Codeforces 483C Diverse Permutation【思维构造】

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#Codeforces 483C

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Permutation p is an ordered set of integers p₁, p₂, ..., p_n, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p₁, p₂, ..., p_n.

Your task is to find such permutation p of length n, that the group of numbers |p₁ - p₂|, |p₂ - p₃|, ..., |p_n - 1 - p_n| has exactly k distinct elements.

Input

The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 10⁵).

Output

Print n integers forming the permutation. If there are multiple answers, print any of them.

Examples

Input

3 2

Output

1 3 2

Input

3 1

Output

1 2 3

Input

5 2

Output

1 3 2 4 5

Note

By |x| we denote the absolute value of number x.

题目大意：

让你构造出来一个长度为N的全排列，使得|p₁ - p₂|, |p₂ - p₃|, ..., |p_n - 1 - p_n| 一共有K种数字。

思路：

在纸上多列出几种情况，不难发现我们目标结果可以为：

A1，A2，A3，A4，A5，A6,.........................

1，A1+k，A2-(k-1)，A3+(k-2).....................当k变成0的时候，再之后我们设定Ai=Ai-1+1即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int ans[600000];
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        int maxn=1;
        int tmp=k;
        ans[1]=1;
        for(int i=2;i<=n;i++)
        {
            if(i<=k+1)
            {
                if(i%2==0)ans[i]=ans[i-1]+tmp;
                else ans[i]=ans[i-1]-tmp;
                tmp--;
                maxn=max(maxn,ans[i]);
            }
            if(i==k+2)ans[i]=maxn+1;
            if(i>k+2)ans[i]=ans[i-1]+1;
        }
        for(int i=1;i<=n;i++)printf("%d ",ans[i]);
        printf("\n");
    }
}