結果
問題 | No.2198 Concon Substrings (COuNt-CONstruct Version) |
ユーザー | 👑 p-adic |
提出日時 | 2023-01-20 23:39:41 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 7,397 bytes |
コンパイル時間 | 3,055 ms |
コンパイル使用メモリ | 217,920 KB |
実行使用メモリ | 10,148 KB |
最終ジャッジ日時 | 2024-06-23 11:36:13 |
合計ジャッジ時間 | 6,910 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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テストケース
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ソースコード
// #define _GLIBCXX_DEBUG #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 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 , MAX_I , LENGTH , MODULO ) \ ll ANSWER[LENGTH]; \ ll ANSWER_INV[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; \ } \ POWER_MOD( FACTORIAL_MAX_INV , ANSWER[MAX_I] , MODULO - 2 , MODULO ); \ ANSWER_INV[MAX_I] = FACTORIAL_MAX_INV; \ FOREQINV( i , MAX_I - 1 , 0 ){ \ ANSWER_INV[i] = ( FACTORIAL_MAX_INV *= i + 1 ) %= 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 ); } int main() { CEXPR( ll ,bound , 1000000000000000000 ); CIN_ASSERT( M , 0 , bound ); if( M == 0 ){ COUT( "yukicoder" ); } ll n = 1; ll c , M_q , M_r , o; while( true ){ if( M % n == 0 ){ M_q = M / n; if( M_q <= 100000000 ){ o = 1; while( o * o <= M_q ){ M_r = M_q % o; // M = con if( M_r == 0 ){ c = M_q / o; FOR( i , 0 , c ){ cout << "c"; } FOR( i , 0 , o ){ cout << "o"; } FOR( i , 0 , n ){ cout << "n"; } QUIT; } // M = ((c+1)o+1)n if( M_r == 1 ){ c = M_q / o - 1; if( c >= 0 ){ cout << "co"; FOR( i , 0 , c ){ cout << "c"; } FOR( i , 0 , o ){ cout << "o"; } FOR( i , 0 , n ){ cout << "n"; } QUIT; } } c++; } } } // M = ((c+1)o+1)(n+1)+(c+1) c = M % ( n + 1 ) - 1; if( c >= 0 ){ M_q = M / ( n + 1 ); if( M_q % ( c + 1 ) == 1 ){ o = M_q / ( c + 1 ); cout << "co"; FOR( i , 0 , c ){ cout << "c"; } FOR( i , 0 , o ){ cout << "o"; } FOR( i , 0 , n ){ cout << "n"; } cout << "on"; QUIT; } } n++; } QUIT; }