結果
問題 | No.2134 $\sigma$-algebra over Finite Set |
ユーザー | 👑 p-adic |
提出日時 | 2022-11-25 22:04:52 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 189 ms / 2,000 ms |
コード長 | 6,182 bytes |
コンパイル時間 | 3,175 ms |
コンパイル使用メモリ | 218,496 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-10-07 07:38:55 |
合計ジャッジ時間 | 4,188 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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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 | 2 ms
5,248 KB |
testcase_01 | AC | 184 ms
5,248 KB |
testcase_02 | AC | 189 ms
5,248 KB |
testcase_03 | AC | 2 ms
5,248 KB |
testcase_04 | AC | 2 ms
5,248 KB |
testcase_05 | AC | 2 ms
5,248 KB |
testcase_06 | AC | 8 ms
5,248 KB |
testcase_07 | AC | 9 ms
5,248 KB |
testcase_08 | AC | 2 ms
5,248 KB |
testcase_09 | AC | 2 ms
5,248 KB |
testcase_10 | AC | 21 ms
5,248 KB |
testcase_11 | AC | 21 ms
5,248 KB |
testcase_12 | AC | 51 ms
5,248 KB |
testcase_13 | AC | 95 ms
5,248 KB |
testcase_14 | AC | 11 ms
5,248 KB |
testcase_15 | AC | 14 ms
5,248 KB |
testcase_16 | AC | 12 ms
5,248 KB |
testcase_17 | AC | 2 ms
5,248 KB |
testcase_18 | AC | 2 ms
5,248 KB |
ソースコード
#pragma GCC optimize ( "O3" ) #pragma GCC target ( "avx" ) #include<bits/stdc++.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 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 RETURN( ANSWER ) cout << ( ANSWER ) << "\n"; QUIT #define DOUBLE( PRECISION , ANSWER ) cout << fixed << setprecision( PRECISION ) << ( ANSWER ) << "\n"; QUIT #define MIN( A , B ) ( A < B ? A : B ) #define MAX( A , B ) ( A < B ? B : A ) #define RESIDUE( A , P ) ( A >= 0 ? A % P : P - ( - A - 1 ) % P - 1 ) #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 = ( 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; \ } \ } \ // 通常の二分探索 #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 ){ \ VARIABLE_FOR_BINARY_SEARCH_L = ANSWER; \ 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 #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; } int main() { CEXPR( ll , P , 998244353 ); CEXPR( ll , bound_N , 1000 ); CIN_ASSERT( N , 1 , bound_N ); CEXPR( ll , bound_M , 1000 ); CIN_ASSERT( M , 0 , bound_M ); CEXPR( ll , length , 1001 ); bitset<length> base[bound_M] = {}; ll two = 2; bitset<length>& base0 = base[0]; base0[0] = 1; REPEAT( N ){ base0 <<= 1; base0[0] = 1; } base0[0] = 0; int count = 1; int count_copy; bitset<length> current; bitset<length> div; FOR( i , 0 , M ){ CIN_ASSERT( li , 1 , N ); current = 0; FOR( j , 0 , li ){ CIN_ASSERT( aij , 1 , N ); current[aij] = 1; } count_copy = count; FOR( j , 0 , count_copy ){ bitset<length>& basej = base[j]; div = current & base[j]; if( div != 0 && div != basej ){ basej &= ~div; base[count] = div; count++; } } } POWER_MOD( answer , two , count , P ); RETURN( answer ); }