結果

問題 No.2111 Sum of Diff
ユーザー tonyu0tonyu0
提出日時 2022-10-28 23:38:29
言語 C++17
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 65 ms / 2,000 ms
コード長 3,139 bytes
コンパイル時間 835 ms
コンパイル使用メモリ 99,964 KB
実行使用メモリ 4,696 KB
最終ジャッジ日時 2023-09-20 06:56:55
合計ジャッジ時間 2,929 ms
ジャッジサーバーID
（参考情報） 		judge14 / judge11
このコードへのチャレンジ（β）

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
4,376 KB
testcase_01 AC 1 ms
4,376 KB
testcase_02 AC 65 ms
4,564 KB
testcase_03 AC 44 ms
4,380 KB
testcase_04 AC 9 ms
4,380 KB
testcase_05 AC 65 ms
4,676 KB
testcase_06 AC 65 ms
4,528 KB
testcase_07 AC 16 ms
4,380 KB
testcase_08 AC 22 ms
4,380 KB
testcase_09 AC 5 ms
4,380 KB
testcase_10 AC 6 ms
4,376 KB
testcase_11 AC 31 ms
4,376 KB
testcase_12 AC 30 ms
4,376 KB
testcase_13 AC 20 ms
4,384 KB
testcase_14 AC 53 ms
4,376 KB
testcase_15 AC 62 ms
4,696 KB
testcase_16 AC 27 ms
4,380 KB
testcase_17 AC 65 ms
4,624 KB
testcase_18 AC 20 ms
4,376 KB
testcase_19 AC 48 ms
4,380 KB
testcase_20 AC 65 ms
4,632 KB
testcase_21 AC 2 ms
4,376 KB
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 # 

#include <algorithm>
#include <chrono>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <vector>
using namespace std;
using ll = long long;
#define rep(i, j, n) for (ll i = j; i < (n); ++i)
#define rrep(i, j, n) for (ll i = (n)-1; j <= i; --i)
#define all(a) a.begin(), a.end()
template <typename T>
std::ostream &operator<<(std::ostream &os, std::vector<T> &a) {
  for (size_t i = 0; i < a.size(); ++i) os << (i > 0 ? " " : "") << a[i];
  return os << '\n';
}
template <typename T>
std::istream &operator>>(std::istream &is, std::vector<T> &a) {
  for (T &x : a) { is >> x; }
  return is;
}
[[maybe_unused]] static constexpr long long MOD = 998244353;
//[[maybe_unused]] static constexpr long long MOD = 1e9 + 7;
[[maybe_unused]] static constexpr int INF = 0x3f3f3f3f;
[[maybe_unused]] static constexpr long long INFL = 0x3f3f3f3f3f3f3f3fLL;

template <long long mod = 1000000007>
class modint {
public:
  long long x;
  constexpr modint(const long long x = 0) noexcept : x(x % mod) {}
  constexpr long long &value() noexcept { return x; }
  constexpr const long long &value() const noexcept { return x; }
  constexpr modint operator+(const modint rhs) const noexcept {
    return modint(*this) += rhs;
  }
  constexpr modint operator-(const modint rhs) const noexcept {
    return modint(*this) -= rhs;
  }
  constexpr modint operator*(const modint rhs) const noexcept {
    return modint(*this) *= rhs;
  }
  constexpr modint operator/(const modint rhs) const noexcept {
    return modint(*this) /= rhs;
  }
  constexpr modint &operator+=(const modint rhs) noexcept {
    if ((x += rhs.x) >= mod) x -= mod;
    return *this;
  }
  constexpr modint &operator-=(const modint rhs) noexcept {
    if ((x += mod - rhs.x) >= mod) x -= mod;
    return *this;
  }
  constexpr modint &operator*=(const modint rhs) noexcept {
    x = x * rhs.x % mod;
    return *this;
  }
  constexpr modint &operator/=(const modint rhs) noexcept {
    *this *= rhs.inverse();
    return *this;
  }
  constexpr bool operator==(const modint &rhs) const { return x == rhs.x; }
  constexpr bool operator!=(const modint &rhs) const { return x != rhs.x; }
  modint inverse() const {
    long long a = x, b = mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return modint(u + mod);
  }

  modint pow(long long exp) const {
    modint res(1), mul(x);
    while (exp > 0) {
      if (exp & 1) res *= mul;
      mul *= mul;
      exp >>= 1;
    }
    return res;
  }
  operator long long() const { return x; }
  friend ostream &operator<<(ostream &os, const modint &rhs) {
    return os << rhs.x;
  }
  friend istream &operator>>(istream &is, modint &a) {
    long long t;
    is >> t;
    a = modint<mod>(t);
    return is;
  }
};

using mint = modint<MOD>;

int main() {
  cin.tie(0)->sync_with_stdio(0);
  int n;
  cin >> n;
  vector<mint> a(n);
  cin >> a;
  mint ans;
  rep(i, 0, n) {
    int l = i, r = n - i - 1;
    ans -= a[i] * (mint(2).pow(l) - mint(1));
    ans += a[i] * (mint(2).pow(r) - mint(1));
  }
  cout << ans;
}
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