给定一个长度为$n$的数组$a$，$1\le a_i\le n$，对于每个$a_i$，我们要找到一个$l\le i,r\ge i$，

使得，我们对区间$[l,r]$升序后，值为$a_i$的数与中位数相隔最远，输出这个最远距离。

我们分两种情况讨论：

• $a_i$在中位数的左边，也就是$a_i\le$中位数，我们考虑把$a_j\ge a_i$的设置为$1$，其他都设置为$-1$，

  则我们就是要对每个$i$找到一个区间$[l,r]$，使得区间和最大，假设这个区间和为$x$，则答案即为$\frac{x}{2}$

• $a_i$在中位数的右边，也就是$a_i\ge$中位数，我们考虑把$a_j\le a_i$的设置为$1$，其他设置为$-1$，

  则我们就是要对每个$i$找到一个区间$[l,r]$，使得区间和最大，假设这个区间和为$x$，则答案即为$\frac{x+1}{2}-1$。

综上我们对这两个答案取$max$即可，至于对值的维护，我们可以考虑用主席树维护区间左、右最大和即可。

#include <bits/stdc++.h>using namespace std;const int N =2e5 + 10;int root[N], ls[N << 5], rs[N << 5], num;int a[N], ans[N], n;vector<int> P[N];struct Res {int lsum, sum, rsum;
}T[N << 5];Res operator + (Res a, Res b) {return {max(a.lsum, a.sum + b.lsum), a.sum + b.sum, max(a.rsum + b.sum, b.rsum)};
}void push_up(int rt) {T[rt] = T[ls[rt]] + T[rs[rt]];
}void build(int &rt, int l, int r) {rt = ++num;if (l == r) {T[rt] = {1, 1, 1};return ;}int mid = l + r >> 1;build(ls[rt], l, mid);build(rs[rt], mid + 1, r);push_up(rt);
}void update(int &rt, int pre, int l, int r, int x, int v) {rt = ++num, ls[rt] = ls[pre], rs[rt] = rs[pre];if (l == r) {T[rt] = {v, v, v};return ;}int mid = l + r >> 1;if (x <= mid) {update(ls[rt], ls[pre], l, mid, x, v);}else {update(rs[rt], rs[pre], mid + 1, r, x, v);}push_up(rt);
}Res query(int rt, int l, int r, int L, int R) {if (l >= L && r <= R) {return T[rt];}int mid = l + r >> 1;if (L <= mid && R > mid) {return query(ls[rt], l, mid, L, R) + query(rs[rt], mid + 1, r, L, R);}else if (L <= mid) {return query(ls[rt], l, mid, L, R);}else {return query(rs[rt], mid + 1, r, L, R);}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);P[a[i]].push_back(i);}build(root[1], 1, n);for (int i = 2; i <= n; i++) {root[i] = root[i - 1];for (auto it : P[i - 1]) {update(root[i], root[i], 1, n, it, -1);}}for (int i = 1; i <= n; i++) {int lsum = 0, rsum = 0, sum = 1;if (i != 1) {lsum = query(root[a[i]], 1, n, 1, i - 1).rsum;}if (i != n) {rsum = query(root[a[i]], 1, n, i + 1, n).lsum;}if (lsum > 0) {sum += lsum;}if (rsum > 0) {sum += rsum;}ans[i] = max(ans[i], sum >> 1);}for (int i = 1; i <= num; i++) {ls[i] = rs[i] = 0;}for (int i = 1; i <= n; i++) {root[i] = 0;}num = 0;build(root[n], 1, n);for (int i = n - 1; i >= 1; i--) {root[i] = root[i + 1];for (auto it : P[i + 1]) {update(root[i], root[i], 1, n, it, -1);}}for (int i = 1; i <= n; i++) {int lsum = 0, rsum = 0, sum = 1;if (i != 1) {lsum = query(root[a[i]], 1, n, 1, i - 1).rsum;}if (i != n) {rsum = query(root[a[i]], 1, n, i + 1, n).lsum;}if (lsum > 0) {sum += lsum;}if (rsum > 0) {sum += rsum;}ans[i] = max(ans[i], (sum + 1) / 2 - 1);}for (int i = 1; i <= n; i++) {printf("%d%c", ans[i], i == n ? '\n' : ' ');}return 0;
}