結果

問題 No.2130 分配方法の数え上げ mod 998244353
ユーザー 👑 p-adic
提出日時 2022-10-02 23:46:13
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 4 ms / 2,000 ms
コード長 1,815 bytes
コンパイル時間 653 ms
コンパイル使用メモリ 67,320 KB
最終ジャッジ日時 2025-02-07 20:43:16
ジャッジサーバーID
（参考情報） 		judge2 / judge3
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ファイルパターン 結果
other AC * 38
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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 <string>
#include <stdio.h>
#include <stdint.h>
using namespace std;
using ll = long long;
#define UNTIE ios_base::sync_with_stdio( false ); cin.tie( nullptr ) 
#define CIN( LL , A ) LL A; cin >> A 
#define TYPE_OF( VAR ) remove_const<remove_reference<decltype( VAR )>::type >::type
#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 , VAR , EXPONENT_REF , MODULO )           \
  TYPE_OF( VAR ) ANSWER = 1;                        \
  TYPE_OF( VAR ) VARIABLE_FOR_SQUARE_FOR_POWER = VAR;           \
  while( EXPONENT_REF != 0 ){                       \
    if( EXPONENT_REF % 2 == 1 ){                    \
      ANSWER = ( ANSWER * VARIABLE_FOR_SQUARE_FOR_POWER ) % MODULO; \
    }                                   \
    VARIABLE_FOR_SQUARE_FOR_POWER = ( VARIABLE_FOR_SQUARE_FOR_POWER * VARIABLE_FOR_SQUARE_FOR_POWER ) % MODULO; \
    EXPONENT_REF /= 2;                          \
  }         
#include <cassert>
#define MAIN main
int MAIN()
{
  constexpr const ll bound_N = 1000000000000000000;
  CIN( ll , N );
  assert( 1 <= N && N <= bound_N );
  constexpr const int bound_M = 100000;
  CIN( int , M );
  assert( 1 <= M && M <= bound_M );
  if( N < M ){
    RETURN( 0 );
  }
  constexpr const ll P = 998244353;
  constexpr const ll two = 2;
  ll N_copy = N % ( P - 1 );
  POWER( answer , two , N_copy , P );
  ll inverse[bound_M];
  inverse[1] = 1;
  N %= P;
  ll combination = N;
  answer--;
  if( M != 1 ){
    answer = ( answer + P - N ) % P;
  }
  FOR( i , 2 , M ){
    inverse[i] = P - ( inverse[P % i] * ( P / i ) ) % P;
    combination = ( combination * ( N + 1 + P - i ) ) % P;
    combination = ( combination * inverse[i] ) % P;
    answer = ( answer + P - combination ) % P;
  }
  RETURN( answer );
}
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