#824474 (C++17) No.2111 Sum of Diff

• 問題: No.2111 Sum of Diff
• ユーザー: mikam
• 提出日時: 2022-12-10 14:25:33
• 言語: C++17 (gcc 12.3.0 + boost 1.83.0)
• 結果: AC
• 実行時間: 59 ms / 2,000 ms
• コード長: 3,131 bytes
• コンパイル時間: 4,430 ms
• コンパイル使用メモリ: 264,184 KB
• 実行使用メモリ: 5,248 KB
• 最終ジャッジ日時: 2024-10-14 23:40:42
• 合計ジャッジ時間: 7,239 ms
• ジャッジサーバーID （参考情報）: judge2 / judge5

入力	結果	実行時間; 実行使用メモリ
testcase_00	AC	2 ms; 5,248 KB
testcase_01	AC	2 ms; 5,248 KB
testcase_02	AC	59 ms; 5,248 KB
testcase_03	AC	40 ms; 5,248 KB
testcase_04	AC	9 ms; 5,248 KB
testcase_05	AC	59 ms; 5,248 KB
testcase_06	AC	58 ms; 5,248 KB
testcase_07	AC	15 ms; 5,248 KB
testcase_08	AC	20 ms; 5,248 KB
testcase_09	AC	5 ms; 5,248 KB
testcase_10	AC	7 ms; 5,248 KB
testcase_11	AC	28 ms; 5,248 KB
testcase_12	AC	27 ms; 5,248 KB
testcase_13	AC	18 ms; 5,248 KB
testcase_14	AC	48 ms; 5,248 KB
testcase_15	AC	57 ms; 5,248 KB
testcase_16	AC	25 ms; 5,248 KB
testcase_17	AC	58 ms; 5,248 KB
testcase_18	AC	19 ms; 5,248 KB
testcase_19	AC	44 ms; 5,248 KB
testcase_20	AC	59 ms; 5,248 KB
testcase_21	AC	3 ms; 5,248 KB

ソースコード

#include <atcoder/all>
using namespace atcoder;
#include <bits/stdc++.h>
using namespace std;
// #include <boost/multiprecision/cpp_int.hpp>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i,a,b) for (int i = (int)(a); i < (int)(b); i++)
#define all(v) v.begin(),v.end()
#define inc(x,l,r) ((l)<=(x)&&(x)<(r)) 
#define Unique(x) sort(all(x)), x.erase(unique(all(x)), x.end())
typedef long long ll;
#define int ll
using ld = long double;
using vi = vector<int>;
using vs = vector<string>;
using P = pair<int,int>;
using vp = vector<P>;
// using Bint = boost::multiprecision::cpp_int;
template<typename T> using priority_queue_greater = priority_queue<T, vector<T>, greater<T>>;
template<typename T> ostream &operator<<(ostream &os,const vector<T> &v){rep(i,v.size())os<<v[i]<<(i+1!=v.size()?" ":"");return os;}
template<typename T> istream &operator>>(istream& is,vector<T> &v){for(T &in:v)is>>in;return is;}
template<class... T> void _IN(T&... a){(cin>> ... >> a);}
template<class T> void _OUT(T& a){cout <<a<< '\n';}
template<class T,class... Ts> void _OUT(const T&a, const Ts&... b){cout<< a;(cout<<...<<(cout<<' ',b));cout<<'\n';}
#define INT(...) int __VA_ARGS__; _IN(__VA_ARGS__)
#define STR(...) string __VA_ARGS__; _IN(__VA_ARGS__)
#define pcnt __builtin_popcountll
int sign(int x){return (x>0)-(x<0);}
int ceil(int x,int y){assert(y!=0);if(sign(x)==sign(y))return (x+y-1)/y;return -((-x/y));}
int floor(int x,int y){assert(y!=0);if(sign(x)==sign(y))return x/y;if(y<0)x*=-1,y*=-1;return x/y-(x%y<0);}
int abs(int x,int y){return abs(x-y);}
bool ins(string s,string t){return s.find(t)!=string::npos;}
P operator+ (const P &p, const P &q){ return P{p.first+q.first,p.second+q.second};}
P operator- (const P &p, const P &q){ return P{p.first-q.first,p.second-q.second};}
template<typename T1,typename T2> ostream &operator<< (ostream &os, const pair<T1,T2> &p){os << p.first <<" "<<p.second;return os;}
ostream &operator<< (ostream &os, const modint1000000007 &m){os << m.val();return os;}
istream &operator>> (istream &is, modint1000000007 &m){ll in;is>>in;m=in;return is;}
ostream &operator<< (ostream &os, const modint998244353 &m){os << m.val();return os;}
istream &operator>> (istream &is, modint998244353 &m){ll in;is>>in;m=in;return is;}
template<typename T1,typename T2> bool chmax(T1 &a, const T2 b) {if (a < b) {a = b; return true;} else return false; }
template<typename T1,typename T2> bool chmin(T1 &a, const T2 b) {if (a > b) {a = b; return true;} else return false; }
void yesno(bool ok,string y="Yes",string n="No"){ cout<<(ok?y:n)<<endl;}
int di[]={-1,0,1,0,-1,-1,1,1};
int dj[]={0,1,0,-1,-1,1,-1,1};
const int INF = 8e18;
//using mint = modint1000000007;
using mint = modint998244353;

signed main() {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    INT(n);
    vi a(n);cin>>a;
    mint ans = 0;
    rep(i,n){
        ans += a[i] * (mint(2).pow(n-i-1)-1);
        ans -= a[i] * (mint(2).pow(i)-1);
    }
    
    cout << ans << endl;
    return 0;
}