結果

問題 No.2445 奇行列式
コンテスト
ユーザー 👑 p-adic
提出日時 2023-03-25 22:00:03
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 947 ms / 3,000 ms
コード長 2,705 bytes
コンパイル時間 3,197 ms
コンパイル使用メモリ 103,800 KB
最終ジャッジ日時 2025-02-11 18:06:09
ジャッジサーバーID
（参考情報） 		judge5 / judge3
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ファイルパターン 結果
other AC * 20
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 #
raw source code 

#pragma GCC optimize ( "O3" )
#pragma GCC optimize( "unroll-loops" )
#pragma GCC target ( "sse4.2,fma,avx2,popcnt,lzcnt,bmi2" )
#include <iostream>
#include <stdio.h>
#include <stdint.h>
#include <cassert>
#include <vector>
using namespace std;

using ll = long long;

#define MAIN main
#define TYPE_OF( VAR ) remove_const<remove_reference<decltype( VAR )>::type >::type
#define UNTIE ios_base::sync_with_stdio( false ); cin.tie( nullptr )
#define CEXPR( LL , BOUND , VALUE ) constexpr const LL BOUND = VALUE
#define CIN( LL , A ) LL A; cin >> A
#define ASSERT( A , MIN , MAX ) assert( ( MIN ) <= A && A <= ( MAX ) )
#define CIN_ASSERT( A , MIN , MAX ) CIN( TYPE_OF( MAX ) , A ); ASSERT( A , MIN , MAX )
#define GETLINE( S ) string S; getline( cin , S )
#define FOR( VAR , INITIAL , FINAL_PLUS_ONE ) for( TYPE_OF( FINAL_PLUS_ONE ) VAR = INITIAL ; VAR < FINAL_PLUS_ONE ; VAR ++ )
#define FOREQ( VAR , INITIAL , FINAL ) for( TYPE_OF( FINAL ) VAR = INITIAL ; VAR <= FINAL ; VAR ++ )
#define FOR_ITR( ARRAY , ITR , END ) for( auto ITR = ARRAY .begin() , END = ARRAY .end() ; ITR != END ; ITR ++ )
#define QUIT return 0
#define COUT( ANSWER ) cout << ( ANSWER ) << "\n"
#define RETURN( ANSWER ) COUT( ANSWER ); QUIT

int MAIN()
{
  UNTIE;
  CEXPR( int , bound_N , 20 );
  CIN_ASSERT( N , 1 , bound_N );
  CEXPR( ll , bound_B , 1000000000 );
  CIN_ASSERT( B , 1 , bound_B );
  CEXPR( ll , bound_Aij , 1000000000000000000 );
  ll A[bound_N][bound_N];
  FOR( i , 0 , N ){
    ll ( &Ai )[bound_N] = A[i];
    FOR( j , 0 , N ){
      CIN_ASSERT( Aij , 0 , bound_Aij );
      Ai[j] = Aij %= B;
    }
  }
  vector<int> D[bound_N+1] = {};
  CEXPR( int , bound_power_N , 1 << bound_N );
  vector<int> S[bound_power_N] = {};
  int power_N = 1 << N;
  FOR( d , 0 , power_N ){
    int d_copy = d;
    vector<int>& Sd = S[d];
    FOR( j , 0 , N ){
      if( d_copy % 2 == 1 ){
	Sd.push_back( j );
      }
      d_copy /= 2;
      if( d_copy == 0 ){
	break;
      }
    }
    D[Sd.size()].push_back( d );
  }
  ll det[2][bound_power_N] = {};
  ll ( &e )[bound_power_N] = det[0];
  ll ( &o )[bound_power_N] = det[1];
  e[0] = 1;
  o[0] = 0;
  FOREQ( n , 1 , N ){
    vector<int>& Dn = D[n];
    FOR_ITR( Dn , itr_Dn , end_Dn ){
      int& d = *itr_Dn;
      vector<int>& Sd = S[d];
      ll& ed = e[d] = 0;
      ll& od = o[d] = 0;
      ll ( &Ai )[bound_N] = A[N - n];
      bool even = true;
      FOR_ITR( Sd , itr_Sd , end_Sd ){
	int& j = *itr_Sd;
	ll& Aij = Ai[j];
	int d_sub = d - ( 1 << j );
	if( even ){
	  ( ed += Aij * e[d_sub] ) %= B;
	  ( od += Aij * o[d_sub] ) %= B;
	} else {
	  ( ed += Aij * o[d_sub] ) %= B;
	  ( od += Aij * e[d_sub] ) %= B;
	}
	even = ! even;
      }
    }
  }
  RETURN( o[power_N - 1] );
}
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