描述

传送门：hdu-2795

At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions, changes in the dining room menu, and other important information.

On September 1, the billboard was empty. One by one, the announcements started being put on the billboard.

Each announcement is a stripe of paper of unit height. More specifically, the i-th announcement is a rectangle of size 1 * wi.

When someone puts a new announcement on the billboard, she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.

If there is no valid location for a new announcement, it is not put on the billboard (that’s why some programming contests have no participants from this university).

Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed.

Input

There are multiple cases (no more than 40 cases).

The first line of the input file contains three integer numbers, h, w, and n (1 <= h,w <= $10^9$; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements.

Each of the next n lines contains an integer number wi (1 <= wi <=$ 10^9 $) - the width of i-th announcement.

Output

For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h, starting with the top row. If an announcement can’t be put on the billboard, output “-1” for this announcement.

Examples

intput
1
2
3
4
5
6
3 5 5
2
4
3
3
3
output
1
2
3
4
5
1
2
1
3
-1

思路

线段树，对每行建树，叶子节点为此行剩余的宽度，维护区间最大值。

单点修改，单点查询。

每次贴广告就更新一次，如果根节点左右都放不下就返回-1。

代码

#include<bits/stdc++.h>
using namespace std;
#define CRL(a) memset(a,0,sizeof(a))
#define INF 0xfffffff
typedef unsigned long long LL;
typedef  long long ll;
const double Pi = acos(-1);
const double e = exp(1.0);
const int mod =1e9+7;
const int N=2*1e6+5;

int tr[N<<2];

void Built(int root,int l,int r,int c)
{
    if(l==r) tr[root]=c;
    else
    {
        int mid=l+r>>1;
        Built(root<<1,l,mid,c);
        Built(root<<1|1,mid+1,r,c);

        tr[root]=max(tr[root<<1],tr[root<<1|1]);
    }
}

int Update(int root,int l,int r,int c)
{
    int tem;
    if(l==r)
    {
        if(tr[root]<c) return -1;
        tr[root]-=c;
        return l;
    }
    else
    {
        int mid=l+r>>1;
        if(tr[root<<1]>=c)  tem=Update(root<<1,l,mid,c);
        else if(tr[root<<1|1]>=c) tem=Update(root<<1|1,mid+1,r,c);
        else
            return -1;
        tr[root]=max(tr[root<<1],tr[root<<1|1]);
    }
    return tem;
}

int main()
{
    int h,w,n,x;
    ios::sync_with_stdio(false);
    while(cin>>h>>w>>n)
    {
        h= h>n? n:h;
        Built(1,1,h,w);
        while(n--)
        {
            cin>>x;
            cout<<Update(1,1,h,x)<<endl;
        }
    }
    return 0;
}