CodeForces 483C Diverse Permutation

Diverse Permutation

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

Permutationp 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 Input

Input
3 2

Output
1 3 2

Input
3 1

Output
1 2 3

Input
5 2

Output
1 3 2 4 5

Hint

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

题意：有一个p1,   p2,   …,   pn（代表1–n）的序列，将里面的序列重新排列，使其满足 |p1 - p2|, |p2 - p3|, …, |pn - 1 - pn|出现k种植，k种值的意思就是所有的|pi - pi-1|的值必须包含1–k种的每一个值。
这下就可以想如果要出现k，就可以用k+1减去1，要得k-1，用k+1减去2，因为是绝对值，所以可以写成，1，k+1，2。得到k-2就用2减去k(绝对值不用管顺序)，就变成1,k+1,2,k，重复这个过程就能得到满足要求的序列。相当于在1到k+1,从左边1和右边k+1开始，一边取一个数字重新排成一个序列，1，k+1, 2 , k , 3 , k-1 …… 然后后面还有n-k-1个数字，因为已经满足题目所给要求，所以只需要顺序输出（相邻的数字差值为1）。

#include"stdio.h"
#include"iostream"
#include"algorithm"
#include"string.h"

using namespace std;

bool vis[100010];

int main(void)
{
    int n,k;
    while(scanf("%d%d",&n,&k) !=EOF)
    {
        memset(vis,false,sizeof(vis));
        printf("1");
        int x = 1;
        int t = 1;
        for(int i = k;i > 0;i--)
        {
            if(t == 1)
            {
                printf(" %d",x+i);
                x = x+i;
            }
            else
            {
                printf(" %d",x-i);
                x = x-i;
            }
            t = 1-t;
        }
        for(int i = k+2;i <= n;i++)
            printf(" %d",i);
        printf("\n");
    }
    return 0;
}