第十九次codeforces竞技结束 #275 Div 1

A. 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 n 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.

吾辈的Div1的第一场，只做出来了A题（也就是Div2的C），19Min出的名次为

我采取的是来回弹的方式~ 咱们1、1+k、2、k、3、k-1…… 这样是不是就是k、k-1、k-2……了呀^_^，等k一路到1之后，把没用到的数字按顺序排一遍~

这样为啥成呢~

首先，这样一定可以保证k一路到1的绝对值都存在，请使用数学归纳法~

其次，为啥按顺序把剩下的写出来不会超出k呢？ 因为我们是从1和k+1两侧向中间走，所以最后剩一个数字和他后头那个数字的差距是k/2，依旧可行。

代码如下：

#include <cmath> 
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))

bool cmp(const int a, const int b)
{
	return a > b;
}

bool usd[100086];

int main()
{
	int n,k,f=1,now;	
	scanf("%d%d",&n,&k);
	memset(usd,false,sizeof usd);
	for(int i=1;i<=k;i++)
	{
		if(f) now=(i+1)/2;
		else now=2+k-i/2;
		f=1-f;
		if(i-1)	printf(" %d",now);
		else printf("%d",now);
		usd[now]=true;
	}
	//cout<<"then"<<endl;
	for(int i=1;i<=n;i++)
	{
		if(usd[i]==false)
		printf(" %d",i);
	}
	return 0;
}