JZOJ 3469. 【NOIP2013模拟联考7】数列(sequence)

\(\text{Solution}\)

明显的 \(\text{K-D Tree}\) 基操题
提前给出了数列，那么考虑提前建好树，省去重构
但还是要开 \(O\)

\(\text{Code}\)

#pragma GCC optimize(3)
#pragma GCC optimize("inline")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse3","sse2","sse")
#pragma GCC diagnostic error "-std=c++14"
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#pragma GCC optimize("fast-math","unroll-loops","no-stack-protector","inline")
#include <cstdio>
#include <iostream>
#include <algorithm>
#define LL long long
#define re register
using namespace std;

const int N = 5e4 + 5;
const LL P = 536870912;
int n, m, cur, x0, y0, x1, y1, rt, L1[N], L2[N], R1[N], R2[N], siz[N], ls[N], rs[N], ds[N], gs[N];
LL sum[N], tg1[N], tg2[N], v1, v2;
struct node{int x, y; LL v;}s[N];
inline bool cmpx(int a, int b){return s[a].x < s[b].x;}
inline bool cmpy(int a, int b){return s[a].y < s[b].y;}

inline void read(int &x)
{
	x = 0; char ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
}

inline void update(int p, int o)
{
	L1[p] = min(L1[p], L1[o]), R1[p] = max(R1[p], R1[o]);
	L2[p] = min(L2[p], L2[o]), R2[p] = max(R2[p], R2[o]);
}
inline void maintain(int p)
{
	siz[p] = siz[ls[p]] + siz[rs[p]] + 1, sum[p] = sum[ls[p]] + sum[rs[p]] + s[p].v;
	L1[p] = R1[p] = s[p].x, L2[p] = R2[p] = s[p].y;
	if (ls[p]) update(p, ls[p]); if (rs[p]) update(p, rs[p]);
}

int build(int l, int r)
{
	if (l > r) return 0;
	int mid = (l + r) >> 1;
	double av1 = 0, av2 = 0, s1 = 0, s2 = 0;
	for(re int i = l; i <= r; i++) av1 += s[gs[i]].x, av2 += s[gs[i]].y;
	av1 /= (r - l + 1), av2 /= (r - l + 1);
	for(re int i = l; i <= r; i++)
		s1 += (av1 - s[gs[i]].x) * (av1 - s[gs[i]].x), s2 += (av2 - s[gs[i]].y) * (av2 - s[gs[i]].y);
	if (s1 > s2) nth_element(gs + l, gs + mid, gs + r + 1, cmpx), ds[gs[mid]] = 1;
	else nth_element(gs + l, gs + mid, gs + r + 1, cmpy), ds[gs[mid]] = 2;
	ls[gs[mid]] = build(l, mid - 1), rs[gs[mid]] = build(mid + 1, r), maintain(gs[mid]);
	return gs[mid];
}

inline void add1(int p, LL v)
{
	s[p].v = s[p].v * v % P, sum[p] = sum[p] * v % P, tg1[p] = tg1[p] * v % P, tg2[p] = tg2[p] * v % P;
}
inline void add2(int p, LL v)
{
	s[p].v = (s[p].v + v) % P, sum[p] = (sum[p] + v * siz[p] % P) % P, tg2[p] = (tg2[p] + v) % P;
}
inline void pushdown(int p)
{
	add1(ls[p], tg1[p]), add1(rs[p], tg1[p]), tg1[p] = 1;
	if (tg2[p]) add2(ls[p], tg2[p]), add2(rs[p], tg2[p]), tg2[p] = 0;
}

void modify(int p)
{
	if (!p || L1[p] > x1 || R1[p] < x0 || L2[p] > y1 || R2[p] < y0) return;
	if (x0 <= L1[p] && R1[p] <= x1 && y0 <= L2[p] && R2[p] <= y1) return add1(p, v1), add2(p, v2);
	if (s[p].x >= x0 && s[p].x <= x1 && s[p].y >= y0 && s[p].y <= y1)
		sum[p] = (sum[p] - s[p].v + P) % P, s[p].v = (s[p].v * v1 % P + v2) % P, sum[p] = (sum[p] + s[p].v) % P;
	pushdown(p), modify(ls[p]), modify(rs[p]), maintain(p);
}

LL query(int p)
{
	if (!p || L1[p] > x1 || R1[p] < x0 || L2[p] > y1 || R2[p] < y0) return 0;
	if (x0 <= L1[p] && R1[p] <= x1 && y0 <= L2[p] && R2[p] <= y1) return sum[p];
	pushdown(p);
	LL res = 0;
	if (s[p].x >= x0 && s[p].x <= x1 && s[p].y >= y0 && s[p].y <= y1) res = s[p].v;
	return (res + query(ls[p]) + query(rs[p])) % P;
}

int main()
{
	freopen("sequence.in", "r", stdin), freopen("sequence.out", "w", stdout);
	read(n), read(m);
	for(re int i = 1; i <= n; i++) ++cur, read(s[cur].y), s[cur].x = i, gs[++gs[0]] = i, tg1[i] = 1;
	rt = build(1, n);
	for(int opt; m; m--)
	{
		read(opt);
		if (opt == 0) read(x0), read(x1), read(y0), read(y1), v1 = y0, v2 = y1, y0 = 1, y1 = n, modify(rt);
		else if (opt == 1) read(y0), read(y1), read(x0), read(x1), v1 = x0, v2 = x1, x0 = 1, x1 = n, modify(rt);
		else if (opt == 2) read(x0), read(x1), y0 = 1, y1 = n, printf("%lld\n", query(rt));
		else if (opt == 3) read(y0), read(y1), x0 = 1, x1 = n, printf("%lld\n", query(rt));
	}
}