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.

 

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.

 

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.

 

 

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <algorithm>

#define LL long long

using namespace std;

int h, w, n;

//线段树

//区间某点增值，求区间最大值

const int maxn = ;

struct node

{

    int lt, rt;

    int val;

}tree[*maxn];

//向上更新

void pushUp(int id)

{

    tree[id].val = max(tree[id<<].val, tree[id<<|].val);

}

//建立线段树

void build(int lt, int rt, int id)

{

    tree[id].lt = lt;

    tree[id].rt = rt;

    tree[id].val = ;//每段的初值，根据题目要求

    if (lt == rt)

    {

        tree[id].val = w;

        return;

    }

    int mid = (lt + rt) >> ;

    build(lt, mid, id<<);

    build(mid + , rt, id<<|);

    pushUp(id);

}

//查询某段区间内的符合p的

int query(int lt, int rt, int id, int p)

{

    if (tree[id].val < p)

        return -;

    if (tree[id].lt == tree[id].rt)

    {

        tree[id].val -= p;

        return tree[id].lt;

    }

    int tmp;

    tmp = query(lt, rt, id<<, p);

    if (tmp != -)

    {

        pushUp(id);

        return tmp;

    }

    tmp = query(lt, rt, id<<|, p);

    pushUp(id);

    return tmp;

}

void work()

{

    int k, ans;

    int len = min(h, n);

    build(, len, );

    for (int i = ; i < n; ++i)

    {

        scanf("%d", &k);

        ans = query(, len, , k);

        printf("%d\n", ans);

    }

}

int main()

{

    //freopen("test.in", "r", stdin);

    while (scanf("%d%d%d", &h, &w, &n) != EOF)

    {

        work();

    }

    return ;

}