线段树板子题求调

懒标记还不会所以区间修改就用了单点修改的方法(

不需要AC,下方代码样例都过不去(

#include <iostream>

using namespace std;
const int N = (int)1e5 + 5;
int num[N];
struct node
{
    int left;
    int right;
    int sum;
} nodes[4 * N];
int total_node;
int total_ask;
void push_up(int a)
{
    nodes[a].sum = nodes[a << 1].sum + nodes[a * 2 + 1].sum;
    return;
}
void build(int now, int left, int right)
{
    if (left == right)
    {
        nodes[now] = {left, right, num[left]};
    }
    else
    {
        nodes[now] = {left, right};
        int mid = nodes[now].left + nodes[now].right >> 1;
        build(now << 1, left, mid);
        build(now * 1 + 1, mid + 1, right);
        push_up(now);
    }
}
// int query(int now, int left, int right)
// {
//     if (left <= nodes[now].left && right >= nodes[now].right)
//     {

//         return nodes[now].sum;
//     }
//     else
//     {
//         int sum = 0;
//         int mid = nodes[now].left + nodes[now].right >> 1;
//         if (left <= mid)
//         {
//             sum += query(now << 1, left, right);
//         }
//         if (right >= mid + 1)
//         {
//             sum += query(now << 1 | 1, left, right);
//         }
//         return sum;
//     }
// }
int query(int u, int l, int r)
{
    if (l <= nodes[u].left && nodes[u].right <= r)
        return nodes[u].sum;
    int mid = nodes[u].left + nodes[u].right >> 1;
    int sum = 0; // 用 sum 来表示一下我们的总和
    if (mid >= l)
        sum += query(u << 1, l, r);
    if (r >= mid + 1) // 看一下我们当前区间的中点和右边有没有交集
        sum += query(u * 2 + 1, l, r);
    return sum;
}
void modify(int now, int target, int demand)
{
    if (nodes[now].left == nodes[now].right)
    {
        nodes[now].sum += demand;
        return;
    }
    else
    {
        int mid = nodes[now].left + nodes[now].right >> 1;
        if (target <= mid)
        {
            modify(now << 1, target, demand);
        }
        else
        {
            modify(now * 2 + 1, target, demand);
        }
        push_up(now);
    }
}
int main()
{
    cin >> total_node >> total_ask;
    for (int i = 1; i <= total_node; i++)
    {
        cin >> num[i];
    }
    build(1, 1, total_node);
    while (total_ask--)
    {
        int bind_1 = 0;
        cin >> bind_1;
        if (bind_1 == 1)
        {
            int bind_2, bind_3;
            int bind_4;
            cin >> bind_2 >> bind_3 >> bind_4;
            for (int i = bind_2; i <= bind_3; i++)
            {
                modify(1, i, bind_4);
            }
        }
        else
        {
            int bind_5, bind_6;
            cin >> bind_5 >> bind_6;
            int res = query(1, bind_5, bind_6);
            cout << res << endl;
        }
    }
    return 0;
}