結果

問題 No.2939 Sigma Popcount Problem
ユーザー ___Johniel___Johniel
提出日時 2024-10-18 22:49:56
言語 C++23
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 253 ms / 2,000 ms
コード長 3,377 bytes
コンパイル時間 3,172 ms
コンパイル使用メモリ 244,508 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-10-18 22:55:35
合計ジャッジ時間 6,214 ms
ジャッジサーバーID
（参考情報） 		judge6 / judge5
このコードへのチャレンジ
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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
6,816 KB
testcase_01 AC 2 ms
6,816 KB
testcase_02 AC 2 ms
6,940 KB
testcase_03 AC 2 ms
6,940 KB
testcase_04 AC 2 ms
6,940 KB
testcase_05 AC 2 ms
6,944 KB
testcase_06 AC 2 ms
6,940 KB
testcase_07 AC 2 ms
6,940 KB
testcase_08 AC 229 ms
6,944 KB
testcase_09 AC 87 ms
6,944 KB
testcase_10 AC 121 ms
6,940 KB
testcase_11 AC 127 ms
6,940 KB
testcase_12 AC 104 ms
6,944 KB
testcase_13 AC 89 ms
6,944 KB
testcase_14 AC 75 ms
6,940 KB
testcase_15 AC 218 ms
6,940 KB
testcase_16 AC 6 ms
6,944 KB
testcase_17 AC 214 ms
6,944 KB
testcase_18 AC 220 ms
6,940 KB
testcase_19 AC 253 ms
6,944 KB
testcase_20 AC 2 ms
6,944 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 # 

// github.com/Johniel/contests
// yukicoder/2939/main.cpp

#include <bits/stdc++.h>

#define each(i, c) for (auto& i : c)
#define unless(cond) if (!(cond))

using namespace std;

template<typename P, typename Q> ostream& operator << (ostream& os, pair<P, Q> p);
template<typename P, typename Q> istream& operator >> (istream& is, pair<P, Q>& p);
template<typename P, typename Q, typename R> ostream& operator << (ostream& os, tuple<P, Q, R> t) { os << "(" << get<0>(t) << "," << get<1>(t) << "," << get<2>(t) << ")"; return os; }
template<typename P, typename Q, typename R> istream& operator >> (istream& is, tuple<P, Q, R>& t) { is >> get<0>(t) >> get<1>(t) >> get<2>(t); return is; }
template<typename T> ostream& operator << (ostream& os, vector<T> v) { os << "("; for (auto& i: v) os << i << ","; os << ")"; return os; }
template<typename T> istream& operator >> (istream& is, vector<T>& v) { for (auto& i: v) is >> i; return is; }
template<typename T> ostream& operator << (ostream& os, set<T> s) { os << "set{"; for (auto& i: s) os << i << ","; os << "}"; return os; }
template<typename K, typename V> ostream& operator << (ostream& os, map<K, V> m) { os << "map{"; for (auto& i: m) os << i << ","; os << "}"; return os; }
template<typename E, size_t N> istream& operator >> (istream& is, array<E, N>& a) { for (auto& i: a) is >> i; return is; }
template<typename E, size_t N> ostream& operator << (ostream& os, array<E, N>& a) { os << "[" << N << "]{"; for (auto& i: a) os << i << ","; os << "}"; return os; }
template<typename T> ostream& operator << (ostream& os, stack<T> s) { os << "stack{"; while (s.size()) { os << s.top() << ","; s.pop(); } os << "}"; return os; }
template<typename T> ostream& operator << (ostream& os, queue<T> q) { os << "queue{"; while (q.size()) { os << q.front() << ","; q.pop(); } os << "}"; return os; }
template<typename T> ostream& operator << (ostream& os, deque<T> q) { os << "deque{"; for (int i = 0; i < q.size(); ++i) os << q[i] << ","; os << "}"; return os; }
template<typename T> ostream& operator << (ostream& os, priority_queue<T> q) { os << "heap{"; while (q.size()) { os << q.top() << ","; q.pop(); } os << "}"; return os; }
template<typename P, typename Q> ostream& operator << (ostream& os, pair<P, Q> p) { os << "(" << p.first << "," << p.second << ")"; return os; }
template<typename P, typename Q> istream& operator >> (istream& is, pair<P, Q>& p) { is >> p.first >> p.second; return is; }

template<typename T> inline T setmax(T& a, T b) { return a = std::max(a, b); }
template<typename T> inline T setmin(T& a, T b) { return a = std::min(a, b); }

__attribute__((constructor)) static void _____(void) { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.setf(ios_base::fixed); cout.precision(15); return ; }

using lli = long long int;
using ull = unsigned long long;
using str = string;
template<typename T> using vec = vector<T>;

// constexpr lli mod = 1e9 + 7;
constexpr lli mod = 998244353;

lli fn(lli n)
{
  lli total = 0;
  for (int i = 0; i < 60; ++i) {
    lli t = 1LL << (i + 1);
    lli cycles = (n + 1) / t;
    lli ones = cycles * (1LL << i);
    lli remainder = (n + 1) % t;
    ones += max(0LL, remainder - (1LL << i));
    total += ones;
  }
  return total;
}

int main(int argc, char *argv[])
{
  { int _; cin >> _; }
  lli n;
  while (cin >> n) {
    cout << fn(n) << endl;
  }
  return 0;
}
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