結果

問題 No.28 末尾最適化
ユーザー fumiphysfumiphys
提出日時 2019-08-18 15:24:14
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 942 ms / 5,000 ms
コード長 4,553 bytes
コンパイル時間 1,783 ms
コンパイル使用メモリ 177,288 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-10-01 14:07:22
合計ジャッジ時間 2,990 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
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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
5,248 KB
testcase_01 AC 942 ms
5,248 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 # 

// includes
#include <bits/stdc++.h>

// macros
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define vrep(v, i) for(int i = 0; i < (v).size(); i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define FI first
#define SE second
#define bit(n) (1LL<<(n))
#define INT(n) int n; cin >> n;
#define LL(n) ll n; cin >> n;
#define DOUBLE(n) double n; cin >> n;
using namespace std;

template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}

//  types
using ll = long long int;
using P = pair<int, int>;
using Pli = pair<ll, int>;
using Pil = pair<int, ll>;
using Pll = pair<ll, ll>;
using Pdd = pair<double, double>;
using cd = complex<double>;

// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e8 + 9;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

// solve
template <typename T>
vector<T> list_prime(T n){
  vector<T> res;
  vector<bool> i_prime = vector<bool>(n+1, true);
  i_prime[0] = i_prime[1] = false;
  for(ll i = 2; i <= n; i++){
    if(i_prime[i]){
      res.push_back(i);
      for(ll j = 2; j * i <= n; j++){
        i_prime[i * j] = false;
      }
    }
  }
  return res;
}


int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  auto p = list_prime(36);
  INT(q);
  rep(i_, q){
    LL(s); INT(n); INT(k); INT(b);
    vector<ll> x(n + 1); x[0] = s;
    FOR(i, 1, n + 1)x[i] = 1 + (x[i-1]*x[i-1] % mod + 12345LL * x[i-1] % mod) % mod;
    vector<vector<int>> cnt(sz(p), vector<int>(40, 0));
    rep(i, n + 1){
      rep(j, sz(p)){
        ll tmp = x[i];
        int c = 0;
        while(tmp % p[j] == 0){
          c++;
          tmp /= p[j];
        }
        cnt[j][c]++;
      }
    }
    int res = inf;
    rep(j, sz(p)){
      if(b % p[j] == 0){
        int u = 0;
        int bt = b;
        while(bt % p[j] == 0){
          u++;
          bt /= p[j];
        }
        int tmp = 0;
        int curr = 0;
        rep(i, sz(cnt[j])){
          if(curr + cnt[j][i] <= k){
            tmp += i * cnt[j][i];
            curr += cnt[j][i];
          }else{
            tmp += i * (k - curr);
            break;
          }
        }
        res = min(res, tmp / u);
      }
    }
    cout << res << endl;
  }
  return 0;
}
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