結果

問題 No.10 ＋か×か
ユーザー fumiphysfumiphys
提出日時 2019-08-16 14:56:28
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 29 ms / 5,000 ms
コード長 3,749 bytes
コンパイル時間 1,479 ms
コンパイル使用メモリ 167,104 KB
実行使用メモリ 10,436 KB
最終ジャッジ日時 2023-08-21 06:26:28
合計ジャッジ時間 2,310 ms
ジャッジサーバーID
（参考情報） 		judge11 / judge15
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
5,460 KB
testcase_01 AC 2 ms
5,396 KB
testcase_02 AC 2 ms
5,416 KB
testcase_03 AC 29 ms
10,392 KB
testcase_04 AC 21 ms
8,680 KB
testcase_05 AC 3 ms
5,612 KB
testcase_06 AC 29 ms
10,436 KB
testcase_07 AC 22 ms
8,776 KB
testcase_08 AC 7 ms
6,360 KB
testcase_09 AC 11 ms
7,348 KB
testcase_10 AC 2 ms
5,364 KB
testcase_11 AC 2 ms
5,468 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 = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

// solve
bool dp[51][1000010];

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  INT(n); INT(tot); vector<int> a(n); cin >> a;
  dp[n][tot] = true;
  rrep(i, n){
    rep(j, tot + 1){
      if(j - a[i] >= 0)dp[i][j - a[i]] |= dp[i+1][j];
      if(j % a[i] == 0)dp[i][j / a[i]] |= dp[i+1][j];
    }
  }
  assert(dp[1][a[0]]);
  string res = "";
  int curr = a[0];
  for(int i = 1; i < n; i++){
    if(curr + a[i] <= tot && dp[i+1][curr + a[i]]){
      curr += a[i];
      res += "+";
    }else{
      curr *= a[i];
      res += "*";
    }
  }
  cout << res << endl;
  return 0;
}
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