P3834 【模板】可持久化线段树 2（主席树）

这是一道主席树的模板题，

#include<iostream>
#include<set>
#include<map>
#include<algorithm>
#include<vector>
#include<cmath>
#include<climits>
#include<cstring>
#define int long long
const int N = 1e6+5;
using namespace std;
char* p1, * p2, buf[100000];
#define nc() (p1==p2 && (p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
int read()
{
	int x = 0, f = 1;
	char ch = nc();
	while (ch < 48 || ch>57)
	{
		if (ch == '-')
			f = -1;
		ch = nc();
	}
	while (ch >= 48 && ch <= 57)
		x = x * 10 + ch - 48, ch = nc();
	return x * f;
}
int id = 0;
vector<int>v;
int a[N];
struct tree {
	int ch[2];
	int s = 0;
}tr[N*22];
int root[N];
void build(int& x, int l, int r) {
	x = ++id;
	if (l == r)return;
	int m = (l + r) >> 1;
	build(tr[x].ch[0], l, m);
	build(tr[x].ch[1], m + 1, r);
}
void insert(int x, int& y, int l, int r, int v) {
	y = ++id;
	tr[y] = tr[x];
	tr[y].s++;
	if (l == r)return;
	int m = (l + r) >> 1;
	if (v <= m)insert(tr[x].ch[0], tr[y].ch[0], l, m, v);
	else insert(tr[x].ch[1], tr[y].ch[1], m + 1, r, v);
}
int query(int x, int y, int l, int r, int k) {
	if (l == r)return l;
	int m = (l + r) >> 1;
	int s = tr[tr[y].ch[0]].s - tr[tr[x].ch[0]].s;
	if (k <= s)return query(tr[x].ch[0], tr[y].ch[0], l, m, k);
	else return query(tr[x].ch[1], tr[y].ch[1], m+1, r, k-s);
}
int getid(int x) {
	return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		v.push_back(a[i]);
	}
	sort(v.begin(), v.end());
	v.erase(unique(v.begin(), v.end()), v.end());
	//build(root[0], 1, v.size());
	for (int i = 1; i <= n; i++) {
		insert(root[i - 1], root[i], 1, v.size(), getid(a[i]));
	}
	for (int i = 1; i <= m; i++) {
		int l, r, k;
		cin >> l >> r >> k;
		int id = query(root[l - 1], root[r], 1, v.size(), k) - 1;
		cout << v[id] << endl;
	}
	return 0;
}