使用封装了的线段树完成的例子

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 200000 + 5;
int D;
struct SegTree
{
    int sum[MAXN * 4], maxx[MAXN * 4];
    void build(int o, int l, int r)
    {
        if (l == r)
        {
            sum[o] = maxx[o] = 0;
            return;
        }
        int mid = (l + r) >> 1;
        build(o * 2, l, mid);
        build(o * 2 + 1, mid + 1, r);
        sum[o] = sum[o * 2] + sum[o * 2 + 1];
        maxx[o] = max(maxx[o * 2], maxx[o * 2 + 1]);
    }
    int query_max(int o, int l, int r, int ql, int qr)
    {
        if (l > qr || r < ql)
            return -1;
        if (ql <= l && r <= qr)
            return maxx[o];
        int mid = (l + r) >> 1;
        return max(query_max(o * 2, l, mid, ql, qr), query_max(o * 2 + 1, mid + 1, r, ql, qr));
    }
    int query_sum(int o, int l, int r, int ql, int qr)
    {
        if (l > qr || r < ql)
            return 0;
        if (ql <= l && r <= qr)
            return sum[o];
        int mid = (l + r) >> 1;
        return query_sum(o * 2, l, mid, ql, qr) + query_sum(o * 2 + 1, mid + 1, r, ql, qr);
    }
    void update(int o, int l, int r, int x, int t)
    {
        if (l == r)
        {
            maxx[o] = sum[o] = t;
            return;
        }
        int mid = (l + r) >> 1;
        if (x <= mid)
            update(o * 2, l, mid, x, t); // 左右分别更新
        else
            update(o * 2 + 1, mid + 1, r, x, t);
        sum[o] = sum[o * 2] + sum[o * 2 + 1];
        maxx[o] = max(maxx[o * 2], maxx[o * 2 + 1]);
    }
} st;
int m, cnt, t, n;
int a[MAXN];
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> m >> D;
    n = m;
    st.build(1, 1, n);
    cnt = t = 0;
    while (m--)
    {
        char op;
        int x;
        cin >> op >> x;
        if (op == 'A')
        {
            cnt++;
            st.update(1, 1, n, cnt, (t + x) % D);
        }
        else if (op == 'Q')
        {
            t = st.query_max(1, 1, n, cnt - x + 1, cnt);
            cout << t << "\n";
        }
    }
    return 0;
}