結果

問題 No.2215 Slide Subset Sum
ユーザー 👑 p-adicp-adic
提出日時 2023-02-11 10:11:48
言語 C++17
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 325 ms / 3,000 ms
コード長 7,870 bytes
コンパイル時間 3,865 ms
コンパイル使用メモリ 226,900 KB
実行使用メモリ 322,988 KB
最終ジャッジ日時 2023-09-22 08:54:02
合計ジャッジ時間 13,153 ms
ジャッジサーバーID
（参考情報） 		judge14 / judge13
このコードへのチャレンジ（β）

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 AC 49 ms
4,376 KB
testcase_01 AC 50 ms
4,376 KB
testcase_02 AC 163 ms
4,380 KB
testcase_03 AC 159 ms
4,376 KB
testcase_04 AC 244 ms
9,800 KB
testcase_05 AC 2 ms
4,380 KB
testcase_06 AC 1 ms
4,376 KB
testcase_07 AC 1 ms
4,376 KB
testcase_08 AC 2 ms
4,376 KB
testcase_09 AC 1 ms
4,380 KB
testcase_10 AC 1 ms
4,380 KB
testcase_11 AC 2 ms
4,380 KB
testcase_12 AC 2 ms
4,376 KB
testcase_13 AC 2 ms
4,380 KB
testcase_14 AC 2 ms
4,376 KB
testcase_15 AC 1 ms
4,380 KB
testcase_16 AC 2 ms
4,380 KB
testcase_17 AC 311 ms
148,756 KB
testcase_18 AC 295 ms
128,040 KB
testcase_19 AC 271 ms
7,696 KB
testcase_20 AC 264 ms
163,264 KB
testcase_21 AC 305 ms
39,328 KB
testcase_22 AC 266 ms
175,652 KB
testcase_23 AC 317 ms
289,172 KB
testcase_24 AC 306 ms
290,460 KB
testcase_25 AC 293 ms
258,024 KB
testcase_26 AC 272 ms
177,048 KB
testcase_27 AC 10 ms
4,380 KB
testcase_28 AC 16 ms
12,032 KB
testcase_29 AC 20 ms
9,972 KB
testcase_30 AC 80 ms
4,376 KB
testcase_31 AC 117 ms
107,588 KB
testcase_32 AC 25 ms
7,372 KB
testcase_33 AC 198 ms
26,096 KB
testcase_34 AC 138 ms
74,116 KB
testcase_35 AC 21 ms
5,164 KB
testcase_36 AC 61 ms
49,480 KB
testcase_37 AC 325 ms
322,988 KB
testcase_38 AC 29 ms
14,852 KB
testcase_39 AC 51 ms
4,380 KB
testcase_40 AC 38 ms
4,376 KB
testcase_41 AC 1 ms
4,380 KB
testcase_42 AC 325 ms
164,496 KB
testcase_43 AC 33 ms
4,380 KB
testcase_44 AC 33 ms
4,380 KB
testcase_45 AC 292 ms
242,748 KB
testcase_46 AC 290 ms
242,680 KB
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 # 

#pragma GCC optimize ( "O3" )
#pragma GCC optimize( "unroll-loops" )
#pragma GCC target ( "sse4.2,fma,avx2,popcnt,lzcnt,bmi2" )
#include <bits/stdc++.h>
using namespace std;

using uint = unsigned int;
using ll = long long;

#define ATT __attribute__( ( target( "sse4.2,fma,avx2,popcnt,lzcnt,bmi2" ) ) )
#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 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 GETLINE( A ) string A; getline( cin , A )
#define GETLINE_SEPARATE( A , SEPARATOR ) string A; getline( cin , A , SEPARATOR )
#define FOR( VAR , INITIAL , FINAL_PLUS_ONE ) for( TYPE_OF( FINAL_PLUS_ONE ) VAR = INITIAL ; VAR < FINAL_PLUS_ONE ; VAR ++ )
#define FOREQ( VAR , INITIAL , FINAL ) for( TYPE_OF( FINAL ) VAR = INITIAL ; VAR <= FINAL ; VAR ++ )
#define FOREQINV( VAR , INITIAL , FINAL ) for( TYPE_OF( INITIAL ) VAR = INITIAL ; VAR >= FINAL ; VAR -- )
#define FOR_ITR( ARRAY , ITR , END ) for( auto ITR = ARRAY .begin() , END = ARRAY .end() ; ITR != END ; ITR ++ )
#define REPEAT( HOW_MANY_TIMES ) FOR( VARIABLE_FOR_REPEAT , 0 , HOW_MANY_TIMES )
#define QUIT return 0
#define COUT( ANSWER ) cout << ( ANSWER ) << "\n";
#define RETURN( ANSWER ) COUT( ANSWER ); QUIT
#define DOUBLE( PRECISION , ANSWER ) cout << fixed << setprecision( PRECISION ) << ( 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;				\
    }									\
  }									\

#define POWER_MOD( ANSWER , ARGUMENT , EXPONENT , MODULO )		\
  TYPE_OF( ARGUMENT ) ANSWER{ 1 };					\
  {									\
    TYPE_OF( ARGUMENT ) ARGUMENT_FOR_SQUARE_FOR_POWER = ( MODULO + ( ( ARGUMENT ) % MODULO ) ) % 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;				\
    }									\
  }									\

#define FACTORIAL_MOD( ANSWER , ANSWER_INV , INVERSE , MAX_I , LENGTH , MODULO ) \
  static ll ANSWER[LENGTH];						\
  static ll ANSWER_INV[LENGTH];						\
  static ll INVERSE[LENGTH];						\
  {									\
    ll VARIABLE_FOR_PRODUCT_FOR_FACTORIAL = 1;				\
    ANSWER[0] = VARIABLE_FOR_PRODUCT_FOR_FACTORIAL;			\
    FOREQ( i , 1 , MAX_I ){						\
      ANSWER[i] = ( VARIABLE_FOR_PRODUCT_FOR_FACTORIAL *= i ) %= MODULO; \
    }									\
    ANSWER_INV[0] = ANSWER_INV[1] = INVERSE[1] = VARIABLE_FOR_PRODUCT_FOR_FACTORIAL = 1; \
    FOREQ( i , 2 , MAX_I ){						\
      ANSWER_INV[i] = ( VARIABLE_FOR_PRODUCT_FOR_FACTORIAL *= INVERSE[i] = MODULO - ( ( ( MODULO / i ) * INVERSE[MODULO % i] ) % MODULO ) ) %= MODULO; \
    }									\
  }									\

// 通常の二分探索（単調関数-目的値が一意実数解を持つ場合にそれを超えない最大の整数を返す）
#define BS( ANSWER , MINIMUM , MAXIMUM , EXPRESSION , TARGET )		\
  ll ANSWER = MAXIMUM;							\
  {									\
    ll VARIABLE_FOR_BINARY_SEARCH_L = MINIMUM;				\
    ll VARIABLE_FOR_BINARY_SEARCH_U = ANSWER;				\
    ll VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH = ( TARGET ) - ( EXPRESSION ); \
    if( VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH == 0 ){		\
      VARIABLE_FOR_BINARY_SEARCH_L = ANSWER;				\
    } else {								\
      ANSWER = ( VARIABLE_FOR_BINARY_SEARCH_L + VARIABLE_FOR_BINARY_SEARCH_U ) / 2; \
    }									\
    while( VARIABLE_FOR_BINARY_SEARCH_L != ANSWER ){			\
      VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH = ( TARGET ) - ( EXPRESSION ); \
      if( VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH == 0 ){		\
	break;								\
      } else {								\
	if( VARIABLE_FOR_DIFFERENCE_FOR_BINARY_SEARCH > 0 ){		\
	  VARIABLE_FOR_BINARY_SEARCH_L = ANSWER;			\
	} else {							\
	  VARIABLE_FOR_BINARY_SEARCH_U = ANSWER;			\
	}								\
	ANSWER = ( VARIABLE_FOR_BINARY_SEARCH_L + VARIABLE_FOR_BINARY_SEARCH_U ) / 2; \
      }									\
    }									\
  }									\
									\


// 二進法の二分探索（単調関数-目的値が一意実数解を持つ場合にそれを超えない最大の整数を返す）
#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;			\
  }									\
									\


template <typename T> inline T Absolute( const T& a ){ return a > 0 ? a : -a; }
template <typename T> inline T Residue( const T& a , const T& p ){ return a >= 0 ? a % p : p - 1 - ( ( - ( a + 1 ) ) % p ); }

inline CEXPR( ll , P , 998244353 );
int K = 1;
void red( vector<ll>& f )
{
  int size = f.size();
  while( size > K ){
    size--;
    ll& f_curr = f[size - K];
    if( ( f_curr += f[size] ) >= P ){
      f_curr -= P;
    }
    f.pop_back();
  }
  return;
}
void mul( vector<ll>& f , int& Ai )
{
  int size = f.size();
  f.resize( size + Ai );
  FOREQINV( i , size - 1 , 0 ){
    ll& f_curr = f[i + Ai];
    if( ( f_curr += f[i] ) >= P ){
      f_curr -= P;
    }
  }
  red( f );
  return;
}

ATT int main()
{
  UNTIE;
  CIN( int , N );
  CIN( int , M );
  cin >> K;
  vector<int> A( M );
  FOR( i , 0 , M ){
    cin >> A[i];
  }
  vector<vector<ll> > a( M );
  a.back() = vector<ll>( 1 , 1 );
  FOREQINV( i , M - 1 , 1 ){
    mul( a[i-1] = a[i] , A[i] );
  }
  vector<ll> temp( a[0] );
  mul( temp , A[0] );
  CEXPR( ll , P_minus , P - 1 );
  ll answer = temp[0];
  COUT( answer > 0 ? --answer : P_minus );
  temp = vector<ll>( 1 , 1 );
  int i = 0;
  N -= M;
  REPEAT( N ){
    CIN( int , Ai );
    A[i] = Ai;
    mul( temp , Ai );
    vector<ll>& f = a[i];
    int size_temp = temp.size();
    int size_f = f.size();
    ( answer = temp[0] * f[0] ) %= P;
    FOR( j , K - size_f + 1 , size_temp ){
      ( answer += temp[j] * f[K - j] ) %= P;
    }
    COUT( answer > 0 ? --answer : P_minus );
    if( ++i == M ){
      a.back() = vector<ll>( 1 , 1 );
      FOREQINV( i , M - 1 , 1 ){
	mul( a[i-1] = a[i] , A[i] );
      }
      temp = vector<ll>( 1 , 1 );
      i = 0;
    }
  }
  QUIT;
}
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