线段树
建树O(n),查询/修改O(logn)

ll a[N];

struct Node
{
    int l, r; //区间范围
    ll sum;   //询问值
    ll lazy;  //懒标记
} tr[N * 4];  //若存在无法通过递归得到的信息应当一并写入Node

void pushup(int u) //由子节点信息推算父节点信息
{
    tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}

void pushdown(int u) //下推懒标记
{
    auto &root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
    if (root.lazy)
    {
        left.lazy += root.lazy, left.sum += (ll)(left.r - left.l + 1) * root.lazy;
        right.lazy += root.lazy, right.sum += (ll)(right.r - right.l + 1) * root.lazy;
        root.lazy = 0;
    }
}

void build(int u, int l, int r)
{
    if (l == r)
    {
        tr[u] = {l, r, a[l], 0};
    }
    else
    {
        tr[u] = {l, r};
        int mid = l + r >> 1;
        build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }
}

void modify(int u, int l, int r, int d) //修改区间[l,r]信息
{
    if (l <= tr[u].l && tr[u].r <= r) //节点区间已被完全包含在[l, r]中
    {
        tr[u].sum += (ll)(tr[u].r - tr[u].l + 1) * d;
        tr[u].lazy += d;
    }
    else
    {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) //分治处理左右子树
        {
            modify(u << 1, l, r, d);
        }
        if (r > mid)
        {
            modify(u << 1 | 1, l, r, d);
        }
        pushup(u);
    }
}

ll query(int u, int l, int r) //在区间[l,r]中进行查询
{
    if (l <= tr[u].l && tr[u].r <= r) //节点区间已被完全包含在[l, r]中
    {
        return tr[u].sum;
    }
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    ll sum = 0;
    if (l <= mid)
    {
        sum += query(u << 1, l, r);
    }
    if (r > mid)
    {
        sum += query(u << 1 | 1, l, r);
    }
    return sum;
}

主席树(可持久化线段树)
查询/修改O(logn)
例题
给定长度为 N 的整数序列 A，下标为 1∼N。
现在要执行 M 次操作，每次操作为给出三个整数 l,r,k，求 A[l],A[l+1],…,A[r] (即A的下标区间 [l,r])中第k小的数是多少。

int a[N];
int root[N], idx; //[1, i]区间的主席树的根节点
vector<int> nums;

struct Node
{
    int l, r;
    int cnt;
} tr[N << 5];

int find(int x)
{
    return lower_bound(nums.begin(), nums.end(), x) - nums.begin();
}

int build(int l, int r) //建树
{
    int p = ++idx;
    if (l == r)
    {
        return p;
    }
    int mid = l + r >> 1;
    tr[p].l = build(l, mid), tr[p].r = build(mid + 1, r);
    return p;
}

int insert(int p, int l, int r, int x)
{
    int q = ++idx;
    tr[q] = tr[p]; //复制
    if (l == r)
    {
        tr[q].cnt++;
        return q;
    }
    int mid = l + r >> 1;
    if (x <= mid) //加新点
    {
        tr[q].l = insert(tr[p].l, l, mid, x);
    }
    else
    {
        tr[q].r = insert(tr[p].r, mid + 1, r, x);
    }
    tr[q].cnt = tr[tr[q].l].cnt + tr[tr[q].r].cnt;
    return q;
}

int query(int q, int p, int l, int r, int k)
{
    if (l == r)
    {
        return l;
    }
    int cnt = tr[tr[q].l].cnt - tr[tr[p].l].cnt; //求出[l,r]的值
    int mid = l + r >> 1;
    if (k <= cnt) //按需递归
    {
        return query(tr[q].l, tr[p].l, l, mid, k);
    }
    else
    {
        return query(tr[q].r, tr[p].r, mid + 1, r, k - cnt);
    }
}

int main()
{
    std::ios::sync_with_stdio(0), std::cin.tie(0), std::cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        nums.push_back(a[i]);
    }
    sort(nums.begin(), nums.end());
    nums.erase(unique(nums.begin(), nums.end()), nums.end()); //去重
    root[0] = build(0, nums.size() - 1);
    for (int i = 1; i <= n; i++) //初始化
    {
        root[i] = insert(root[i - 1], 0, nums.size() - 1, find(a[i]));
    }
    while (m--)
    {
        int l, r, k;
        cin >> l >> r >> k;
        cout << nums[query(root[r], root[l - 1], 0, nums.size() - 1, k)] << endl;
    }
    return 0;
}