結果

問題 No.2979 直角三角形の個数
ユーザー Kude
提出日時 2024-12-03 14:52:01
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2,945 ms / 4,000 ms
コード長 4,411 bytes
コンパイル時間 3,307 ms
コンパイル使用メモリ 278,256 KB
実行使用メモリ 37,888 KB
最終ジャッジ日時 2024-12-03 14:52:15
合計ジャッジ時間 13,318 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 26
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>
namespace {
#pragma GCC diagnostic ignored "-Wunused-function"
#include<atcoder/all>
#pragma GCC diagnostic warning "-Wunused-function"
using namespace std;
using namespace atcoder;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; i--)
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
template<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } else return false; }
template<class T> bool chmin(T& a, const T& b) { if (b < a) { a = b; return true; } else return false; }
using ll = long long;
using P = pair<int,int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;
pair<vector<int>, vector<int>> primes_lpf(const int n) {
  vector<int> primes; primes.reserve(n / 10);
  vector<int> lpf(n + 1);
  for (int i = 2; i <= n; i += 2) lpf[i] = 2;
  for (int i = 3; i <= n; i += 6) lpf[i] = 3;
  if (2 <= n) primes.push_back(2);
  if (3 <= n) primes.push_back(3);
  // 5 * x <= n, x <= floor(n / 5)
  const int n5 = n / 5;
  int x = 5;
  char add_next = 2;
  for (; x <= n5; x += add_next, add_next ^= 0x2 ^ 4) {
    int px = lpf[x];
    if (px == 0) {
      lpf[x] = px = x;
      primes.push_back(x);
    }
    for (int i = 2;; ++i) {
      int q = primes[i];
      int y = q * x;
      if (y > n) break;
      lpf[y] = q;
      if (q == px) break;
    }
  }
  for (; x <= n; x += add_next, add_next ^= 0x2 ^ 4) {
    if (lpf[x] == 0) {
      lpf[x] = x;
      primes.push_back(x);
    }
  }
  return {move(primes), move(lpf)};
}
constexpr int PSIZE = 1010000;
auto [primes, lpf] = primes_lpf(PSIZE);
template <integral T>
T isqrt(T x) {
  assert (x >= 0);
  if (x < 1 << 24) return sqrt((float)x);
  else if (x < 1LL << 52) return sqrt(x);
  else return sqrtl(x);
}
ll qcount(ll q) {
  // x<y<=min(2x-1, [q/x])
  // sum rhs - lhs over x<=2x-1, x<=[q/x] <=> 1<=x<=isqrt(q)
  // 2x-1<[q/x] <=> 2x<=[q/x] <=> 2x<=q/x <=> x<=isqrt(q/2)
  // (0, x1], (x1, x2]
  int x1 = isqrt(q / 2), x2 = isqrt(q);
  // sum rhs
  // 1+3+...+2x1-1 = x1^2
  ll res = (ll)x1 * x1;
  for (int i = x1 + 1; i <= x2; i++) res += q / i;
  // sum lhs
  // 1+2+...+x2
  res -= (ll)x2 * (x2 + 1) / 2;
  return res;
}
template <class T>
vector<T> list_quotients(T n) {
  vector<T> q;
  int sq = isqrt(n);
  // floor(n/sq)==sq
  // iff floor((n-sq^2)/sq)==0
  // iff n-sq^2<sq
  bool dup = n < T(sq + 1) * sq;
  q.reserve(2 * sq - dup);
  for (int i = 1; i <= sq; i++) q.emplace_back(n / i);
  for (int i = sq - dup; i >= 1; i--) q.emplace_back(i);
  return q;
}
} int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll N;
  cin >> N;
  N /= 2;
  auto qs = list_quotients(N);
  const int sq = isqrt(N), sz = qs.size();
  auto get_idx = [&](ll x) {
    int idx = x <= sq ? sz - (int)x : int(N / x) - 1;
    assert(0 <= idx && idx < sz);
    assert(qs[idx] == x);
    return idx;
  };
  VI coeff(sq + 1);
  coeff[1] = 1;
  for (int i = 2; i <= sq; i++) {
    int j = i / lpf[i];
    if (lpf[i] != lpf[j]) coeff[i] = -coeff[j];
  }
  // for (int e2 = 0; N >> e2; e2++) {
  //   ll add = 0;
  //   ll n = N >> e2;
  //   for (int i = 1; (ll)i * i <= n; i++) add += coeff[i] * ans_memo[get_idx(n / ((ll)i * i))];
  //   ans += e2 % 2 == 0 ? add : -add;
  // }
  VI mul(sz);
  for (int i = 1; (ll)i * i <= N; i += 2) {
    ll v = N / ((ll)i * i);
    mul[get_idx(v)] += coeff[i];
    // ans += coeff[i] * ans_memo[get_idx(v)];
    v /= 2;
    if (v) {
      // ans -= coeff[i] * ans_memo[get_idx(v)];
      mul[get_idx(v)] -= coeff[i];
    }
  }
  // int cnt = 0;
  // rep(i, sz) cnt += mul[i] != 0;
  // cout << cnt <<endl;
  // exit(0);
  // VL memo(sz);
  // rep(i, sz) memo[i] = qcount(qs[i]);
  // exit(0);
  // VL ans_memo(sz);
  VL mul2(sz);
  rep(i, sz) if (mul[i]) {
    ll N = qs[i];
    // ll ans = 0;
    for (ll q = 1, q_pre = 0;;) {
      ll ub = N / q;
      if (ub == 0) break;
      ll add = q - q_pre;
      // ans += add * memo[get_idx(ub)];
      mul2[get_idx(ub)] += add * mul[i];
      if (q == N) break;
      q_pre = q;
      q = N / (N / (q + 1));
    }
    // ans_memo[i] = ans;
  }
  // rep(i, sz) cout << qs[i] << ' ' << memo[i] << '\n';
  ll ans = 0;
  rep(i, sz) if (mul2[i]) ans += mul2[i] * qcount(qs[i]);
  cout << ans << '\n';
  // int cnt = 0;
  // rep(i, sz) cnt += mul2[i] != 0;
  // cout << cnt << endl;
  // cout << sz << endl;
}
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