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| #include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
using namespace std;
struct PairHash {inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }};
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);
using ll = long long;
using ld = long double;
const ld PI = 3.14159265358979323846;
const ll INF = 1e18;
const ld EPS = 1e-9;
const ll MAX_N = 101010;
const ll mod = 1e9 + 7;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<pll, ll, PairHash> umpll;
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define sz(x) ll((x).size())
#define pyes cout<<"Yes\n";
#define pno cout<<"No\n";
#define pneg1 cout<<"-1\n";
#define CASE(x) cout<<"Case #"<<x<<": ";
template <typename T>
void print(T &&t) { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<" ";cout<<"\n";}
ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
ll __lcm(ll x, ll y) { return x * y / __gcd(x,y); }
ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }
vll solve(ll n, ll k) {
vll res(n*2);
ll now = 2 * n;
if(k) {
res[0] = 2 * n - k, res[1] = 2 * n;
}
rep(i,k ? 2 : 0, 2 * n) {
while(now == res[0] or now == res[1]) now--;
res[i] = now--;
}
return res;
}
int main() {
Code By Sumfi
cout.precision(12);
ll tc = 1;
rep(i,1,tc+1) {
ll n,k;
cin>>n>>k;
printv(solve(n,k));
}
return 0;
}
|
