'C')
        {
            scanf("%d", &d);
            modify(, l, r, d);
        }
        else printf("%lld\n", query(, l, r));
    }
    return ;
}