CF896C Willem, Chtholly and Seniorious

题意

维护一个序列 \(s\),有以下操作。

  • 区间加。
  • 区间覆盖。
  • \(l\)\(r\) 的第 \(k\) 小元素。
  • \(l\)\(r\) 的每个元素的 \(x\) 次方之和膜 \(y\)

输入由给定种子 随机 生成。

Sol

珂朵莉树。

本质上就是拿 \(set\) 乱搞。

考虑每次操作对于颜色段的影响。

每次操作分裂出两个区间,然后暴力做就可以了。

注意要开 \(int128\)

Code

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <set>
#define int __int128
using namespace std;
#ifdef ONLINE_JUDGE

#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;

#endif
int read() {
	int p = 0, flg = 1;
	char c = getchar();
	while (c < '0' || c > '9') {
		if (c == '-') flg = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9') {
		p = p * 10 + c - '0';
		c = getchar();
	}
	return p * flg;
}
void write(int x) {
	if (x < 0) {
		x = -x;
		putchar('-');
	}
	if (x > 9) {
		write(x / 10);
	}
	putchar(x % 10 + '0');
}

#define fi first
#define se second

const int N = 1e5 + 5;

namespace Mth {

int pow_(int x, int k, int p) {
	int ans = 1;
	while (k) {
		if (k & 1) ans = ans * x % p;
		x = x * x % p;
		k >>= 1;
	}
	return ans;
}

}

namespace Chtholly {

void Mod(int &x, int y) {
	if (x >= y) x -= y;
	if (x < 0) x += y;
}

class Node {

public:

	int l, r;
	mutable int v;

	Node(int _l, int _r, int _v) : l(_l), r(_r), v(_v) {}

	bool operator <(const Node &t) const { return l < t.l; }

};

set <Node> Cht;

auto split(int x, int n) {
	if (x > n) return Cht.end();
	auto it = --Cht.upper_bound(Node(x, 0, 0));
	if (it -> l == x) return it;
	int l = it -> l, r = it -> r, v = it -> v;
	Cht.erase(it); Cht.insert(Node(l, x - 1, v));
	return Cht.insert(Node(x, r, v)).fi;
}

void assign(int l, int r, int v, int n) {
	auto itr = split(r + 1, n), itl = split(l, n);
	Cht.erase(itl, itr);
	Cht.insert(Node(l, r, v));
}

void modify(int l, int r, int v, int n) {
	auto itr = split(r + 1, n), itl = split(l, n);
	for (auto it = itl; it != itr; it++)
		it -> v += v;
}

int rnk(int l, int r, int k, int n) {
	auto itr = split(r + 1, n), itl = split(l, n);
	vector <Node> isl;
	for (auto it = itl; it != itr; it++)
		isl.push_back(*it);
	sort(isl.begin(), isl.end(), [](Node x, Node y) { return x.v < y.v; });
	int cnt = 0;
	for (auto x : isl) {
		cnt += x.r - x.l + 1;
		if (cnt < k) continue;
		return x.v;
	}
	return -1;
}

int query(int l, int r, int x, int y, int n) {
	auto itr = split(r + 1, n), itl = split(l, n);
	int ans = 0;
	for (auto it = itl; it != itr; it++) {
		ans += (it -> r - it -> l + 1) * (Mth::pow_(it -> v, x, y)) % y;
		Mod(ans, y);
	}
	return ans;
}

}

array <int, N> s;

/* int rnd(int& seed) { */
	/* int res = seed; */
	/* seed = (seed * 7 + 13) % (int)(1e9 + 7); */
	/* return res; */
/* } */

signed main() {
	int n = read(), m = read(), seed = read(), maxn = read();

	auto rnd = [](auto& seed) {
		int res = seed;
		seed = (seed * 7 + 13) % (int)(1e9 + 7);
		return res;
	};

	for (int i = 1; i <= n; i++) {
		s[i] = (rnd(seed) % maxn) + 1;
		Chtholly::Cht.insert(Chtholly::Node(i, i, s[i]));
	}

	for (int i = 1; i <= m; i++) {
		int op = (rnd(seed) % 4) + 1,
			l = (rnd(seed) % n) + 1,
			r = (rnd(seed) % n) + 1;
		if (l > r) swap(l, r);

		int x = 0, y = 0;
		if (op == 3) x = (rnd(seed) % (r - l + 1)) + 1;
		else x = (rnd(seed) % maxn) + 1;
		if (op == 4) y = (rnd(seed) % maxn) + 1;

		switch (op) {

		case 1:
			Chtholly::modify(l, r, x, n);
			break;
		case 2:
			Chtholly::assign(l, r, x, n);
			break;
		case 3:
			write(Chtholly::rnk(l, r, x, n)), puts("");
			break;
		default:
			write(Chtholly::query(l, r, x, y, n)), puts("");
			/* break; */

		}
	}
	return 0;
}
posted @ 2023-12-08 11:54  cxqghzj  阅读(20)  评论(0编辑  收藏  举报