結果

問題 No.1061 素敵な数列
コンテスト
ユーザー Sumitacchan
提出日時 2020-05-23 00:31:48
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 5,019 bytes
コンパイル時間 2,088 ms
コンパイル使用メモリ 171,692 KB
実行使用メモリ 10,496 KB
最終ジャッジ日時 2024-10-06 00:36:57
合計ジャッジ時間 6,693 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 2 TLE * 1 -- * 30
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
using namespace std;
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename T> using gpp_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename L> using gpp_map = tree<T, L, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using gpp_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;*/
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define bit_exist(x, n) ((x >> n) & 1)
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) { cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl; }
#define mdebug(m) { cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} }
#define Return(ans) { cout << (ans) << endl; return 0; }
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p){ cout << '(' << p.first << ',' << p.second << ')'; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using Pdi = pair<double, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;
//#define Add(x, y) x = (x + (y)) % mod
//#define Mult(x, y) x = (x * (y)) % mod

void calcans(int i0, int N, bool f, vec &ans){
    if(N <= 6){
        if(f){
            vec p(3 * N);
            REP(i, 3 * N) p[i] = i;
            do{
                bool g = true;
                REP(i, N){
                    int a = p[3 * i], b = p[3 * i + 1], c = p[3 * i + 2];
                    if(!(a < b && b < c && llabs((c - b) - (b - a)) == i)) g = false;
                }
                if(g) break;
            }while(next_permutation(all(p)));
            REP(i, 3 * N) ans[i0 + p[i]] = i / 3;
        }else{
            vec p(3 * N);
            p[0] = 0;
            FOR(i, 1, 3 * N) p[i] = i + 1;
            do{
                bool g = true;
                REP(i, N){
                    int a = p[3 * i], b = p[3 * i + 1], c = p[3 * i + 2];
                    if(!(a < b && b < c && llabs((c - b) - (b - a)) == i)) g = false;
                }
                if(g) break;
            }while(next_permutation(all(p)));
            REP(i, 3 * N) ans[i0 + p[i]] = i / 3;
        }

        return;
    }

    if(N % 2 == 1){
        int n = (N + 1) / 2;
        int M = i0 + 3 * N - 1;
        if(!f) M++;
        REP(i, n){
            ans[M - (2 * i)] = N - 1 - i;
            ans[M - (2 * i + 1)] = N - 1 - i;
            ans[M - (2 * n + i)] = N - 1 - i;
        }
        calcans(i0, N - n, f, ans);
    }else if(f){
        int n = N / 2;
        REP(i, n - 1){
            ans[i0 + i] = N - n + i;
            ans[i0 + n + 2 * i + 1] = N - n + i;
            ans[i0 + n + 2 * i + 2] = N - n + i;
        }
        ans[i0 + n - 1] = N - 1;
        ans[i0 + n] = N - 1;
        ans[i0 + 3 * n] = N - 1;
        calcans(i0 + 3 * n - 1, N - n, false, ans);
    }else{
        int n = N / 2;
        ans[i0] = N - 1;
        ans[i0 + N] = N - 1;
        ans[i0 + N + 1] = N - 1;
        REP(i, n - 1){
            ans[i0 + 2 * i + 2] = N - 2 - i;
            ans[i0 + 2 * i + 3] = N - 2 - i;
            ans[i0 + N + 2 + i] = N - 2 - i;
        }
        calcans(i0 + 3 * n, N - n, true, ans);
    }
}

signed main(){

    int N; cin >> N;
    if(N == 2){
        Return(-1);
    }
    vec ans(3 * N, -1);
    calcans(0, N, true, ans);
    REP(i, 3 * N) cout << ans[i] << " ";
    cout << endl;

    return 0;
}
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