【数据结构 树状数组】JZOJ_6342 Tiny Counting

题意

求四元组$(a,b,c,d)$，满足$a<b;c<d;S_a<S_b;S_c>S_d$，$abcd$不相等。

思路

若能相等，答案即顺序对$\ast$逆序对，否则考虑以下重复点的四种情况：
$a=c$
$(a=c)<b,d$且$S_d<S_{a=c}<S_b$

$a=d$
$c<(a=d)<b$且$S_{a=d}<S_b,S_c$

$b=c$
$a<(b=c)<d$且$S_{b=c}>S_{a,d}$

$b=d$
$a,c<(b=d)$且$S_a<S_{b=d}<S_c$
用树状数组求解。（我真是超级无敌大傻逼，连逆序对都不会求）

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

int n;
long long ans, p, q;
long long t[100001], a[100001], b[100001], rs[100001], rb[100001], ls[100001], lb[100001];

void add(int x, int val) {
	for (; x <= n; x += x & -x)
		t[x] += val;
}

long long query(int x) {
	long long res = 0;
	for (; x; x -= x & -x)
		res += t[x];
	return res;
}

void init() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &a[i]), b[i] = a[i];
	std::sort(b + 1, b + n + 1);
	int m = std::unique(b + 1, b + n + 1) - b - 1;
	for (int i = 1; i <= n; i++)
		a[i] = std::lower_bound(b + 1, b + m + 1, a[i]) - b;
}

int main() {
	init();
	for (int i = n; i >= 1; i--)
		p += rs[i] = query(a[i] - 1), rb[i] = query(n) - query(a[i]), add(a[i], 1);
	memset(t, 0, sizeof(t));
	for (int i = 1; i <= n; i++)
		q += ls[i] = query(a[i] - 1), lb[i] = query(n) - query(a[i]), add(a[i], 1);
	ans = p * q;
	for (int i = 1; i <= n; i++)
		ans -= rb[i] * rs[i] + rb[i] * lb[i] + rs[i] * ls[i] + lb[i] * ls[i];
	printf("%lld", ans);
}