結果

問題 No.2446 完全列
ユーザー 👑 p-adicp-adic
提出日時 2023-05-15 08:31:55
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 3,706 bytes
コンパイル時間 856 ms
コンパイル使用メモリ 72,572 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-06-06 20:27:12
合計ジャッジ時間 1,615 ms
ジャッジサーバーID
(参考情報)
judge4 / judge5
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 1 ms
5,248 KB
testcase_01 AC 2 ms
5,376 KB
testcase_02 AC 2 ms
5,376 KB
testcase_03 AC 2 ms
5,376 KB
testcase_04 AC 2 ms
5,376 KB
testcase_05 AC 2 ms
5,376 KB
testcase_06 AC 2 ms
5,376 KB
testcase_07 AC 2 ms
5,376 KB
testcase_08 AC 2 ms
5,376 KB
testcase_09 AC 2 ms
5,376 KB
testcase_10 AC 2 ms
5,376 KB
testcase_11 AC 2 ms
5,376 KB
testcase_12 AC 2 ms
5,376 KB
testcase_13 AC 2 ms
5,376 KB
testcase_14 AC 2 ms
5,376 KB
testcase_15 AC 2 ms
5,376 KB
testcase_16 AC 2 ms
5,376 KB
testcase_17 AC 1 ms
5,376 KB
testcase_18 AC 2 ms
5,376 KB
testcase_19 AC 2 ms
5,376 KB
testcase_20 AC 2 ms
5,376 KB
testcase_21 AC 2 ms
5,376 KB
testcase_22 AC 2 ms
5,376 KB
testcase_23 AC 2 ms
5,376 KB
testcase_24 AC 2 ms
5,376 KB
testcase_25 AC 2 ms
5,376 KB
testcase_26 AC 1 ms
5,376 KB
testcase_27 AC 2 ms
5,376 KB
testcase_28 AC 2 ms
5,376 KB
testcase_29 AC 2 ms
5,376 KB
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 #

// 中国剰余定理による解法
#include <iostream>
#include <stdio.h>
#include <stdint.h>
#include <cassert>
using namespace std;

using ll = long long;

#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 FOREQ( VAR , INITIAL , FINAL ) for( TYPE_OF( FINAL ) VAR = INITIAL ; VAR <= FINAL ; VAR ++ ) 
#define QUIT return 0 
#define RETURN( ANSWER ) cout << ( ANSWER ) << "\n"; 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;				\
    }									\
  }									\

inline CEXPR( int , bound_size , 15 );
inline CEXPR( int , bound_r , 9 );

int Rank( const ll ( &A )[bound_size][bound_size] , const int& L , const int& M )
{
  constexpr ll P[bound_r] = {10000019,10000079,10000103,10000121,10000139,10000141,10000169,10000189,10000223};
  int rank = 0;
  FOR( r , 0 , bound_r ){
    const ll& P_r = P[r];
    ll A_copy[bound_size][bound_size];
    FOR( i , 0 , L ){
      const ll ( &A_i )[bound_size] = A[i];
      ll ( &A_copy_i )[bound_size] = A_copy[i];
      FOR( j , 0 , M ){
	A_copy_i[j] = A_i[j];
      }
    }
    int i_min = 0;
    int i_curr;
    int j_curr = 0;
    while( i_min < L && j_curr < M ){
      i_curr = i_min;
      while( i_curr < L ? A_copy[i_curr][j_curr] % P_r == 0 : false ){
	i_curr++;
      }
      if( i_curr < L ){
	swap( A_copy[i_min] , A_copy[i_curr] );
	ll ( &A_copy_i_min )[bound_size] = A_copy[i_min];
	POWER_MOD( inv , A_copy_i_min[j_curr] , P_r - 2 , P_r );
	FOR( j , j_curr , M ){
	  ( A_copy_i_min[j] *= inv ) %= P_r;
	}
	FOR( i , i_min + 1 , L ){
	  ll ( &A_copy_i )[bound_size] = A_copy[i];
	  ll A_copy_i_j_curr = A_copy_i[j_curr];
	  FOR( j , j_curr , M ){
	    ll& A_copy_ij = A_copy_i[j] -= A_copy_i_j_curr * A_copy_i_min[j];
	  }
	}
	i_min++;
      }
      j_curr++;
    }
    rank < i_min ? rank = i_min : rank;
  }
  return rank;
}

int main()
{
  UNTIE;
  CIN_ASSERT( L , 1 , bound_size );
  CIN_ASSERT( M , 1 , bound_size / L );
  CIN_ASSERT( N , 1 , bound_size / M );
  CEXPR( ll , bound , 6 );
  ll A[bound_size][bound_size];
  FOR( i , 0 , L ){
    ll ( &Ai )[bound_size] = A[i];
    FOR( j , 0 , M ){
      CIN_ASSERT( Aij , - bound , bound );
      Ai[j] = Aij;
    }
  }
  ll B[bound_size][bound_size];
  FOR( j , 0 , M ){
    ll ( &Bj )[bound_size] = B[j];
    FOR( k , 0 , N ){
      CIN_ASSERT( Bjk , - bound , bound );
      Bj[k] = Bjk;
    }
  }
  ll sum = 0;
  FOR( i , 0 , L ){
    ll ( &Ai )[bound_size] = A[i];
    FOR( k , 0 , N ){
      FOR( j , 0 , M ){
	sum += Ai[j] * B[j][k];
      }
      if( sum != 0 ){
	RETURN( "No" );
      }
    }
  }
  int rankA = Rank( A , L , M );
  int rankB = Rank( B , M , N );
  RETURN( rankB == M - rankA ? "Yes" : "No" );
}

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