結果

問題 No.1611 Minimum Multiple with Double Divisors
コンテスト
ユーザー stoq
提出日時 2021-07-24 12:34:24
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,967 bytes
コンパイル時間 2,273 ms
コンパイル使用メモリ 235,512 KB
最終ジャッジ日時 2025-01-23 09:31:33
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 2
other AC * 9 WA * 28
権限があれば一括ダウンロードができます
コンパイルメッセージ
In function ‘bool chmin(T&, T) [with T = long long int]’,
    inlined from ‘void solve()’ at main.cpp:189:10:
main.cpp:74:3: warning: ‘ans’ may be used uninitialized [-Wmaybe-uninitialized]
   74 |   if (a > b)
      |   ^~
main.cpp: In function ‘void solve()’:
main.cpp:162:6: note: ‘ans’ was declared here
  162 |   ll ans;
      |      ^~~

ソースコード

diff # 

#define MOD_TYPE 1

#pragma region Macros

#include <bits/stdc++.h>
using namespace std;

#if 1
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif

#if 1
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
template <typename T>
using extset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#endif

#if 0
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
#endif

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
template <typename T>
using smaller_queue = priority_queue<T, vector<T>, greater<T>>;

constexpr ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
constexpr ld PI = acos(-1.0);
constexpr ld EPS = 1e-7;
constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};

#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)
#define repi(i, n) REPI(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define possible(n) cout << ((n) ? "possible" : "impossible") << "\n"
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\n"
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << "\n";

struct io_init
{
  io_init()
  {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(30) << setiosflags(ios::fixed);
  };
} io_init;
template <typename T>
inline bool chmin(T &a, T b)
{
  if (a > b)
  {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool chmax(T &a, T b)
{
  if (a < b)
  {
    a = b;
    return true;
  }
  return false;
}
inline ll CEIL(ll a, ll b)
{
  return (a + b - 1) / b;
}
template <typename A, size_t N, typename T>
inline void Fill(A (&array)[N], const T &val)
{
  fill((T *)array, (T *)(array + N), val);
}
template <typename T, typename U>
constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept
{
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
constexpr ostream &operator<<(ostream &os, pair<T, U> &p) noexcept
{
  os << p.first << " " << p.second;
  return os;
}
#pragma endregion

// --------------------------------------

const int MAX_N = 2e5;
ll can_div[MAX_N] = {};

void init_prime()
{
  can_div[1] = -1;
  for (ll i = 2; i < MAX_N; i++)
  {
    if (can_div[i] != 0)
      continue;
    for (ll j = i + i; j < MAX_N; j += i)
      can_div[j] = i;
  }
}

struct init_prime_
{
  init_prime_() { init_prime(); };
} init_prime_;

inline bool is_prime(ll n)
{
  if (n <= 1)
    return false;
  return !can_div[n];
}

void factorization(int n, unordered_map<ll, int> &res)
{
  if (n <= 1)
    return;
  if (!can_div[n])
  {
    ++res[n];
    return;
  }
  ++res[can_div[n]];
  factorization(n / can_div[n], res);
}

vector<ll> primes;

void solve()
{
  ll x;
  cin >> x;
  ll ans;
  for (auto p : primes)
  {
    if (x % p != 0)
    {
      ans = x * p;
      break;
    }
  }
  for (auto p : primes)
  {
    ll t = x;
    int cnt = 0;
    while (t % p == 0)
    {
      t /= p;
      cnt++;
    }
    rep(i, cnt * 2 + 1)
    {
      if (t > ll(1e17))
      {
        t = LINF;
        break;
      }
      t *= p;
    }
    chmin(ans, t);
  }
  cout << ans << "\n";
}

int main()
{
  REP(i, 2, 1000)
  {
    if (is_prime(i))
      primes.push_back(i);
  }
  int testcase;
  cin >> testcase;
  while (testcase--)
    solve();
}
0