hdu 1890 Robotic Sort //splay tree

/*
这题因为不是对数列第K个，而是对初始化标记的序号操作，所以标记的处理很不一般
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 100000 + 123;
const int INF = 1 << 28;
struct number
{
    int v, ps;
}a[MAXN];

int s[MAXN], root, m, cnt, No[MAXN];

struct node
{
    int p, c[2], sz;
    bool d, rev;
}T[MAXN];

bool cmp(number a, number b)
{
    if(a.v != b.v) return a.v < b.v;
    return a.ps < b.ps;
}

void sc(int _p, int _c, int _d)
{
    T[_p].c[_d] = _c;
    T[_c].p = _p;
    T[_c].d = _d;
}

void upd(int x)
{
    T[x].sz = T[T[x].c[0]].sz + T[T[x].c[1]].sz + 1;
}

void rot(int x)
{
    int y = T[x].p;
    bool d = T[x].d;
    if(y == root)
    {
        root = x;
        T[root].p = 0;
    }
    else sc(T[y].p, x, T[y].d);
    sc(y, T[x].c[!d], d);
    sc(x, y, !d);
    upd(y);
}

void splay(int x, int r)
{
    int p;
    while((p = T[x].p) != r)
    {
        if(T[p].p == r) rot(x);
        else
        {
            if(T[x].d == T[p].d)
            {
                rot(p);
                rot(x);
            }
            else
            {
                rot(x);
                rot(x);
            }
        }
    }
    upd(x);
}

void newnode(int n)
{
    T[n].c[0] = T[n].c[1] = 0;
    T[n].sz = 1;
    T[n].rev = 0;
}

int mkt(int l, int r)
{
    if(l > r) return 0;
    int mid= (l + r) >> 1;

    newnode(++cnt);

    int n0 = cnt, l_r = mkt(l, mid - 1), r_r = mkt(mid + 1, r);

    s[mid] = n0;

    sc(n0, l_r, 0);
    sc(n0, r_r, 1);
    upd(n0);
    return n0;
}

void prepare()
{
    T[0].sz = T[0].c[0] = T[0].c[1] = T[0].p = 0;
    T[0].rev = 0;

    cnt = 0;
    root = mkt(0, m + 1);
    T[root].p = 0;
}

void rev_opr(int x)
{
    T[x].rev = !T[x].rev;
    int c0 = T[x].c[0], c1 = T[x].c[1];
    sc(x, c0, 1);
    sc(x, c1, 0);
}

void dm(int x)
{
    if(T[x].rev)
    {
        rev_opr(T[x].c[0]);
        rev_opr(T[x].c[1]);
        T[x].rev = 0;
    }
}

int Find_Kth(int k)
{
   int i = root, s0;
   while(i)
   {
       dm(i);
       s0 = T[T[i].c[0]].sz + 1;
       if(k == s0)  break;
       else  if(k < s0) i = T[i].c[0];
       else
       {
           k -= s0;
           i = T[i].c[1];
       }
   }
   return i;
}

int succ(int x)
{
    int i, j = 0, len0 = 0;
    i = x;
    while (i)
    {
        No[len0++] = i;
        i = T[i].p;
    }
    for(int i=len0-1; i>=0; i--) dm(No[i]);

    if (T[x].c[1])
    {
        i = T[x].c[1];
        while (dm(i), j = T[i].c[0]) i = j;
        return i;
    }
    else
    {
        for (int i=len0-1; i>=1; i--)
            if (No[i - 1] == T[No[i]].c[0]) j = No[i];
        return j;
    }
}

void dm_all(int x)
{
    int i = x, len0 = 0;
    while (i)
    {
        No[len0++] = i;
        i = T[i].p;
    }
    for(int i = len0 - 1; i>=0; i--) dm(No[i]);
}

int reve(int l, int r)
{
    dm_all(r);
    splay(l, 0);
    splay(r, l);
    rev_opr(T[r].c[0]);
    return T[T[r].c[0]].sz;
}

int succ()
{
    int i = T[root].c[1], j;
    if(!i) return 0;
    while(j = T[i].c[0]) i = j;
    return i;
}

int main()
{
    while(1)
    {
        scanf("%d", &m);
        if(!m)  break;

        for(int i = 1; i <= m; i++)
        {
            scanf("%d", &a[i].v);
            a[i].ps = i;
        }

        a[0].v = a[m + 1].v = INF;
        a[0].ps = 0, a[m + 1].ps = m + 1;

        sort(a + 1, a + 1 + m, cmp);

        prepare();

        for(int i = 1; i <= m; i++)
        {
            int l = Find_Kth(i);

            /*dm_all(s[a[i].ps]); 超时，这么写，可能因为伸展次数过多
            splay(s[a[i].ps], 0);
            int r = succ(); */
            int r = succ(s[a[i].ps]);
            printf("%d%c", reve(l, r) + i - 1, i == m ? '\n' : ' ');
        }
    }
    return 0;
}