[ABC242G] Range Pairing Query

莫队板子。若 $i$ 出现 $c_i$ 次，我们能要求的是 $\sum \limits_{i \in \mathbb{N}} \lfloor \frac{c_i}{2} \rfloor$。

简单莫队维护即可，复杂度 $O(q \sqrt n)$。

#include <bits/stdc++.h>
using namespace std;

const int N = 1e6 + 5;

int bel[N], ans[N];

struct Query
{
	int id, l, r;
	bool operator<(const Query& g) const
	{
		return (bel[l] ^ bel[g.l] ? bel[l] < bel[g.l] : (bel[l] & 1 ? r < g.r : r > g.r));
	}
}p[N];

int n, q, a[N], cnt[N], res;

inline void add(int x)
{
	res -= cnt[a[x]] / 2;
	cnt[a[x]]++;
	res += cnt[a[x]] / 2;
}

inline void del(int x)
{
	res -= cnt[a[x]] / 2;
	cnt[a[x]]--;
	res += cnt[a[x]] / 2;
}

int main()
{
	scanf("%d", &n);
	int s = (int)sqrt(n);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]), bel[i] = i / s;
	scanf("%d", &q);
	for (int i = 1; i <= q; i++)
	{
		int l, r;
		scanf("%d%d", &l, &r);
		p[i] = { i, l, r };
	}
	sort(p + 1, p + q + 1);
	int nl(1), nr(0);
	for (int i = 1; i <= q; i++)
	{
		int l = p[i].l, r = p[i].r;
		while (nl > l) add(--nl);
		while (nr < r) add(++nr);
		while (nl < l) del(nl++);
		while (nr > r) del(nr--); 
		ans[p[i].id] = res;
	}
	for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
	return 0; 
}