题解！

采用了一位大佬的思路，使用树状数组：

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int MAXN = 1e5 + 10;
int n, a[MAXN], s[MAXN];
int cl[MAXN], cr[MAXN];
int lowbit(int x) { return x & (-x); }
void add_l(int x, int y) {
	for(int i = x; i <= n; i += lowbit(i))
		cl[i] += y;
}
void add_r(int x, int y) {
	for(int i = x; i <= n; i += lowbit(i))
		cr[i] += y;
}
int query_l(int x) {
	int ans = 0;
	for(int i = x; i >= 1; i -= lowbit(i))
		ans += cl[i];
	return ans;
}
int query_r(int x) {
	int ans = 0;
	for(int i = x; i >= 1; i -= lowbit(i))
		ans += cr[i];
	return ans;
}
int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; i ++) {
		int x;
		scanf("%d", &x);
		a[x] = i;
	}
	for(int i = 1, l = 1, r = n, k; i <= n; i ++) {
		if(i % 2 == 1) 
		{
			k = l, l ++;
			int g = query_r(a[k]);
			int p = query_l(n) - query_l(a[k]);
			if(k > a[k])
				printf("%d\n", abs(abs(k - a[k]) - p + g));
			else printf("%d\n", abs(abs(k - a[k]) + p - g));
			add_l(a[k], 1);
		}
		else {
			k = r, r --;
			int g = query_r(a[k]);
			int p = query_l(n) - query_l(a[k]);
			if(k > a[k])
				printf("%d\n", abs(abs(k - a[k]) - p + g));
			else printf("%d\n", abs(abs(k - a[k]) + p - g));
			add_r(a[k], 1);
		}
	}
	return 0;
}