結果

問題 No.2274 三角彩色
コンテスト
ユーザー 👑 p-adic
提出日時 2023-02-19 09:01:13
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 32 ms / 2,000 ms
コード長 3,649 bytes
コンパイル時間 11,628 ms
コンパイル使用メモリ 289,048 KB
最終ジャッジ日時 2025-02-10 19:03:52
ジャッジサーバーID
（参考情報） 		judge4 / judge3
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ファイルパターン 結果
other AC * 22
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 # 

#pragma GCC optimize ( "O3" )
#pragma GCC optimize( "unroll-loops" )
#pragma GCC target ( "sse4.2,fma,avx2,popcnt,lzcnt,bmi2" )
#include <bits/stdc++.h>
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 FOR( VAR , INITIAL , FINAL_PLUS_ONE ) for( TYPE_OF( FINAL_PLUS_ONE ) VAR = INITIAL ; VAR < FINAL_PLUS_ONE ; VAR ++ ) 
#define REPEAT( HOW_MANY_TIMES ) FOR( VARIABLE_FOR_REPEAT ## HOW_MANY_TIMES , 0 , HOW_MANY_TIMES ) 
#define QUIT return 0 
#define COUT( ANSWER ) cout << ( ANSWER ) << "\n"; 
#define RETURN( ANSWER ) COUT( ANSWER ); QUIT 

#define POWER_MOD( ANSWER , ARGUMENT , EXPONENT , MODULO )		\
  ll ANSWER{ 1 };					\
  {									\
    ll ARGUMENT_FOR_SQUARE_FOR_POWER = ( MODULO + ( ( ARGUMENT ) % MODULO ) ) % MODULO; \
    TYPE_OF( EXPONENT ) EXPONENT_FOR_SQUARE_FOR_POWER = ( EXPONENT );	\
    while( EXPONENT_FOR_SQUARE_FOR_POWER != 0 ){			\
      if( EXPONENT_FOR_SQUARE_FOR_POWER % 2 == 1 ){			\
	ANSWER = ( ANSWER * ARGUMENT_FOR_SQUARE_FOR_POWER ) % MODULO;	\
      }									\
      ARGUMENT_FOR_SQUARE_FOR_POWER = ( ARGUMENT_FOR_SQUARE_FOR_POWER * ARGUMENT_FOR_SQUARE_FOR_POWER ) % MODULO; \
      EXPONENT_FOR_SQUARE_FOR_POWER /= 2;				\
    }									\
  }									\

template <typename T> inline T Residue( const T& a , const T& p ){ return a >= 0 ? a % p : p - 1 - ( ( - ( a + 1 ) ) % p ); }

template <int bound_L , int bound_M>
int Rank( bitset<bound_M> ( &A )[bound_L] , const int& L , const int& M )
{

  int i_min = 0;
  int i_curr;
  int j_curr = 0;

  while( i_min < L && j_curr < M ){

    i_curr = i_min;

    while( A[i_curr][j_curr] == 0 && i_curr < L ){
      
      i_curr++;

    }

    if( i_curr < L ){

      swap( A[i_min] , A[i_curr] );
      const bitset<bound_M>& A_i_min = A[i_min];
      i_min++;

      while( ++i_curr < L ){

	bitset<bound_M>& A_i_curr = A[i_curr];
	
	if( A_i_curr[j_curr] == 1 ){

	  A_i_curr ^= A_i_min;

	}

      }

    }

    j_curr++;

  }

  return i_min;
  
}

int MAIN()
{
  UNTIE;
  CEXPR( ll , bound_N , 1000000000000000000 );
  CIN_ASSERT( N , 3 , bound_N );
  N--;
  CEXPR( int , bound_M , 400 );
  CIN_ASSERT( M , 3 , bound_M );
  CEXPR( ll , bound_B , 1000000000 );
  CIN_ASSERT( B , 2 , bound_B );
  CEXPR( int , bound_Q , 1000 );
  CIN_ASSERT( Q , 1 , bound_Q );
  CEXPR( int , bound_V , bound_M * 2 );
  int V = 0;
  map<ll,int> vertex{};
  bitset<bound_V> d1[bound_M] = {};
  FOR( m , 0 , M ){
    bitset<bound_V>& d1m = d1[m];
    CIN_ASSERT( i , 0 , N );
    d1m[ vertex.count( i ) == 0 ? vertex[i] = V++ : vertex[i] ] = 1;
    CIN_ASSERT( j , i + 1 , N );
    d1m[ vertex.count( j ) == 0 ? vertex[j] = V++ : vertex[j] ] = 1;
  }
  int M_minus = M - 1;
  bitset<bound_M> d2[bound_Q] = {};
  FOR( q , 0 , Q ){
    bitset<bound_M>& d2q = d2[q];
    CIN_ASSERT( m0 , 0 , M_minus );
    d2q[m0] = 1;
    CIN_ASSERT( m1 , m0 + 1 , M_minus );
    d2q[m1] = 1;
    CIN_ASSERT( m2 , m1 + 1 , M_minus );
    d2q[m2] = 1;
    assert( ( d1[m0] ^ d1[m1] ^ d1[m2] ) == 0 );
  }
  int rank1 = Rank<bound_M,bound_V>( d1 , M , V );
  int rank2 = Rank<bound_Q,bound_M>( d2 , Q , M );
  POWER_MOD( ker1 , 2 , M - rank1 , B );
  POWER_MOD( im2 , 2 , rank2 , B );
  cout << Residue<ll>( ker1 - im2 , B ) << " " << im2 << "\n";
  QUIT;
}
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