結果

問題 No.2132 1 or X Game
ユーザー 👑 p-adicp-adic
提出日時 2022-11-25 23:40:49
言語 C++17
(gcc 13.2.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 5,964 bytes
コンパイル時間 2,981 ms
コンパイル使用メモリ 209,188 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-04-10 04:41:07
合計ジャッジ時間 10,222 ms
ジャッジサーバーID
(参考情報)
judge3 / judge5
このコードへのチャレンジ(β)

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 2 ms
6,812 KB
testcase_01 WA -
testcase_02 WA -
testcase_03 WA -
testcase_04 WA -
testcase_05 WA -
testcase_06 WA -
testcase_07 WA -
testcase_08 WA -
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
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ソースコード

diff #

#pragma GCC optimize ( "O3" )
#pragma GCC target ( "avx" )
#include<bits/stdc++.h>
using namespace std;

using uint = unsigned int;
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 GETLINE( A ) string A; getline( cin , A ) 
#define GETLINE_SEPARATE( A , SEPARATOR ) string A; getline( cin , A , SEPARATOR ) 
#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 FOREQINV( VAR , INITIAL , FINAL ) for( TYPE_OF( INITIAL ) VAR = INITIAL ; VAR >= FINAL ; VAR -- ) 
#define FOR_ITR( ARRAY , ITR , END ) for( auto ITR = ARRAY .begin() , END = ARRAY .end() ; ITR != END ; ITR ++ ) 
#define REPEAT( HOW_MANY_TIMES ) FOR( VARIABLE_FOR_REPEAT , 0 , HOW_MANY_TIMES ) 
#define QUIT return 0 
#define RETURN( ANSWER ) cout << ( ANSWER ) << "\n"; QUIT 
#define DOUBLE( PRECISION , ANSWER ) cout << fixed << setprecision( PRECISION ) << ( ANSWER ) << "\n"; QUIT 
#define MIN( A , B ) ( A < B ? A : B )
#define MAX( A , B ) ( A < B ? B : A )
#define RESIDUE( A , P ) ( A >= 0 ? A % P : P - ( - A - 1 ) % P - 1 ) 

#define POWER( ANSWER , ARGUMENT , EXPONENT )				\
  TYPE_OF( ARGUMENT ) ANSWER{ 1 };					\
  {									\
    TYPE_OF( ARGUMENT ) ARGUMENT_FOR_SQUARE_FOR_POWER = ( ARGUMENT );	\
    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 *= ARGUMENT_FOR_SQUARE_FOR_POWER;			\
      }									\
      ARGUMENT_FOR_SQUARE_FOR_POWER *= ARGUMENT_FOR_SQUARE_FOR_POWER;	\
      EXPONENT_FOR_SQUARE_FOR_POWER /= 2;				\
    }									\
  }									\


#define POWER_MOD( ANSWER , ARGUMENT , EXPONENT , MODULO )		\
  TYPE_OF( ARGUMENT ) ANSWER{ 1 };					\
  {									\
    TYPE_OF( ARGUMENT ) ARGUMENT_FOR_SQUARE_FOR_POWER = ( ARGUMENT ) % 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;				\
    }									\
  }									\



// 通常の二分探索
#define BS( ANSWER , MINIMUM , MAXIMUM , EXPRESSION , TARGET )		\
  ll ANSWER = MAXIMUM;							\
  {									\
    ll VARIABLE_FOR_BINARY_SEARCH_L = MINIMUM;				\
    ll VARIABLE_FOR_BINARY_SEARCH_U = ANSWER;				\
    ll VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH = ( TARGET ) - ( EXPRESSION ); \
    if( VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH == 0 ){		\
      VARIABLE_FOR_BINARY_SEARCH_L = ANSWER;				\
    } else {								\
      ANSWER = ( VARIABLE_FOR_BINARY_SEARCH_L + VARIABLE_FOR_BINARY_SEARCH_U ) / 2; \
    }									\
    while( VARIABLE_FOR_BINARY_SEARCH_L != ANSWER ){			\
      VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH = ( TARGET ) - ( EXPRESSION ); \
      if( VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH == 0 ){		\
	VARIABLE_FOR_BINARY_SEARCH_L = ANSWER;				\
	break;								\
      } else {								\
	if( VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH > 0 ){		\
	  VARIABLE_FOR_BINARY_SEARCH_L = ANSWER;			\
	} else {							\
	  VARIABLE_FOR_BINARY_SEARCH_U = ANSWER;			\
	}								\
	ANSWER = ( VARIABLE_FOR_BINARY_SEARCH_L + VARIABLE_FOR_BINARY_SEARCH_U ) / 2; \
      }									\
    }									\
  }									\
									\


// 二進法の二分探索
#define BS2( ANSWER , MINIMUM , MAXIMUM , EXPRESSION , TARGET )		\
  ll ANSWER = MINIMUM;							\
  {									\
    ll VARIABLE_FOR_POWER_FOR_BINARY_SEARCH_2 = 1;			\
    ll VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH = ( MAXIMUM ) - ANSWER; \
    while( VARIABLE_FOR_POWER_FOR_BINARY_SEARCH_2 <= VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH ){ \
      VARIABLE_FOR_POWER_FOR_BINARY_SEARCH_2 *= 2;			\
    }									\
    VARIABLE_FOR_POWER_FOR_BINARY_SEARCH_2 /= 2;			\
    ll VARIABLE_FOR_ANSWER_FOR_BINARY_SEARCH_2 = ANSWER;		\
    while( VARIABLE_FOR_POWER_FOR_BINARY_SEARCH_2 != 0 ){		\
      ANSWER = VARIABLE_FOR_ANSWER_FOR_BINARY_SEARCH_2 + VARIABLE_FOR_POWER_FOR_BINARY_SEARCH_2; \
      VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH = ( TARGET ) - ( EXPRESSION ); \
      if( VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH == 0 ){		\
	VARIABLE_FOR_ANSWER_FOR_BINARY_SEARCH_2 = ANSWER;		\
	break;								\
      } else if( VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH > 0 ){	\
	VARIABLE_FOR_ANSWER_FOR_BINARY_SEARCH_2 = ANSWER;		\
      }									\
      VARIABLE_FOR_POWER_FOR_BINARY_SEARCH_2 /= 2;			\
    }									\
    ANSWER = VARIABLE_FOR_ANSWER_FOR_BINARY_SEARCH_2;			\
  }									\
									\


template <typename T> inline T Absolute( const T& a ){ return a > 0 ? a : - a; }

int main()
{
  CEXPR( ll , P , 998244353 );
  CEXPR( int , bound_T , 200000 );
  CIN_ASSERT( T , 1 , bound_T );
  CEXPR( ll , bound_N , 1000000000000000000 );
  CEXPR( ll , bound_X , 1000000000000000000 );
  REPEAT( T ){
    CIN_ASSERT( N , 1 , bound_N );
    CIN_ASSERT( X , 1 , bound_X );
    // X % 2 == 0 かつ n / Xが奇数 -> n % 2 == 0
    // X % 2 == 0 かつ n / Xが偶数 -> n % 2 == 1
    // X % 2 == 1 -> n % 2 == 1
    if( X % 2 == 0 ){
      ll N_q = N / X;
      ll N_r = N % X;
      ll N_m = N - N_r;
      cout << ( N_m / 2 + ( N_q % 2 == 0 ? ( N_r + 1 ) / 2 : N_r / 2 ) ) % P << "\n";
    } else {
      cout << ( ( N + 1 ) / 2 ) % P << "\n";
    }
  }
  QUIT;
}
0