全RE求调

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;

const int N = 501000;

int a[N];
ll c[N]; // c[1...1000] = a[1...0001] + ... + a[1...1000] (二进制)
int n, mx = 0;
ll ans = 0;

// 此处以求和为例
ll query(int x) { // 1, 2, ..., x
	ll s = 0;
	for (; x; x -= x & (-x)) { // 1...1000 -> 1...0000 (二进制)
		s += c[x];
	}
	return s;
}

// 单点修改
void modify(int x, ll s) { // a[x] += s
	for (; x <= n; x += x & (-x)) {
		c[x] += s;
	}
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		mx = max(mx, a[i]);
	}
	for (int i = 1; i <= n; i++) {
		a[i] = mx + 1 - a[i];
	}
	for (int i = 1; i <= n; i++) {
		ans += query(a[i] - 1);
		modify(a[i], 1);
	}
	printf("%lld", ans);
}