B. Permutation：题目

很容易就发现，按顺序正好等于0。把i和i+2换一次可以加2，所以k是多少就换几次

#include <bits/stdc++.h>
using namespace std;
#define int long long
vector<int> a((int)6e5);
vector<int> b((int)6e5), c[(int)6e5];
const int mod = 1e9 + 7;
signed main()
{int n, k;cin >> n >> k;for (int i = 1; i <= n * 2; i++)a[i] = i;for (int i = 2; i <= n * 2; i += 4){if (k == 0)break;swap(a[i], a[i + 2]);k--;}for (int i = 1; i <= n * 2; i++)cout << a[i] << " ";
}