poj 2182 Lost Cows

Lost Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9646		Accepted: 6207

Description

N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood 'watering hole' and drank a few too many beers before dinner. When it was time to line up for their evening meal, they did not line up in the required ascending numerical order of their brands. 

Regrettably, FJ does not have a way to sort them. Furthermore, he's not very good at observing problems. Instead of writing down each cow's brand, he determined a rather silly statistic: For each cow in line, he knows the number of cows that precede that cow in line that do, in fact, have smaller brands than that cow. 

Given this data, tell FJ the exact ordering of the cows. 

Input

* Line 1: A single integer, N 

* Lines 2..N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on. 

Output

* Lines 1..N: Each of the N lines of output tells the brand of a cow in line. Line #1 of the output tells the brand of the first cow in line; line 2 tells the brand of the second cow; and so on.

Sample Input

5
1
2
1
0

Sample Output

2
4
5
3
1

Source

USACO 2003 U S Open Orange

敲一遍代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<stack>
using namespace std;
#define size 8000
struct Node{
    int l,r,snum;
};
Node node[size*4+5];
int fir[size+5],brand[size+5];
void build(int root,int l,int r){
    node[root].l=l;
    node[root].r=r;
    node[root].snum=r-l+1;
    if(l==r){
        return;
    }
    int mid=(l+r)/2;
    build(root*2,l,mid);
    build(root*2+1,mid+1,r);
}
int query(int root,int k){
    node[root].snum--;
    if(node[root].l==node[root].r){
        return node[root].l;
    }
    if(node[root*2].snum>=k){
        return query(root*2,k);//在根为root的树中可以使这个数成为第k个
    }
    else{
        return query(root*2+1,k-node[root*2].snum);
    }
}
int main(){
    //freopen("D:\INPUT.txt","r", stdin);
    int n;
    while(scanf("%d",&n)!=EOF){
        build(1,1,n);
        int i=2;
        fir[1]=0;
        for(;i<=n;i++) scanf("%d",&fir[i]);
        for(i=n;i>=1;i--){
            brand[i]=query(1,fir[i]+1);
        }
        for(i=1;i<=n;i++){
            cout<<brand[i]<<endl;
        }
    }
    return 0;
}