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

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

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

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

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

用n个数1~n，每一个数仅仅能用一次。组成差值的绝对值有k个数。为1~k。

构造题，我是这样构造的，取前k+1个数。第一个数取1，先+k。后一个数-(k-1),在后一个数+k-2.......这样从两头往

中间靠拢。既取完了k+1个数。又构造了1~k的差值绝对值，至于k+1后的嘛，每次+1即可了。

代码：

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cstring>

using namespace std;

const int maxn=100000+1000;

int ans[maxn];

int main()

{

    int n,k;

    ans[1]=1;

    scanf("%d%d",&n,&k);

    if(k==1)

    {

        for(int i=1;i<=n;i++)

        ans[i]=i;

    }

    else

    {

      for(int i=2;i<=k+1;i++)

      {

        if(i%2)

        ans[i]=ans[i-1]-(k-i+2);

        else

        ans[i]=ans[i-1]+(k-i+2);

      }

      int cur=1;

      for(int i=k+2;i<=n;i++)

      {

          ans[i]=k+1+cur;

          cur++;

      }

    }

    for(int i=1;i<n;i++)

    printf("%d ",ans[i]);

    printf("%d\n",ans[n]);

    return 0;

}