Codeforces 482A diverse permutation greedy

Portal: CF 482A

A. Diverse permutation time limit per test 1 second memory limit per test megabytes input standard input output Standa RD output

Permutation P is an ordered set of integers p1, p2, ..., PN, consisting of n distinct positive integers not larger t Han 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 the space-separated positive integers n, K (1≤k < n≤105). Output

Print n integers forming the permutation. If There is 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.

A full arrangement of length n, which causes the absolute value of the difference between two adjacent numbers to appear in total K values

Greedy strategy problem, the number of the former K constantly spaced from both ends, the remaining number of n-k constitute a difference of 1 arithmetic progression can be

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 RA Nge (k,n):
		print (", x,end=")
		x = x + 1
print ()

/******************************************************
 * File Name:   a.cpp
 * Author:      kojimai
 * Create time:2014 October 25 Saturday 12:44 05 seconds
******************************************************/

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
# Include<iostream>
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<<endl;
	return 0;
}