結果

問題 No.2396 等差二項展開
ユーザー 👑 p-adicp-adic
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
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
other AC * 31
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

#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 main
inline 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] );
}
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