poj 2182 Lost Cows

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

这题可以直接用循环做，也可以用线段树，开一个结构体，记录l,r,sum,len(表示此区域内还有几个数），先找左子树，如果左子树含有的数大于要找的数的编号，那么在左子树寻找，否则在右子树寻找，并减去左子树的个数。

#include<stdio.h>
#define maxn 8005
int a[maxn],c[maxn];
struct node
{
	int len,l,r,sum;
}b[4*maxn];

void build(int l,int r,int i)
{
	int mid;
	b[i].l=l;
	b[i].r=r;
	b[i].len=r-l+1;
	b[i].sum=0;
	if(b[i].l==b[i].r)
	{
		b[i].sum=l;
		return;
	}
	mid=(l+r)/2;
	build(l,mid,i*2);
	build(mid+1,r,i*2+1);
}

int question(int num,int i)
{
	int mid;
	b[i].len--;
	if(b[i].l==b[i].r)
	return b[i].sum;
	if(b[i*2].len>=num)
	return question(num,i*2);
	else 
	return question(num-b[i*2].len,i*2+1);
}

int main()
{
	int m,n,i;
	while(scanf("%d",&n)!=EOF)
	{
		build(1,n,1);
		for(i=1;i<=n-1;i++)
		{
			scanf("%d",&a[i]);
		}
		for(i=n-1;i>=1;i--)
		{
			c[i]=question(a[i]+1,1);
		}
		c[0]=question(1,1);
		for(i=0;i<=n-1;i++)
		{
			printf("%d\n",c[i]);
		}
	}
	return 0;
}