P5048 [Ynoi2019 模拟赛] Yuno loves sqrt technology III

题意

给定序列 \(s\)，每次询问 \(l, r\) 的区间众数的出现次数。

强制在线。空间：\(62.5MB\)。

Sol

蒲公英卡常卡空间版。

考虑优化那个 \(n \times m\) 的数组。

我们要求 \(l, r\) 之中某个数的个数。

乍一看不好弄，仔细想想就会发现，如果我们知道当前的最优答案。在长常数时间内就能知道当前散块是否比最优答案更优。

考虑将每个数按照下标扔进 \(n\) 个 \(vector\)。

直接暴力查询 \(p_x + ans\) 即可。

不难发现，该算法的时间复杂度是正确的。

Code

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <queue>
#include <vector>
#include <cassert>
#define pii pair <int, int>
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 rg register
#define il inline

#define fi first
#define se second

const int N = 5e5 + 10, M = 1505;

array <int, N> s, h;

namespace Blk {

int bsi = 433;

il int calc(rg int x) {
	return (x - 1) / bsi + 1;
}

array <int, N> cur;
queue <int> q;
int tot;

il void add(rg int x, rg int y) {
	if (!x) return;
	cur[x] += y;
	q.push(x);
	if (cur[x] > cur[tot]) tot = x;
	else if (cur[x] == cur[tot]) tot = min(tot, x);
}

il void clear() {
	while (!q.empty())
		cur[q.front()] = 0, q.pop();
	tot = 0;
}

array <array <pii, M>, M> isl;
array <vector <int>, N> suf, pre;

array <int, N> idx, dfn;

il void init(rg int n) {
	for (rg int i = 1; i <= calc(n); i++) {
		for (rg int j = i; j <= calc(n); j++) {
			for (rg int k = (j - 1) * bsi + 1; k <= min(j * bsi, n); k++)
				add(s[k], 1);
			isl[i][j] = make_pair(tot, cur[tot]);
		}
		clear();
	}
	for (rg int i = 1; i <= n; i++)
		suf[s[i]].push_back(i), dfn[i] = suf[s[i]].size() - 1;
	for (rg int i = n; i >= 1; i--)
		pre[s[i]].push_back(i), idx[i] = pre[s[i]].size() - 1;
}

il int query(rg int x, rg int y) {
	if (abs(calc(x) - calc(y)) <= 1) {
		for (rg int i = x; i <= y; i++)
			add(s[i], 1);
		rg int ans = cur[tot]; clear();
		return ans;
	}
	rg int ans, tot; tie(tot, ans) = isl[calc(x) + 1][calc(y) - 1];
	for (rg int i = x; i <= calc(x) * bsi; i++) {
		while ((int)suf[s[i]].size() > ans + dfn[i] && suf[s[i]][ans + dfn[i]] <= y)
			ans++, tot = s[i];
		/* assert(ans + dfn[i] - 1 >= 0); */
		/* if ((int)suf[s[i]].size() > ans + dfn[i] - 1 && suf[s[i]][ans + dfn[i] - 1] <= y) */
			/* tot = min(tot, s[i]); */
	}
	for (rg int i = (calc(y) - 1) * bsi + 1; i <= y; i++) {
		while ((int)pre[s[i]].size() > ans + idx[i] && pre[s[i]][ans + idx[i]] >= x)
			ans++, tot = s[i];
		/* assert(ans + idx[i] - 1 >= 0); */
		/* if ((int)pre[s[i]].size() > ans + idx[i] - 1 && pre[s[i]][ans + idx[i] - 1] >= x) */
			/* tot = min(tot, s[i]); */
	}
	/* write(ans), puts("@"); */
	return ans;
}


}

signed main() {
	/* freopen("P4168_2.in", "r", stdin); */
	rg int n = read(), m = read();
	for (rg int i = 1; i <= n; i++)
		s[i] = h[i] = read();
	/* sort(h.begin() + 1, h.begin() + 1 + n); */
	/* rg int k = unique(h.begin() + 1, h.begin() + 1 + n) - h.begin() - 1; */
	/* for (rg int i = 1; i <= n; i++) */
		/* s[i] = lower_bound(h.begin() + 1, h.begin() + 1 + k, s[i]) - h.begin(); */
	rg int lst = 0;
	Blk::init(n);
	while (m--) {
		rg int l = read(), r = read();
		l = (l ^ lst), r = (r ^ lst);
		/* l = (l + lst - 1) % n + 1, r = (r + lst - 1) % n + 1; */
		if (l > r) swap(l, r);
		write(lst = Blk::query(l, r)), puts("");
	}

	return 0;
}