简单数学变换+线段树

简单数据结构签到题不解释
本来应该贴板子的,鉴于最近写代码太少了,而且由于要用两个线段树,平时板子都是一个的。以及板子在队友那。就当熟悉写代码,自己写了一下。

#include <bits/stdc++.h>
using namespace std;

#define dual(i, n) (n) + 1 - (i)

typedef long long ll;
int n, q;
const int maxn = 1e+5 + 5;
typedef ll arr[maxn];
arr a, pa;

struct seg_tree
{
    // a,nds坐标都是从1开始记
    ll *a;
    int n;
    struct nd
    {
        ll sum;
    };
    vector<nd> nds;
    seg_tree(ll *a0, int n0)
    {
        a = a0;
        n = n0;
        nds.resize(4 * n);
        build(1, 1, n);
    }
    void build(int i, int l, int r)
    {
        if (l == r)
        {
            nds[i].sum = a[l];
            return;
        }
        int m = l + (r - l) / 2;
        build(2 * i, l, m);
        build(2 * i + 1, m + 1, r);
        nds[i].sum = nds[2 * i].sum + nds[2 * i + 1].sum;
    }

    void change(int i, ll x)
    {
        ll delta = x - a[i];
        a[i] = x;
        add(i, delta);
    }

    void add(int i, ll x)
    {
        int l = 1;
        int r = n;
        int u = 1;
        int m;
        while (true)
        {
            nds[u].sum += x;
            if (l == r)
                break;
            m = l + (r - l) / 2;
            if (i <= m)
            {
                r = m;
                u = 2 * u;
            }
            else
            {
                l = m + 1;
                u = 2 * u + 1;
            }
        }
    }

    ll query(int l, int r, int u, int ul, int ur)
    {
        // ensure ul<=l<=r<=ur
        if (ul == l && r == ur)
            return nds[u].sum;
        int um = ul + (ur - ul) / 2;
        if (r <= um) // left
            return query(l, r, 2 * u, ul, um);
        else if (l >= um + 1) // right
            return query(l, r, 2 * u + 1, um + 1, ur);
        else
            return query(l, um, 2 * u, ul, um)                // left
                   + query(um + 1, r, 2 * u + 1, um + 1, ur); // right
    }

    inline ll query(int l,int r) {
        return query(l,r,1,1,n);
    }
};

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> q;
    for (int i = n; i >= 1; --i)
    {
        cin >> a[i];
        pa[i] = a[i] * i;
    }
    seg_tree s(a, n);
    seg_tree ps(pa, n);
    int t, l, r, k;
    ll x, px;
    for (int i = 0; i < q; ++i)
    {
        cin >> t;
        if (1 == t)
        {
            // query
            cin >> r >> l;
            l = dual(l, n);
            r = dual(r, n);
            cout << ps.query(l, r) - (ll)(l - 1) * s.query(l, r) << "\n";
        }
        else
        {
            // a[k] = x
            cin >> k >> x;
            k = dual(k, n);
            px = x * k;
            s.change(k, x);
            ps.change(k, px);
        }
    }
    cout.flush();
    return 0;
}