CF 482A(Diverse Permutation-相邻距离不同数为k的1~n全排列构造)

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本文介绍了一个算法问题，目标是在给定长度n和差异元素数量k的情况下找到一个整数排列p，使得相邻元素之间的绝对差值恰好包含k个不同的数值。

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

1 10 2 9 3 8 |7 6 5 4 3 2

不同的。。。全为1的

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
#define MAXK (100000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,k;
int main()
{
//	freopen("CF482A.in","r",stdin);
//	freopen(".out","w",stdout);
	
	cin>>n>>k;k--;
	int l=1,r=n,b=1;
	while(l<=r)
	{
		
		if (b) printf("%d",l++);
		else printf("%d",r--);
		if (k) b^=1,k--; 
		
		if (l<=r) putchar(' ');
	}
	cout<<endl;
	return 0;
}