题目大意：有$n$个数，有$m$次$4$种操作：

1.  1 l r x ：将$[l,r]$区间所有数加上$x$
2.  2 l r x ：将$[l,r]$区间所有数变成$x$
3.  3 l r k ：输出$[l,r]$区间第$k$大
4.  4 l r x p ：输出$\sum\limits_{i=l}^rs_i^x\pm

$n,m\leqslant10^5$

题解：珂朵莉树模板题

卡点：$split$中写错，写成 it==s.end() 

 

C++ Code：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <set>
#include <vector>
#define mul(a, b) (static_cast<long long> (a) * (b) % mod)
#define _mul(a, b) (a = static_cast<long long> (a) * (b) % mod)

inline int pw(int base, int p, int mod) {
	static int res;
	for (res = 1; p; p >>= 1, _mul(base, base)) if (p & 1) _mul(res, base);
	return res;
}

int n, m, seed, vmax;
int rnd() {
	static const int mod = 1e9 + 7;
	static int t; t = seed;
	seed = (7ll * seed + 13) % mod;
	return t;
}

namespace ODT {
	struct node {
		int l, r; mutable long long v;
		inline bool operator < (const node &rhs) const { return l < rhs.l; }
	} ;
	typedef std::set<node>::iterator SIT;
	typedef std::pair<long long, int> PLI;
	std::set<node> s;
	SIT split(int pos) {
		SIT it = s.lower_bound((node) { pos, 0, 0 });
		if (it != s.end() && it -> l == pos) return it;
		--it; const int l = it -> l, r = it -> r;
		const long long v = it -> v;
		s.erase(it), s.insert((node) { l, pos - 1, v });
		return s.insert((node) { pos, r, v }).first;
	}
	void assign(int l, int r, int v) {
		SIT R = split(r + 1), L = split(l);
		s.erase(L, R), s.insert((node) { l, r, v });
	}
	void add(int l, int r, int v) {
		SIT R = split(r + 1), L = split(l);
		for (SIT it = L; it != R; ++it) it -> v += v;
	}
	long long rnk(int l, int r, int k) {
		static std::vector<PLI> v; v.clear();
		SIT R = split(r + 1), L = split(l);
		for (SIT it = L; it != R; ++it)
			v.push_back(std::make_pair(it -> v, it -> r - it -> l + 1));
		std::sort(v.begin(), v.end());
		for (PLI i : v) {
			k -= i.second;
			if (k <= 0) return i.first;
		}
	}
	int query(int l, int r, int x, int mod) {
		SIT R = split(r + 1), L = split(l);
		int ans = 0;
		for (SIT it = L; it != R; ++it)
			ans = (ans + mul(pw(it -> v % mod, x, mod), it -> r - it -> l + 1)) % mod;
		return ans;
	}
}

int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	std::cin >> n >> m >> seed >> vmax;
	for (int i = 1, x; i <= n; ++i) {
		x = (rnd() % vmax) + 1;
		ODT::s.insert((ODT::node) { i, i, x });
	}
	while (m --> 0) {
		static int op, l, r, x, y;
		op = (rnd() % 4) + 1;
		l = (rnd() % n) + 1;
		r = (rnd() % n) + 1;
		if (l > r) std::swap(l, r);
		if (op == 3) x = rnd() % (r - l + 1) + 1;
		else x = rnd() % vmax + 1;
		if (op == 4) y = rnd() % vmax + 1;
		switch (op) {
			case 1: ODT::add(l, r, x); break;
			case 2: ODT::assign(l, r, x); break;
			case 3: std::cout << ODT::rnk(l, r, x) << '\n'; break;
			case 4: std::cout << ODT::query(l, r, x, y) << '\n';
		}
	}
	return 0;
}