結果

問題 No.2274 三角彩色
ユーザー 👑 p-adic
提出日時 2023-02-19 09:01:13
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 32 ms / 2,000 ms
コード長 3,649 bytes
コンパイル時間 11,628 ms
コンパイル使用メモリ 289,048 KB
最終ジャッジ日時 2025-02-10 19:03:52
ジャッジサーバーID
(参考情報)
judge4 / judge3
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 * 22
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

#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 ll = long long;
#define MAIN main
#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 REPEAT( HOW_MANY_TIMES ) FOR( VARIABLE_FOR_REPEAT ## HOW_MANY_TIMES , 0 , HOW_MANY_TIMES )
#define QUIT return 0
#define COUT( ANSWER ) cout << ( ANSWER ) << "\n";
#define RETURN( ANSWER ) COUT( ANSWER ); QUIT
#define POWER_MOD( ANSWER , ARGUMENT , EXPONENT , MODULO ) \
ll ANSWER{ 1 }; \
{ \
ll 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; \
} \
} \
template <typename T> inline T Residue( const T& a , const T& p ){ return a >= 0 ? a % p : p - 1 - ( ( - ( a + 1 ) ) % p ); }
template <int bound_L , int bound_M>
int Rank( bitset<bound_M> ( &A )[bound_L] , const int& L , const int& M )
{
int i_min = 0;
int i_curr;
int j_curr = 0;
while( i_min < L && j_curr < M ){
i_curr = i_min;
while( A[i_curr][j_curr] == 0 && i_curr < L ){
i_curr++;
}
if( i_curr < L ){
swap( A[i_min] , A[i_curr] );
const bitset<bound_M>& A_i_min = A[i_min];
i_min++;
while( ++i_curr < L ){
bitset<bound_M>& A_i_curr = A[i_curr];
if( A_i_curr[j_curr] == 1 ){
A_i_curr ^= A_i_min;
}
}
}
j_curr++;
}
return i_min;
}
int MAIN()
{
UNTIE;
CEXPR( ll , bound_N , 1000000000000000000 );
CIN_ASSERT( N , 3 , bound_N );
N--;
CEXPR( int , bound_M , 400 );
CIN_ASSERT( M , 3 , bound_M );
CEXPR( ll , bound_B , 1000000000 );
CIN_ASSERT( B , 2 , bound_B );
CEXPR( int , bound_Q , 1000 );
CIN_ASSERT( Q , 1 , bound_Q );
CEXPR( int , bound_V , bound_M * 2 );
int V = 0;
map<ll,int> vertex{};
bitset<bound_V> d1[bound_M] = {};
FOR( m , 0 , M ){
bitset<bound_V>& d1m = d1[m];
CIN_ASSERT( i , 0 , N );
d1m[ vertex.count( i ) == 0 ? vertex[i] = V++ : vertex[i] ] = 1;
CIN_ASSERT( j , i + 1 , N );
d1m[ vertex.count( j ) == 0 ? vertex[j] = V++ : vertex[j] ] = 1;
}
int M_minus = M - 1;
bitset<bound_M> d2[bound_Q] = {};
FOR( q , 0 , Q ){
bitset<bound_M>& d2q = d2[q];
CIN_ASSERT( m0 , 0 , M_minus );
d2q[m0] = 1;
CIN_ASSERT( m1 , m0 + 1 , M_minus );
d2q[m1] = 1;
CIN_ASSERT( m2 , m1 + 1 , M_minus );
d2q[m2] = 1;
assert( ( d1[m0] ^ d1[m1] ^ d1[m2] ) == 0 );
}
int rank1 = Rank<bound_M,bound_V>( d1 , M , V );
int rank2 = Rank<bound_Q,bound_M>( d2 , Q , M );
POWER_MOD( ker1 , 2 , M - rank1 , B );
POWER_MOD( im2 , 2 , rank2 , B );
cout << Residue<ll>( ker1 - im2 , B ) << " " << im2 << "\n";
QUIT;
}
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