Permutation CodeForces 359B

Permutation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

input

1 0

output

1 2

input

2 1

output

3 2 1 4

input

4 0

output

2 7 4 6 1 3 5 8

Note

Record |x| represents the absolute value of number x. 

In the first sample |1 - 2| - |1 - 2| = 0.

In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.

In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.

当两个相邻的数是递增的时候，这两个数对K是没有影响的。

所以只要找到几个递减的数对，他们的差之和等于K就行了。

#include

#include

#include

using namespace std;

int main()

{

   int n, k;

   int sum = 0;

    scanf("%d%d", &n, &k);

   int l = 1;

   int r = 2 * n;

   int mid = 0;

    

   while ((r - l) <= k)

    {

        k = k - (r - l);

        printf("%d ", r);

        r++;

        printf("%d ", l);

        l++;

        sum++;

       if (sum < n)printf(" ");

       else printf("\n");

    }

   if (k != 0)

    {

        mid = r-k;

        k = 0;

        printf("%d ", r);

        r--;

        printf("%d", mid);

        sum++;

       if (sum < n)printf(" ");

       else printf("\n");

    }

   while (l < r)

    {

       if (l == mid) l++;

       if (r == mid) r--;

        printf("%d %d", l, r);

        sum++;

       if (sum < n)printf(" ");

       else printf("\n");

        l++;

        r--;

    }

    return 0;

    

}