結果

問題 No.2039 Copy and Avoid
ユーザー miscalcmiscalc
提出日時 2022-08-12 23:22:59
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 99 ms / 2,000 ms
コード長 3,194 bytes
コンパイル時間 4,502 ms
コンパイル使用メモリ 274,976 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-09-23 04:01:25
合計ジャッジ時間 5,928 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
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 57 ms
6,812 KB
testcase_01 AC 2 ms
6,944 KB
testcase_02 AC 3 ms
6,944 KB
testcase_03 AC 2 ms
6,944 KB
testcase_04 AC 2 ms
6,944 KB
testcase_05 AC 5 ms
6,940 KB
testcase_06 AC 2 ms
6,940 KB
testcase_07 AC 2 ms
6,940 KB
testcase_08 AC 2 ms
6,944 KB
testcase_09 AC 2 ms
6,940 KB
testcase_10 AC 2 ms
6,940 KB
testcase_11 AC 2 ms
6,944 KB
testcase_12 AC 2 ms
6,944 KB
testcase_13 AC 96 ms
6,944 KB
testcase_14 AC 99 ms
6,944 KB
testcase_15 AC 91 ms
6,940 KB
testcase_16 AC 86 ms
6,944 KB
testcase_17 AC 4 ms
6,944 KB
testcase_18 AC 4 ms
6,944 KB
testcase_19 AC 4 ms
6,944 KB
testcase_20 AC 5 ms
6,944 KB
testcase_21 AC 2 ms
6,944 KB
testcase_22 AC 2 ms
6,940 KB
testcase_23 AC 2 ms
6,940 KB
testcase_24 AC 2 ms
6,940 KB
testcase_25 AC 2 ms
6,940 KB
testcase_26 AC 2 ms
6,940 KB
testcase_27 AC 2 ms
6,944 KB
testcase_28 AC 2 ms
6,944 KB
testcase_29 AC 2 ms
6,944 KB
testcase_30 AC 2 ms
6,940 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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pll = pair<ll, ll>;
using tlll = tuple<ll, ll, ll>;
constexpr ll INF = 1LL << 60;
template<class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
ll safemod(ll A, ll M) {ll res = A % M; if (res < 0) res += M; return res;}
ll divfloor(ll A, ll B) {if (B < 0) {return divfloor(-A, -B);} return (A - safemod(A, B)) / B;}
ll divceil(ll A, ll B) {if (B < 0) {return divceil(-A, -B);} return divfloor(A + B - 1, B);}
ll pow_ll(ll A, ll B) {if (A == 0 || A == 1) {return A;} if (A == -1) {return B & 1 ? -1 : 1;} ll res = 1; for (int i = 0; i < B; i++) {res *= A;} return res;}
ll logfloor(ll A, ll B) {assert(A >= 2); ll res = 0; for (ll tmp = 1; tmp <= B / A; tmp *= A) {res++;} return res;}
ll logceil(ll A, ll B) {assert(A >= 2); ll res = 0; for (ll tmp = 1; tmp < B; tmp *= A) {res++;} return res;}
ll arisum_ll(ll a, ll d, ll n) { return n * a + (n & 1 ? ((n - 1) >> 1) * n : (n >> 1) * (n - 1)) * d; }
ll arisum2_ll(ll a, ll l, ll n) { return n & 1 ? ((a + l) >> 1) * n : (n >> 1) * (a + l); }
ll arisum3_ll(ll a, ll l, ll d) { assert((l - a) % d == 0); return arisum2_ll(a, l, (l - a) / d + 1); }
template<class T> void unique(vector<T> &V) {V.erase(unique(V.begin(), V.end()), V.end());}
template<class T> void sortunique(vector<T> &V) {sort(V.begin(), V.end()); V.erase(unique(V.begin(), V.end()), V.end());}
#define FINALANS(A) do {cout << (A) << '\n'; exit(0);} while (false)
template<class T> void printvec(const vector<T> &V) {int _n = V.size(); for (int i = 0; i < _n; i++) cout << V[i] << (i == _n - 1 ? "" : " ");cout << '\n';}
template<class T> void printvect(const vector<T> &V) {for (auto v : V) cout << v << '\n';}
template<class T> void printvec2(const vector<vector<T>> &V) {for (auto &v : V) printvec(v);}
//*
#include <atcoder/all>
using namespace atcoder;
using mint = modint998244353;
//using mint = modint1000000007;
//using mint = modint;
//*/

int main()
{
  ll N, M, A, B;
  cin >> N >> M >> A >> B;
  vector<ll> C(M);
  for (ll i = 0; i < M; i++)
  {
    cin >> C.at(i);
  }
  sort(C.begin(), C.end());

  vector<ll> D;
  for (ll d = 1; d * d <= N; d++)
  {
    if (N % d == 0)
    {
      D.push_back(d);
      if (d != N / d)
        D.push_back(N / d);
    }
  }
  sort(D.begin(), D.end());

  map<ll, ll> mp;
  mp[1] = 0;
  for (auto x : D)
  {
    if (mp.find(x) == mp.end())
      continue;
    
    ll c = INF;
    for (ll i = 0; i < M; i++)
    {
      if (C.at(i) % x == 0 && x <= C.at(i))
      {
        c = C.at(i);
        break;
      }
    }
    //cerr << x << " " << mp[x] << " " << c << endl;
    for (auto k : D)
    {
      if (k == 1)
        continue;
      ll nx = k * x;
      if (c <= nx)
        break;
      if (N % nx != 0)
        continue;
      ll tmp = mp[x] + (k - 1) * A + B;
      //cerr << " " << nx << " " << tmp << endl;
      if (mp.find(nx) == mp.end())
        mp[nx] = tmp;
      else
        chmin(mp[nx], tmp);
    }
  }
  if (mp.find(N) == mp.end())
    cout << -1 << endl;
  else
    cout << mp[N] - B << endl;
}
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