結果
問題 | No.2396 等差二項展開 |
ユーザー |
👑 |
提出日時 | 2022-11-11 01:41:12 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 917 ms / 6,000 ms |
コード長 | 2,986 bytes |
コンパイル時間 | 841 ms |
コンパイル使用メモリ | 75,524 KB |
最終ジャッジ日時 | 2025-02-08 19:39:01 |
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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ファイルパターン | 結果 |
---|---|
other | AC * 31 |
ソースコード
#include <iostream>#include <vector>#include <string>#include <stdio.h>#include <stdint.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 FOR( VAR , INITIAL , FINAL_PLUS_ONE ) for( TYPE_OF( FINAL_PLUS_ONE ) VAR = INITIAL ; VAR < FINAL_PLUS_ONE ; VAR ++ )#define QUIT return 0#define RETURN( ANSWER ) cout << ( ANSWER ) << "\n"; QUIT#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; \} \} \#include <cassert>#define MAIN maininline CEXPR( uint , bound_L , 100 );class Polynomial{public:vector<ll> m_f;static ll g_M;static uint g_L;static ll g_B;inline Polynomial() : m_f( g_L ) {};inline Polynomial( const ll& c ) : m_f( g_L ) { m_f[0] = c; };inline Polynomial( const Polynomial& g ) : m_f( g.m_f ) {};inline Polynomial& operator*=( const Polynomial& g );};ll Polynomial::g_M = 1;uint Polynomial::g_L = 1;ll Polynomial::g_B = 1;inline Polynomial& Polynomial::operator*=( const Polynomial& g ){vector<ll> answer( g_L * 2 );FOR( i , 0 , g_L ){const ll& fi = m_f[i];FOR( j , 0 , g_L ){( answer[i + j] += fi * g.m_f[j] ) %= g_B;if( answer[i + j] < 0 ){cout << "here" << endl;}}}FOR( k , 0 , g_L ){( answer[k] += answer[ k + g_L ] * g_M ) %= g_B;if( answer[k] < 0 ){cout << "here" << endl;}}m_f = move( answer );return *this;}inline Polynomial operator*( const Polynomial& f , const Polynomial& g ) { return Polynomial( f ).operator*=( g ); }int MAIN(){UNTIE;CEXPR( ll , bound_N , 1000000000000000000 );CIN_ASSERT( N , 1 , bound_N );CIN_ASSERT( M , 1 , bound_N );CEXPR( uint , bound_L , 1000 );CIN_ASSERT( L , 1 , bound_L );Polynomial::g_L = L;CIN_ASSERT( K , 0 , L - 1 );CEXPR( ll , bound_B , 1000000000 );CIN_ASSERT( B , 1 , bound_B );Polynomial::g_B = B;Polynomial::g_M = M % B;Polynomial f{};f.m_f[0] = 1;if( L == 1 ){f.m_f[0] += Polynomial::g_M;} else {f.m_f[1] = 1;}POWER( answer , f , N );RETURN( answer.m_f[K] );}