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.

1~n,乱序排列，告诉每个位置的前面的数字中比它小的数的个数，求每个位置的数字是多少

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 i

  n slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in sl

  ot #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

---

一个比较有趣的题目，线段树里面维护当前区间内还有多少个数，每次从找出原序列最后一个数（因为不会对其他数造成影响）

至于怎么找数？在线段树里找到第k大的数，这是非常简单的

（ps：这题除了用到线段树思想外，就和线段树没有任何关系了……）

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=8e3;
int tree[N*3+10],a[N+10],ans[N+10];
inline int read(){
    int x=0,f=1;char ch=getchar();
    for (;ch<'0'||ch>'9';ch=getchar())    if (ch=='-')    f=-1;
    for (;ch>='0'&&ch<='9';ch=getchar())  x=x*10+ch-'0';
    return x*f;
}
#define ls (p<<1)
#define rs (p<<1|1)
void build(int p,int l,int r){
    tree[p]=r+1-l;
    if (l==r)   return;
    int mid=(l+r)>>1;
    build(ls,l,mid),build(rs,mid+1,r);
}
int query(int p,int l,int r,int t){
    tree[p]--;//记得减掉
    if (l==r)   return l;
    int mid=(l+r)>>1;
    return t<=tree[ls]?query(ls,l,mid,t):query(rs,mid+1,r,t-tree[ls]);//两边判断查询
}
int main(){
    int n=read();
    build(1,1,n);
    for (int i=2;i<=n;i++)   a[i]=read();
    for (int i=n;i;i--)     ans[i]=query(1,1,n,a[i]+1);//a[i]+1是其本身的位置
    for (int i=1;i<=n;i++)   printf("%d\n",ans[i]);
    return 0;
}