题目描述
You are given an array a of length n consisting of zeros. You perform n actions with this array: during the i-th action, the following sequence of operations appears:
Choose the maximum by length subarray (continuous subsegment) consisting only of zeros, among all such segments choose the leftmost one;
Let this segment be [l;r]. If r−l+1 is odd (not divisible by 2) then assign (set) a[l+r2]:=i (where i is the number of the current action), otherwise (if r−l+1 is even) assign (set) a[l+r−12]:=i.
Consider the array a of length 5 (initially a=[0,0,0,0,0]). Then it changes as follows:
Firstly, we choose the segment [1;5] and assign a[3]:=1, so a becomes [0,0,1,0,0];
then we choose the segment [1;2] and assign a[1]:=2, so a becomes [2,0,1,0,0];
then we choose the segment [4;5] and assign a[4]:=3, so a becomes [2,0,1,3,0];
then we choose the segment [2;2] and assign a[2]:=4, so a becomes [2,4,1,3,0];
and at last we choose the segment [5;5] and assign a[5]:=5, so a becomes [2,4,1,3,5].
Your task is to find the array a of length n after performing all n actions. Note that the answer exists and unique.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1≤n≤2⋅105) — the length of a.
It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 (∑n≤2⋅105).
Output
For each test case, print the answer — the array a of length n after performing n actions described in the problem statement. Note that the answer exists and unique.
Example
input
6
1
2
3
4
5
6
output
1
1 2
2 1 3
3 1 2 4
2 4 1 3 5
3 4 1 5 2 6
题目大意
给你一个全为0的数组 要求把 1 到 n 插入数字，插入规则为：第i次选择一个左边最长的全0区间，将区间 (l+r)/2的位置变为 i。
题目分析
用优先队列维护一个区间集合。每次取出长度最大的区间。
将中点赋值，再从中点将这个区间分割为成两个，再放入队列中。
这样求出每个点的值来。
代码如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
int const N=2e5+5;
struct Node{
int l,r;
bool operator< (const Node a) const //定义比较准则
{
if(r-l==a.r-a.l) return l>a.l; //长度相等时，找靠前的
return r-l<a.r-a.l; //否则找长度大的
}
};
int ans[N]; //答案数组
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
memset(ans,0,sizeof ans); //清空数组
priority_queue<Node> heap;
heap.push({1,n});
int i=1; //第i次操作
while(heap.size())
{
Node t=heap.top();
heap.pop();
int len=t.r-t.l+1;
int k=(t.l+t.r)/2;
ans[k]=i; //将中点进行赋值
if(len==2) heap.push({t.r,t.r}); //长度为2，只需要放入后面的那一个点
else if(len>2) //长度大于2，则将区间分成两段
{
heap.push({t.l,k-1});
heap.push({k+1,t.r});
}
i++;
}
for(int i=1;i<=n;i++) //输出答案序列
printf("%d ",ans[i]);
puts("");
}
return 0;
}