結果

問題 No.2142 Segment Zero
コンテスト
ユーザー 👑 p-adic
提出日時 2022-12-02 21:45:15
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 6,355 bytes
コンパイル時間 2,962 ms
コンパイル使用メモリ 214,216 KB
最終ジャッジ日時 2025-02-09 03:42:03
ジャッジサーバーID
（参考情報） 		judge4 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 34 TLE * 1
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:162:11: warning: ‘l_best’ may be used uninitialized [-Wmaybe-uninitialized]
  162 |     r_max = l_best - 2;
      |     ~~~~~~^~~~~~~~~~~~
main.cpp:133:31: note: ‘l_best’ was declared here
  133 |   ll l , D , dif , dif_best , l_best;
      |                               ^~~~~~

ソースコード

diff # 

#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()
{
  UNTIE;
  CEXPR( ll , bound_N , 1000000 );
  CIN_ASSERT( N , 1 , bound_N );
  ll sum = ( N * ( N + 1 ) ) / 2;
  CIN_ASSERT( K , 0 , sum );
  sum -= K;
  // (A_l + ... + A_r) = (r+l)(r-l+1)/2 = (r^2-l^2+r+l)/2
  // (r+l)(r-l+1)/2 <= sum
  // <-> l^2 - l + 2sum - r^2 - r >= 0
  // <-> l <= one/2 - sqrt(one + 4(r^2 + r) - 8sum)/2 || one/2 + sqrt(one + 4(r^2 + r) - 8sum)/2 <= l
  int answer = 0;
  ll r_max = N;
  ll l , D , dif , dif_best , l_best;
  while( sum > 0 ){
    dif_best = 0;
    FOREQ( r , 1 , r_max ){
      D = 1 + 4 * r * ( r + 1 ) - 8 * sum;
      if( D > 0 ){
	l = ( 1.0 + sqrt( D ) ) / 2 - 1;
	FOR( i , 0 , 3 ){
	  if( 1 <= l && l <= r ){
	    dif = ( ( r + l ) * ( r - l + 1 ) ) / 2;
	    if( dif_best < dif && dif <= sum ){
	      l_best = l;
	      dif_best = dif;
	      break;
	    }
	  }
	  ++l;
	}
      } else {
	l = 1;
	dif = ( ( r + l ) * ( r - l + 1 ) ) / 2;
	if( dif_best < dif && dif <= sum ){
	  l_best = l;
	  dif_best = dif;
	}
      }
    }
    answer++;
    sum -= dif_best;
    r_max = l_best - 2;
  }
  RETURN( answer );
}
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