Permutationp is an ordered set of integersp₁,???p₂,???...,???p_n, consisting ofn distinct positive integers not larger thann. We‘ll denote as n the length of permutation p₁,???p₂,???...,???p_n.

Your task is to find such permutationp of lengthn, that the group of numbers|p₁?-?p₂|,?|p₂?-?p₃|,?...,?|p_n?-?1?-?p_n| has exactly k distinct elements.

The single line of the input contains two space-separated positive integersn,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 numberx.

题解：

     本题要求给定1到n的序列，满足相邻两项之差的绝对值不相同的个数为k。由于给定的1?≤?k?<?n?≤?10⁵ 范围较大，所以只能寻找时间复杂度为O(n)的算法。可以想到该序列最多有n-1个不同的相邻差（绝对值），其中一个满足条件的序列是：n，1，n-1，2，n-3，3…………。可以尝试构造满足条件的前k-1，然后后面的顺序填写。

Java源代码：

import java.util.Scanner;

public class Main {

	private int k, n;

	public Main(int tn, int tk) {
		n = tn;
		k = tk;
	}

	public void printMain() {
		boolean flag = true;
		int count = k - 1;
		int i = 1;
		while (count > 0) {
			if (count != k - 1) {
				--count;
				if (count <= 0) {
					flag = false;
					break;
				}
				System.out.print(" ");
			}

			System.out.print(i);
			System.out.print(" " + (n - i + 1));
			++i;
			--count;
		}
		int j = n - i + 1;
		if (count != k - 1 && i <= j)
			System.out.print(" ");
		if (flag) {
			while (i <= j) {
				System.out.print(j);
				--j;
				if (i <= j)
					System.out.print(" ");
			}
		} else {
			while (i <= j) {
				System.out.print(i);
				++i;
				if (i <= j)
					System.out.print(" ");
			}
		}
		System.out.println();
	}

	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		while (input.hasNext()) {
			int tn = input.nextInt();
			int tk = input.nextInt();
			Main AC = new Main(tn, tk);
			AC.printMain();
		}
		input.close();
	}
}

       由于时间和水平的原因，难免有不足的地方，欢迎斧正！