Codeforces Round #209 B. Permutation

构造特定排列算法解析

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本文详细解析了一道关于构造特定排列的问题，通过分析题目要求，给出了一种有效的算法实现方式，利用数学公式和编程技巧，构造出了满足条件的数列。

A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.

Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: .

Input

The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).

Output

Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.

Examples

input

Copy

1 0

output

Copy

1 2

input

Copy

2 1

output

Copy

3 2 1 4

input

Copy

4 0

output

Copy

2 7 4 6 1 3 5 8

题意:

要你用1-2*n的数构造数列，构造的数列满足上面的公式。

思路：

这道题我是没思路，我是看了题解才会的，其实我们只要求一对差的绝对值为k的就好但必须是a2*j-1<aj,这是有公式决定的其余的从大到小构造就好，(k-(-k)=2k);

代码:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define LL long long

using namespace std;

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    if(k==0)
    {
        for(int i=2*n;i>=1;i--)
        {
            if(i!=1)
                printf("%d ",i);
            else
                printf("%d\n",i);
        }
    }
    else
    {
        int j=2*n-k;
        printf("%d %d ",j,2*n);
        for(int k=2*n-1;k>=1;k--)
        {
            if(k!=j)
            {
                if(k!=1)
                    printf("%d ",k);
                else
                    printf("%d\n",k);
            }

        }
    }
    return 0;
}