Lost Cows
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 402 Accepted Submission(s): 203
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.
* 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.
#include<stdio.h>
#define maxn 500004
int a[],s[],n;
struct node
{
int left,right;
int num;
};
node tree[*maxn];
void build(int left,int right,int i)
{
tree[i].left =left;
tree[i].right=right;
tree[i].num =right-left+;
if(tree[i].left ==tree[i].right )
return ;
build(left,(left+right)/,*i);
build((left+right)/+,right,*i+);
}
int insert(int i,int r)
{
tree[i].num--;
if(tree[i].left==tree[i].right)
return tree[i].left ;
if(tree[*i].num>=r)
insert(i*,r);
else
insert(i*+,r-tree[*i].num);
}
int main()
{
int i;
while(~scanf("%d",&n))
{
a[]=;
for(i=;i<=n;i++)
scanf("%d",&a[i]);
build(,n,);
for(i=n;i>=;i--)
s[i]=insert(,a[i]+);
for(i=;i<=n;i++)
printf("%d\n",s[i]); }
return ;
}
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