题目链接

题目AC代码：

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

const int N = 2e5 + 10;

int m, p;

struct Node
{
    int l, r;
    int v; //区间最大值
}tr[4 * N];

void pushup(int u) //由子节点确定父节点
{
    tr[u].v = max(tr[u << 1].v, tr[u << 1 | 1].v);
}

void build(int u, int l, int r)
{
    tr[u] = {l, r};
    if(l == r) return ;
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
}

int query(int u, int l, int r)
{
    if(tr[u].l >= l && tr[u].r <= r) return tr[u].v;

    int mid = tr[u].l + tr[u].r >> 1;
    int v = 0;
    if(l <= mid) v = query(u << 1, l, r);
    if(r > mid) v = max(query(u << 1 | 1, l, r), v);

    return v;
}

void modify(int u, int x, int v)
{
    if(tr[u].l == tr[u].r) tr[u].v = v;
    else 
    {
        int mid = (tr[u].l + tr[u].r) >> 1;
        if(x <= mid) modify(u << 1, x, v);
        else modify(u << 1 | 1, x, v);
        pushup(u);
    }
}

signed main()
{
    cin >> m >> p;
    int n = 0, last = 0;;
    build(1, 1, m);
    while(m--)
    {
        char s[2];
        int x;
        cin >> s >> x;
        if(*s == 'Q')
        {
            last = query(1, n - x + 1, n);
            cout << last << endl;
        }
        else 
        {
            modify(1, n + 1, (last + x) % p);
            n++;
        }
    }

    return 0;
}

线段树维护的区间关系，他将一个一个区间划分为不同的两个区间，层层递归，当然，他的运用范围挺广泛的。

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

const int N = 1e5 + 10;

struct tre
{
    int l, r; //边界
    int mx;   //区间最大值
} tr[4 * N];

int a[N]; //存储数据

void jianshu(int u, int l, int r)
{ //建树
    tr[u].l = l, tr[u].r = r;
    if (l == r)
    {
        tr[u].mx = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    jianshu(u << 1, l, mid), jianshu(u << 1 | 1, mid + 1, r);
    tr[u].mx = max(tr[u << 1].mx, tr[u << 1 | 1].mx);
}

int query(int u, int l, int r)
{ //查询操作
    if (tr[u].l >= l && tr[u].r <= r)
        return tr[u].mx; //全部包含

    int mx = 0;
    int mid = (tr[u].l + tr[u].r) >> 1;
    if (l <= mid)
        mx = query(u << 1, l, r);
    if (r > mid)
        mx = max(mx, query(u << 1 | 1, l, r));

    return mx;
}

signed main()
{
    int n, q;
    cin >> n >> q;

    for (int i = 1; i <= n; i++)
        cin >> a[i];

    jianshu(1, 1, n);

    while (q--)
    {
        int l, r;
        cin >> l >> r;
        cout << l << ' ' << r << "max is :" << query(1, l, r) << endl;
    }

    return 0;
}