結果

問題 No.2537 多重含意
ユーザー 👑 p-adic
提出日時 2023-04-11 22:39:39
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 59 ms / 2,000 ms
コード長 2,126 bytes
コンパイル時間 9,588 ms
コンパイル使用メモリ 277,896 KB
最終ジャッジ日時 2025-02-12 05:10:27
ジャッジサーバーID
（参考情報） 		judge3 / judge1
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ファイルパターン 結果
other AC * 24
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 #
プレゼンテーションモードにする

#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 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;               \
    }                                   \
  }                                 \
int MAIN()
{
  UNTIE;
  CEXPR( int , bound_N , 100000 );
  CIN_ASSERT( N , 1 , bound_N );
  CEXPR( ll , bound_B , 1000000000 );
  CIN_ASSERT( B , 1 , bound_B );
  set<int> imA{};
  ll two = 2;
  POWER_MOD( ambient , two , N , B );
  int size = 0;
  REPEAT( N ){
    CIN_ASSERT( Aj , 1 , N );
    if( imA.count( Aj ) ){
      COUT( ambient );
    } else {
      imA.insert( Aj );
      POWER_MOD( complement , two , N - ( ++size ) , B );
      ll answer = ambient - complement;
      COUT( answer < 0 ? answer += B : answer );
    }
  }
  QUIT;
}
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