結果
問題 | No.2189 六平方和 |
ユーザー |
👑 |
提出日時 | 2022-11-13 16:32:52 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
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 | 752 ms |
コンパイル使用メモリ | 68,360 KB |
最終ジャッジ日時 | 2025-02-08 19:55:38 |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
other | WA * 28 |
ソースコード
#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;}