codeforces 482A Diverse Permutation 贪心

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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 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.

           求一个长度为n的全排列,使得相邻两数之间的差值的绝对值一共出现k个值

           贪心策略题,前k个数不断间隔着从两头取,剩余的n-k个数构成差值为1的等差数列即可

python3:

s = input()
arr = s.split()
n = int(arr[0])
k = int(arr[1])

print(1,end='')
for i in range(2,k+1):
	if i % 2 == 0:
		print('',n - i//2 + 1,end='')
	else:
		print('',(i+1)//2,end='')

if k % 2 == 0:
	x = n - k//2
	for i in range(k,n):
		print('',x,end='')
		x = x - 1
else:
	x = (k + 1)//2 + 1
	for i in range(k,n):
		print('',x,end='')
		x = x + 1
print()

/******************************************************
 * File Name:   a.cpp
 * Author:      kojimai
 * Create Time: 2014年10月25日 星期六 12时44分05秒
******************************************************/

#include
#include
#include
#include
#include
using namespace std;
int judge(int n,int cnt) {
	if(cnt%2==0)
		return n-(cnt/2)+1;
	else
		return (cnt+1)/2;
}
void solve(int x,int n)
{
	if(x%2 == 0)
		for(int l = n - x/2; x < n; l--,x++)
			printf(" %d",l);
	else
		for(int l = (x+1) / 2 + 1; x < n; l++,x++)
			printf(" %d",l);
}
int main()
{
	int n,k;
	scanf("%d%d",&n,&k);
	cout<<1;
	for(int i=2;i<=k;i++)
	{
		printf(" %d",judge(n,i));
	}
	solve(k,n);
	cout<