結果

問題 No.2111 Sum of Diff
ユーザー 👑 p-adic
提出日時 2022-10-28 22:13:03
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 27 ms / 2,000 ms
コード長 1,837 bytes
コンパイル時間 1,934 ms
コンパイル使用メモリ 193,728 KB
最終ジャッジ日時 2025-02-08 14:35:30
ジャッジサーバーID
(参考情報)
judge1 / judge2
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ファイルパターン 結果
sample AC * 2
other AC * 20
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

#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 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 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 )
int main()
{
UNTIE;
constexpr const int bound_N = 200000;
CIN_ASSERT( N , 2 , bound_N );
constexpr const ll P = 998244353;
ll power[bound_N + 1];
ll power_i = 1;
FOREQ( i , 0 , N ){
power[i] = power_i;
power_i = ( power_i + power_i ) % P;
}
constexpr const ll bound = P - 1;
ll answer = 0;
ll Ai;
FOR( i , 0 , N ){
cin >> Ai;
ASSERT( Ai , 0 , bound );
Ai *= power[ N - i - 1 ] - power[i];
answer += RESIDUE( Ai , P );
}
RETURN( answer % P );
}
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