C. Diverse Permutation
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

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.

Sample test(s)
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.

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。

输出任一个方案。

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;
}