Problem Description
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.
 

Sample Input
3 5 5 2 4 3 3 3
 

Sample Output
1 2 1 3 -1
 

Author
hhanger@zju
 


靠左靠上贴长布条的那个题==和之前做的线段树还是有小区别的~~建树是在输入前，故无需更新递归的值，因为之前赋值过了，update里面不能只写当左等于右时才返回！必须再用一个变量

#include <iostream>
#include<cstdio>
#include<cstring>
#define maxn 200000
using namespace std;
struct bill
{
     int l,r,ans;
}num[maxn*4];
int m;
int w,h,n;
void build(int L,int R,int root)
{
     num[root].l=L;
     num[root].r=R;
     num[root].ans=w;
     if(L==R) return;
     build(L,(L+R)/2,root<<1);
     build((L+R)/2+1,R,root<<1|1);
    //
}
int update(int root,int m)
{
     if(num[root].ans<m) return -1;
     if(num[root].l==num[root].r)
     {
          num[root].ans-=m;
          return num[root].l;
     }
     int ret;
     if(num[root<<1].ans>=m) ret=update(root<<1,m);
     else ret=update(root<<1|1,m);
     num[root].ans=max(num[root<<1].ans,num[root<<1|1].ans);
     return ret;
}
int main()
{
    //freopen("data.in.txt","r",stdin);
    while(~scanf("%d%d%d",&h,&w,&n))
    {
         build(1,min(n,h),1);
         for(int i=1;i<=n;i++)
         {
              scanf("%d",&m);
              printf("%d\n",update(1,m));
         }
    }
    return 0;
}

现在看之前的代码真是又短又简单啊==也说明有进步了吧