題意：

有長度為 n 的數組 a ，全為 0，接下來循環 n 次，每次選出一段最長的連續區間 [l, r]（全為 0 ，如果一樣長，就選最左邊的)。

如果 r?l+1 是奇數，那么 $a[\frac{l+r}{2}]=i$;

否則，$a[\frac{l+r?1}{2}]=i$;（i 是第幾輪循環）。

輸出最終的數組 a。

題目：

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) $a[\frac{l+r}{2}]=i$(where i is the number of the current action), otherwise (if r?l+1 is even) assign (set) $a[\frac{l+r?1}{2}]=i$.
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

分析：

我們可以直接暴力去做。即我們每次選出符合條件的 [l, r]，然后對應的給 a[i] 賦值，又得到了兩個新的更小的區間，我們需要存儲下來，并按照上述的規則對所有的區間排序。顯然優先隊列可以完美的滿足我們的要求。優先隊列的BFS。每次處理完一段區間后就把這段區間拆分，丟進優先隊列里就好。注意priority_queue本身是一個大根二叉堆，所以重載運算符時符號要反一下
（或者按照藍書講的 把len換成相反數）。
#AC代碼：

#include<bits/stdc++.h> using namespace std; typedef long long ll; const int M=2e5+10; int s[M]; int t,n,tot; struct node {int l,r,d; }; bool operator<(const node &a,const node &b) {if(a.d<b.d)return 1;else if(a.d==b.d)if(a.l>b.l)return 1;return 0; } int main() {scanf("%d",&t);while(t--){tot=0;memset(s,0,sizeof(s));scanf("%d",&n);priority_queue<node>q;node u,v;u.l=1,u.r=n,u.d=u.r-u.l+1;q.push(u);while(!q.empty()){u=q.top();q.pop();int x=u.l;int y=u.r;int mid=(x+y)>>1;s[mid]=++tot;//printf("%d****\n",s[mid]);v.l=x,v.r=mid-1,v.d=v.r-v.l+1;if(v.l<=v.r)q.push(v);v.l=mid+1,v.r=y,v.d=v.r-v.l+1;if(v.l<=v.r)q.push(v);}for(int i=1;i<=n;i++)printf("%d ",s[i]);printf("\n");}return 0; }

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