結果

問題 No.2189 六平方和
ユーザー 👑 p-adicp-adic
提出日時 2022-11-13 16:32:52
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
(最新)
AC  
(最初)
実行時間 -
コード長 3,898 bytes
コンパイル時間 1,381 ms
コンパイル使用メモリ 71,608 KB
実行使用メモリ 6,824 KB
最終ジャッジ日時 2024-11-26 19:17:32
合計ジャッジ時間 5,204 ms
ジャッジサーバーID
(参考情報)
judge5 / judge3
このコードへのチャレンジ
(要ログイン)

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 WA -
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 -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 WA -
testcase_20 WA -
testcase_21 WA -
testcase_22 WA -
testcase_23 WA -
testcase_24 WA -
testcase_25 WA -
testcase_26 WA -
testcase_27 WA -
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ソースコード

diff #

#include <iostream>
#include <string>
#include <stdio.h>
#include <stdint.h>
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 FOREQ( VAR , INITIAL , FINAL ) for( TYPE_OF( FINAL ) VAR = INITIAL ; VAR <= FINAL ; VAR ++ ) 
#define QUIT return 0 
#define RETURN( ANSWER ) cout << ( ANSWER ) << "\n"; QUIT 

#include <cassert>

#define MAIN main

#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;				\
    }									\
  }									\




// 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;			\
  }									\
									\


int MAIN()
{
  UNTIE;
  CEXPR( ll , bound_N , 1000000000000000000 );
  CIN_ASSERT( N , 1 , bound_N );
  CEXPR( ll , bound_M , 1000 );
  CIN_ASSERT( M , 1 , bound_M );
  CEXPR( ll , bound_B , 1000000000 );
  CIN_ASSERT( B , 1 , bound_B );
  POWER_MOD( power_M , M , N , B );
  BS2( x4 , 0 , power_M , x4 * x4 , power_M );
  ll diff = power_M - x4 * x4;
  BS2( x5 , 0 , diff , x5 * x5 , diff );
  diff -= x5 * x5;
  BS2( sqrt_diff , 0 , diff , sqrt_diff * sqrt_diff , diff );
  ll x02 , x12;
  ll diff0 , diff1 , diff2;
  FOREQ( x0 , 0 , sqrt_diff ){
    x02 = x0 * x0;
    diff0 = diff - x02;
    FOREQ( x1 , 0 , sqrt_diff ){
      x12 = x1 * x1;
      diff1 = diff0 - x12;
      if( diff1 < 0 ){
	break;
      }
      FOREQ( x2 , 0 , sqrt_diff ){
	diff2 = diff1 - x2 * x2;
	if( diff2 < 0 ){
	  break;
	}
	BS2( x3 , 0 , diff2 , x3 * x3 , diff2 );
	if( x3 * x3 == diff2 ){
	  cout << "Yes\n";
	  if( x0 == 0 ){
	    x0 = B;
	  }
	  if( x1 == 0 ){
	    x1 = B;
	  }
	  if( x2 == 0 ){
	    x2 = B;
	  }
	  if( x3 == 0 ){
	    x3 = B;
	  }
	  if( x4 == 0 ){
	    x4 = B;
	  }
	  if( x5 == 0 ){
	    x5 = B;
	  }
	  cout << x0 << " " << x1 << " " << x2 << " " << x3 << " " << x4 << " " << x5;
	  RETURN( "" );
	}
      }
    }
  }
  assert( false );
  QUIT;
}
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