結果
| 問題 | No.2225 Treasure Searching Rod (Easy) |
| ユーザー |
👑  p-adic
|
| 提出日時 | 2023-02-24 21:31:53 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 11 ms / 2,000 ms |
| コード長 | 6,936 bytes |
| コンパイル時間 | 2,638 ms |
| コンパイル使用メモリ | 216,848 KB |
| 最終ジャッジ日時 | 2025-02-10 20:33:26 |
|
ジャッジサーバーID （参考情報） |
judge5 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 23 |
ソースコード
#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; \{ \ll VARIABLE_FOR_BINARY_SEARCH_L = MINIMUM; \ll VARIABLE_FOR_BINARY_SEARCH_U = MAXIMUM + 1; \ANSWER = ( VARIABLE_FOR_BINARY_SEARCH_L + VARIABLE_FOR_BINARY_SEARCH_U ) / 2; \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 );ATT int main(){UNTIE;CIN( int , H );CIN( int , W );CIN( int , K );ll v[2500] = {};FOR( k , 0 , K ){CIN( int , x );CIN( int , y );cin >> v[ ( x - 1 ) * 50 + ( y - 1 ) ];}ll answer = 0;FOR( i , 0 , H ){FOR( j , 0 , W ){FOR( x , 0 , H ){FOR( y , 0 , W ){if( x + y >= i + j && x - y >= i - j ){( answer += v[ x * 50 + y ] ) %= P;}}}}}RETURN( answer );}