Merge Sort

这篇博客介绍了归并排序算法及其调用次数的计算。在给定一个整数n和调用次数k的情况下,需要找到一个n大小的排列数组,使得在排序过程中归并排序恰好被调用k次。文章通过分析指出,由于每次调用后会成对地产生新的调用,因此k必须为奇数。然后提出了一个深度优先搜索的策略来构造满足条件的数组,通过交换中点元素或整体交换两侧元素来保持两部分有序。当调用次数达到k或者区间不能再分时,返回构造的数组。

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Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array a with indices from [l, r) can be implemented as follows:

  1. If the segment [l, r) is already sorted in non-descending order (that is, for any i such that l ≤ i < r - 1 a[i] ≤ a[i + 1]), then end the function call;
  2. Let ;
  3. Call mergesort(a, l, mid);
  4. Call mergesort(a, mid, r);
  5. Merge segments [l, mid) and [mid, r), making the segment [l, r) sorted in non-descending order. The merge algorithm doesn't call any other functions.

The array in this problem is 0-indexed, so to sort the whole array, you need to call mergesort(a, 0, n).

The number of calls of function mergesort is very important, so Ivan has decided to calculate it while sorting the array. For example, if a = {1, 2, 3, 4}, then there will be 1 call of mergesortmergesort(0, 4), which will check that the array is sorted and then end. If a = {2, 1, 3}, then the number of calls is 3: first of all, you call mergesort(0, 3), which then sets mid = 1 and calls mergesort(0, 1) and mergesort(1, 3), which do not perform any recursive calls because segments (0, 1) and (1, 3) are sorted.

Ivan has implemented the program that counts the number of mergesort calls, but now he needs to test it. To do this, he needs to find an array a such that a is a permutation of size n (that is, the number of elements in a is n, and every integer number from [1, n] can be found in this array), and the number of mergesort calls when sorting the array is exactly k.

Help Ivan to find an array he wants!

Input

The first line contains two numbers n and k (1 ≤ n ≤ 100000, 1 ≤ k ≤ 200000) — the size of a desired permutation and the number of mergesort calls required to sort it.

Output

If a permutation of size n such that there will be exactly k calls of mergesort while sorting it doesn't exist, output  - 1. Otherwise output n integer numbers a[0], a[1], ..., a[n - 1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.

Examples

Input

Copy

3 3

Output

Copy

2 1 3 

Input

Copy

4 1

Output

Copy

1 2 3 4 

Input

Copy

5 6

Output

Copy

-1

 

考虑模拟该归并排序的过程。

第一次调用merge_sort(0,n),之后不管怎样哪一部分有序无序的,调用都是成对出现的,故初始的k必须为奇数,若为偶数,一定不可行。

设计dfs函数是一个难点。不能确定每一部分究竟是有序还是无序。

那么我们开始时默认已排好序,如果当前调用次数cnt还不足k的话,就把当前序列搞成无序(但两部分均有序),继续向下递归。

把当前有序序列搞成整体无序但两部分均有序的方法:①把中点m两旁的元素换一下就好了②左右两边整体调换一下也行

最后cnt==k或者区间已经不能再分时返回。

#include<bits/stdc++.h>
using namespace std;

int n,k,a[100000+1000],cnt;

void dfs(int l,int r)   //  [  ,  )
{
	if(r-l==1||cnt==k)return;
	int m=(l+r)/2;
	swap(a[m-1],a[m]);
	cnt+=2;
	dfs(l,m);
	dfs(m,r);
}

int main()
{
//	freopen("input.in","r",stdin);
	scanf("%d%d",&n,&k);
	for(int i=0;i<n;i++)a[i]=i+1;
	cnt++;
	if(!(k&1))cout<<-1<<endl;
	else
	{
		dfs(0,n);
		if(cnt<k)cout<<-1<<endl;
		else for(int i=0;i<n;i++)cout<<a[i]<<" ";
	}
	return 0;
}

 

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