結果

問題 No.2496 LCM between Permutations
ユーザー Astral__
提出日時 2023-10-06 22:16:05
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
RE  
実行時間 -
コード長 9,315 bytes
コンパイル時間 2,248 ms
コンパイル使用メモリ 213,228 KB
最終ジャッジ日時 2025-02-17 05:14:39
ジャッジサーバーID
（参考情報） 		judge3 / judge3
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 * 1
other AC * 26 RE * 2
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コンパイルメッセージ
main.cpp: In function ‘void solve()’:
main.cpp:302:23: warning: ‘tmp’ may be used uninitialized [-Wmaybe-uninitialized]
  302 |     rres.push_back(P(a(i, tmp), i));
      |                      ~^~~~~~~~
main.cpp:287:8: note: ‘tmp’ was declared here
  287 |     ll tmp;
      |        ^~~
main.cpp:318:9: warning: ‘pre’ may be used uninitialized [-Wmaybe-uninitialized]
  318 |     B.at(pre) = p;
      |     ~~~~^~~~~
main.cpp:288:8: note: ‘pre’ was declared here
  288 |     ll pre;
      |        ^~~
main.cpp:237:23: warning: ‘tmp’ may be used uninitialized [-Wmaybe-uninitialized]
  237 |     rres.push_back(P(a(tmp, i), i));
      |                      ~^~~~~~~~
main.cpp:218:6: note: ‘tmp’ was declared here
  218 |   ll tmp;
      |      ^~~
main.cpp:254:9: warning: ‘pre’ may be used uninitialized [-Wmaybe-uninitialized]
  254 |     A.at(pre) = p;
      |     ~~~~^~~~~
main.cpp:219:6: note: ‘pre’ was declared here
  219 |   ll pre;
      |      ^~~

ソースコード

diff # 

#include<bits/stdc++.h>
#define PPque priority_queue<tuple<ll, ll, ll>, vector<tuple<ll, ll, ll>>, greater<tuple<ll, ll, ll>>>
#define Pque  priority_queue<pair<ll, ll>, vector<pair<ll, ll>>, greater<pair<ll, ll>>>
#define pque priority_queue<ll, vector<ll>, greater<ll>>
#define umap unordered_map
#define uset unordered_set
#define rep(i, s, f) for(ll i = s; i <= f; i++)
#define per(i, s, f) for(ll i = s; i >= f; i--)
#define all0(x) (x).begin() ,(x).end()
#define all(x) (x).begin() + 1, (x).end()
#define vvvvl vector<vvvl>
#define vvvi vector<vector<vector<int>>>
#define vvvl vector<vector<vector<ll>>>
#define vvvc vector<vector<vector<char>>>
#define vvvd vector<vector<vector<double>>>
#define vvvm vector<vector<vector<mint>>>
#define vvi vector<vector<int>>
#define vvl vector<vector<ll>>
#define vvs vector<vector<string>>
#define vvc vector<vector<char>>
#define vvp vector<vector<pair<ll, ll>>>
#define vvb vector<vector<bool>>
#define vvd vector<vector<double>>
#define vvm vector<vector<mint>>
#define vp vector<pair<ll, ll>>
#define vi vector<int>
#define vl vector<ll>
#define vs vector<string>
#define vc vector<char>
#define vb vector<bool>
#define vd vector<double>
#define vm vector<mint>
#define P pair<ll, ll>
#define TU tuple<ll, ll, ll>
#define rrr(l, r) mt()%(r-l+1)+l
#define ENDL '\n'

#define ull unsigned long long
typedef long long ll;
using namespace std;

////////////////////////////////////////////////////////////////////////////////////////////////////////////
//これが本当の組み込み関数ってね（笑）

template <typename T>
T or_less(vector<T> &A, T x) { //x以下で最大要素の添字 前提: sort済み 存在しない: -1
  return distance(A.begin(), upper_bound(A.begin(), A.end(), x)-1);
}
template <typename T>
T under(vector<T> &A, T x) { //x未満の最大要素の添字　前提: sort済み 存在しない: -1
  return distance(A.begin(), lower_bound(A.begin(), A.end(), x)-1);
}
template <typename T>
T or_more(vector<T> &A, T x) { //x以上で最小要素の添字　　前提: sort済み 存在しない: N .   //distanceのA.beginは添字を出すために常にA.begin() NG: A.begin() + 1
  return distance(A.begin(), lower_bound(A.begin(), A.end(), x));
}
template <typename T>
T over(vector<T> &A, T x) { //xより大きい最小要素の添字前提: sort済み 存在しない: N
  return distance(A.begin(), upper_bound(A.begin(), A.end(), x));
}
void compress(vector<ll> &A) {//小さい順に順位、大きい順にしたいならreverseはNG最後に変換 //重複無し //最初は番兵
    vector<ll> temp = A;
    sort(temp.begin()+1, temp.end());
    temp.erase(unique(temp.begin()+1, temp.end()), temp.end());
    for (int i = 1; i <= int(A.size()-1); i++) {
        A.at(i) = distance(temp.begin(), lower_bound(temp.begin()+1, temp.end(), A.at(i)));
    }
}
template <typename T>
void UNIQUE(vector<T> &A) {
    sort(all0(A));
  return A.erase(unique(A.begin(), A.end()), A.end());
}
template <typename T>
void rev90(vector<vector<T>> &A, int indexed = 1) {
  reverse(A.begin() + indexed, A.end());
  int n = A.size();
  rep(i, indexed, n-1) {
   rep(j, i+1, n-1) {
      swap(A.at(i).at(j), A.at(j).at(i));
    }
  }
}
void chmin(ll &a, ll b) {
  a = min(a, b);
}
void chmax(ll &a, ll b) {
  a = max(a, b);
}
//////////////////////////////////////////////////////////////////////
//数学系
///////////////////////////////////////////////////////////////////////
ll round(ll x, ll i) {return ll(x + 5 * pow(10, i-1))/ll(pow(10, i)) * ll(pow(10, i));}
vp insu_bunkai(ll N) {
  vp res;
  for (int i = 2; i * i <= N; i++) {
    ll cnt = 0;
    while(N % i == 0) {
      cnt++;
      N /= i;
    }
    if(cnt != 0) res.push_back(P(i, cnt));
  }
  if(N != 1) res.push_back(P(N, 1));
  return res;
}
ll extgcd (ll a, ll b,  ll &x, ll &y) {
  if(b == 0) { x = 1;y = 0;return a;}
  ll d = extgcd(b, a%b, y, x);
  y -= a/b * x;
  return d;
}
template <typename T>
T ceil(T a, T b) {
  assert(b != 0);
  if(a % b == 0) return a / b;
  if((a <= 0 && b < 0) || (a >= 0 && b > 0)) return a/b + 1;
  else return a / b;
}
template <typename T>
T floor(T a, T b) {
  assert(b != 0);
  if(a % b == 0) return a / b;
  if((a <= 0 && b < 0) || (a >= 0 && b > 0)) return a/b;
  else return a/b - 1;
}
ll modpow(ll x, ll y) {
  const ll mod = 9223372036854775807LL;
  if(x > mod) x %= mod;
  if(y == 0) return 1;
  ll res = modpow(x, y >> 1);
  res = res * res % mod;
  if(y & 1) res *= x, res %= mod;
  return res;
}
//////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//グローバル変数を置くところ（情報工学意識高め）
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
const ll int_max = 1001001001;
const ll ll_max = 1001001001001001001LL;
const double pi = 3.141592653589793;
vl dx{0, 1, 0, -1, 0, 1, 1, -1, -1};  // (番兵) → ↓ ← ↑  ※ 右から時計回り　
vl dy{0, 0, -1, 0, 1, 1, -1, -1, 1};
//const ll mod = 1000000007;
//const ll mod = 998244353;


//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
vector<ll> Exhurui(ll N) {
    vector<ll> sosu;
    ll Max = sqrt(N);
    vector<int> hantei(N+1, 1);
    hantei.at(1) = 0;
    for (int i = 2; i <= Max; i++) {
        if (hantei.at(i) == 1) {
            for (int j = 2; j * i <= N; j++) {
                hantei.at(j*i) = 0;
            }
        }
    }

    for (int i = 2; i <= N; i++) {
        if (hantei.at(i) == 1) {
            sosu.push_back(i);
        }
    }

    return sosu;
}

void solve() {
  ll N;
  cin >> N;

  auto primes = Exhurui(N);
  ll p = primes.back();//こいつで聞けばわかる
  auto a = [](ll a, ll b) {
    cout << "? " << a << " " << b << endl;
    ll res;
    cin >> res;
    return res;
  };

  ll idx = 1;
  bool aaa = false;
  for( ; idx <= N; idx++) {
    cout <<"? " << 1 << " " << idx << endl;
    ll res;
    cin >> res;
    if(res % p == 0) {
      if(idx == 1) {
        ll res2 = a(1, 2);
        if(res2 % p == 0) {
          aaa = true;
        }
        else {
          break;
        }
      }
      else {
        break;
      }
    }
  }

  //if(aaa) A[1]が p
  //else B[idx]が p 　ここまでN回以下

  

  if(!aaa) {
  //Bのbiggest prime idxがわかった
  vl A(N+1, -1);
  vl B(N+1, -1);
  B[idx] = p;
  ll tmp;
  ll pre;
  rep(i, 1, N) {
    cout << "? " << i << " " << idx << endl;
    ll res;
    cin >> res;
    A.at(i) = res / p;
    if(A.at(i) == 1) {
      pre = tmp;
      tmp = i;
    }
  }

 //2N回

  //A[tmp]は今1 or pなので、それを特定する
  vp rres;
  rep(i, 1, 3) {
    if(i == idx) continue;
    rres.push_back(P(a(tmp, i), i));
    if(rres.size() >= 2) break;
  }

  //2N + 2回

  if(rres.at(0).first % p == 0 && rres.at(1).first % p == 0) {
    //A[tmp] = p
    A.at(tmp) = p;
    tmp = pre;
    auto [r1, i1] = rres.at(0);
    B.at(i1) = r1 / p;
    auto [r2, i2] = rres.at(1);
    B.at(i2) = r2 / p;
  }
  else {
    //A[tmp] = 1
    A.at(pre) = p;
     auto [r1, i1] = rres.at(0);
    B.at(i1) = r1;
    auto [r2, i2] = rres.at(1);
    B.at(i2) = r2;
  }

    rep(i, 1, N) {
      if(B.at(i) != -1) continue;
    cout << "? " << tmp << " " << i << endl;
    ll res;
    cin >> res;
    //(1, B.at(i)) または (p, B.at(i))
    B.at(i) = res;
  }

  cout <<"! ";
  rep(i, 1, N) cout << A.at(i) << " ";
  rep(i, 1, N) cout << B.at(i) << " ";
  }








  else {
    vl A(N+1, -1);
    vl B(N+1, -1);
    A.at(1) = p;
    //N回質問済み
    ll tmp;
    ll pre;
    rep(i, 1, N) {
      B.at(i) = a(1, i) / p;
      if(B.at(i) == 1) {
        pre = tmp;
        tmp = i;
      }
    }

 //2N回

  //B[tmp]は今1 or pなので、それを特定する
  vp rres;
  rep(i, 2, 3) {
    rres.push_back(P(a(i, tmp), i));
  }

  //2N + 2回

  if(rres.at(0).first % p == 0 && rres.at(1).first % p == 0) {
    //B[tmp] = p
    B.at(tmp) = p;
    tmp = pre;
    auto [r1, i1] = rres.at(0);
    A.at(i1) = r1 / p;
    auto [r2, i2] = rres.at(1);
    A.at(i2) = r2 / p;
  }
  else {
    //B[tmp] = 1
    B.at(pre) = p;

     auto [r1, i1] = rres.at(0);
    A.at(i1) = r1;
    auto [r2, i2] = rres.at(1);
    A.at(i2) = r2;
  }

    rep(i, 1, N) {
      if(A.at(i) != -1) continue;
    cout << "? " << i << " " << tmp << endl;
    ll res;
    cin >> res;
    //(1, B.at(i)) または (p, B.at(i))
    A.at(i) = res;
  }

  cout <<"! ";
  rep(i, 1, N) cout << A.at(i) << " ";
  rep(i, 1, N) cout << B.at(i) << " ";


  }
  exit(0);



  



    

}
//座標を2で割った時の”切り捨て側(左側）”を求めるには 誤:(x / 2) マイナスの時！！！ 正:floor<ll>(x, 2);
//stringでの数字の下から1桁目は 正:S.at(N-1)  誤:S.at(0)
//if(S.at(i) == 1)　← charなのに1...?
// modは取りましたか...?(´・ω・｀)
//sortの比較関数は、 a == b　ならば falseを返す必要がある（そうで無いとRE）



int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  cout << fixed << setprecision(15);
    ll T = 1;
    //cin >> T;
    rep(i, 1, T) {
        solve();
    }
  return 0;
}
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