1 #include <bits/stdc++.h>
  2
  3 using namespace std;
  4
  5 #define low lower_bound
  6 #define upp upper_bound
  7
  8 const int N = 100005;
  9 const int inf = 0x3f3f3f3f;
 10
 11 int n;
 12 int pos[N];
 13 int num[N];
 14 int ans[N];
 15 int stk[N];
 16
 17 struct node
 18 {
 19     int lt, rt, sum;
 20
 21     node (void) :
 22         lt (0), rt (0), sum (0) {};
 23 };
 24
 25 node tree[N << 2];
 26
 27 void build (int p, int l, int r)
 28 {
 29     node &t = tree[p];
 30
 31     t.lt = l;
 32     t.rt = r;
 33
 34     t.sum = r - l + 1;
 35
 36     if (l != r)
 37     {
 38         int mid = (l + r) >> 1;
 39
 40         build (p << 1, l, mid);
 41         build (p << 1 | 1, mid + 1, r);
 42     }
 43 }
 44
 45 void change (int p, int pos, int val)
 46 {
 47     node &t = tree[p];
 48
 49     if (t.lt != t.rt)
 50     {
 51         int mid = (t.lt + t.rt) >> 1;
 52
 53         if (pos <= mid)
 54             change (p << 1, pos, val);
 55         else
 56             change (p << 1 | 1, pos, val);
 57
 58         t.sum = tree[p << 1].sum + tree[p << 1 | 1].sum;
 59     }
 60     else
 61         t.sum = val;
 62 }
 63
 64 int query (int p, int val)
 65 {
 66     node &t = tree[p];
 67
 68     if (t.lt != t.rt)
 69     {
 70         int tmp = tree[p << 1].sum;
 71
 72         if (val <= tmp)
 73             return query (p << 1, val);
 74         else
 75             return query (p << 1 | 1, val - tmp);
 76     }
 77     else
 78         return t.lt;
 79 }
 80
 81 signed main (void)
 82 {
 83     scanf ("%d", &n);
 84
 85     for (int i = 1; i <= n; ++i)
 86         scanf ("%d", pos + i);
 87
 88     build (1, 1, n);
 89
 90     for (int i = n; i >= 1; --i)
 91     {
 92         int t = query (1, pos[i] + 1);
 93         num[t] = i, change (1, t, 0);
 94     }
 95
 96     memset (stk, inf, sizeof(stk));
 97
 98     for (int i = 1; i <= n; ++i)
 99     {
100         *low (stk, stk + i, num[i]) = num[i];
101         ans[num[i]] = low (stk, stk + i, num[i]) - stk;
102     }
103
104     for (int i = 1; i <= n; ++i)
105         ans[i] = max (ans[i], ans[i - 1]);
106
107     for (int i = 1; i <= n; ++i)
108         printf ("%d\n", ans[i] + 1);
109 }