Permutation(构造+思维）

A permutation p is an ordered group of numbers p1,   p2,   ...,   pn, consisting of ndistinct 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

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.

 

题意：找1-n的数，组成

的式子，使得最后的结果为2*k

我们可以来构造数列解决：设前2*k个数对前项比后项多1，2*k到2*n个数前项比后项小1，则式子的结果为

k+（n-k）-|k-（n-k）因为题目中说2*k<=n所以2*k符合题目要求

代码：

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<queue> #include<stack> #include<vector> #include<map> #include<set> #define MAX 300005 using namespace std;typedef long long ll; int a[MAX]; int main() {int n,k;cin>>n>>k;for(int t=1;t<=2*k;t+=2){a[t]=t;a[t+1]=t+1;}for(int t=2*k+1;t<=2*n;t+=2){a[t]=t+1;a[t+1]=t;}for(int t=1;t<=2*n;t++){cout<<a[t]<<" ";}return 0;}

 

转载于:https://www.cnblogs.com/Staceyacm/p/10781800.html

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