/*51nod 1364 最大字典序排列(线段树)problem:给出一个1至N的排列，允许你做不超过K次操作，每次操作可以将相邻的两个数交换，问能够得到的字典序最大的排列是什么？例如：N = 5， {1 2 3 4 5}，k = 6，在6次交换后，能够得到的字典序最大的排列为{5 3 1 2 4}。solve:贪心的思想.  从1~n维护第i个数尽可能大. 就成了在[i+1,k+i]中查找最大值. 然后把找到的数从后面的数组中删除插到前面.因为我的删除是用num来表示,所以每次都要先找到对应区间的边界位置. 然后在计算移动过去需要多少步. 直到操作结束hhh-2016/09/04-11:16:36*/#pragma comment(linker,"/STACK:124000000,124000000")#include <algorithm>#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <vector>#include <math.h>#include <queue>#include <set>#include <map>#define lson  i<<1#define rson  i<<1|1#define ll long long#define clr(a,b) memset(a,b,sizeof(a))#define scanfi(a) scanf("%d",&a)#define scanfs(a) scanf("%s",a)#define scanfl(a) scanf("%I64d",&a)#define scanfd(a) scanf("%lf",&a)#define key_val ch[ch[root][1]][0]#define eps 1e-7#define inf 0x3f3f3f3f3f3f3f3fusing namespace std;const ll mod = 1000000007;const int maxn = 100010;const double PI = acos(-1.0);int n;int a[maxn];struct node{    int l,r,mid;    int pos,num;    int Max;} tree[maxn<<2];void push_up(int i){    if(tree[lson].Max >= tree[rson].Max)    {        tree[i].Max= tree[lson].Max;        tree[i].pos = tree[lson].pos;    }    else    {        tree[i].Max = tree[rson].Max;        tree[i].pos = tree[rson].pos;    }    tree[i].num = tree[lson].num + tree[rson].num;}void build(int i,int l,int r){    tree[i].l = l,tree[i].r = r;    tree[i].mid = (l+r) >>1;    tree[i].num = 0;    if(l == r && l)    {        tree[i].num = 1;        tree[i].Max= a[l];        tree[i].pos = l;        return ;    }    build(lson,l,tree[i].mid);    build(rson,tree[i].mid+1,r);    push_up(i);}int tMax,tpos;void update(int i,int k){    if(tree[i].l >= k && tree[i].r <= k)    {        tree[i].Max = 0;        tree[i].num = 0;        return ;    }    int mid = tree[i].mid;    if(k <= mid)        update(lson,k);    else        update(rson,k);    push_up(i);    return ;}void query(int i,int l,int r){    if(tree[i].l >= l && tree[i].r <= r)    {        if(tMax < tree[i].Max)            tMax = tree[i].Max,tpos = tree[i].pos;        else if(tMax == tree[i].Max)            tpos = min(tree[i].pos,tpos);        return ;    }    int mid = tree[i].mid;    if(l <= mid)        query(lson,l,r);    if(r > mid)        query(rson,l,r);    push_up(i);    return ;}int query_x(int i,int k){    if(tree[i].l == tree[i].r)    {        return tree[i].l;    }    int mid = tree[i].mid;    if(tree[lson].num >= k)       return  query_x(lson,k);    else        return query_x(rson,k-tree[lson].num);    push_up(i);}int query_num(int i,int l,int r){    if(tree[i].l >= l && tree[i].r <= r)    {        return tree[i].num;    }    int ans = 0;    if(l <=tree[i].mid)        ans += query_num(lson,l,r);    if(r > tree[i].mid)        ans += query_num(rson,l,r);    return ans;}int ans[maxn];int main(){    int t;//    freopen("in.txt","r",stdin);    while(scanfi(n) != EOF)    {        scanfi(t);        a[0] = 0;        for(int i = 1; i <= n; i++)        {            scanfi(a[i]);        }        build(1,1,n);        int cnt = 1;        while(t > 0 && cnt <= n)        {            tMax = 0,tpos = n;            if(cnt + t >= n)            {                query(1,1,n);                update(1,tpos);                t -= query_num(1,1,tpos);                printf("%d\n",tMax);                cnt++;                a[tpos] = -1;            }            else            {                int pos = query_x(1,t+1);                query(1,1,pos);                update(1,tpos);                t -= query_num(1,1,tpos);                cnt++;                printf("%d\n",tMax);                a[tpos] = -1;            }        }        for(int i = 1; i <= n; i++)            if(a[i]  !=  -1)                printf("%d\n",a[i]);    }    return 0;}